Chapter 3

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Find the probability that when a couple has three children, at least one of them is a boy. (Assume that boys and girls are equally likely.)

The probability is 7/8 that at least one of the three children is a boy.

In a computer instant messaging survey, respondents were asked to choose the most fun way to flirt, and it found that P(D)=0.550, where D is directly in person. If someone is randomly selected, what does PD represent, and what is its value?

is the probability of randomly selecting someone who does not choose a direct in-person encounter as the most fun way to flirt. 1-0.550=

A Social Security number consists of nine digits in a particular order, and repetition of digits is allowed. After seeing the last four digits printed on a receipt, if you randomly select the other digits, what is the probability of getting the correct Social Security number of the person who was given the receipt? The probability of an even A, P(A), can be found by using the formula given below, where a procedure has n different simple events that are equally likely and if event A can occur in s different ways. P(A)=number of ways A occursnumber of different simple events=sn The multiplication counting rule states that for a sequence of events in which the first event can occur n1 ways, the second event can occur n2 ways, the third event can occur n3 ways, and so on, the total number of outcomes is n1•n2•n3.

1/100,000

Use z scores to compare the given values. Based on sample data, newborn males have weights with a mean of 3244.1 g and a standard deviation of 538.9 g. Newborn females have weights with a mean of 3071.9 g and a standard deviation of 625.8 g. Who has the weight that is more extreme relative to the group from which they came: a male who weighs 1700 g or a female who weighs 1700 g? Since the z score for the male is z=negative 2.87−2.87 and the z score for the female is z=negative 2.19−2.19, the male has the weight that is more extreme.

1700-3028.7 / 587.5 =-2.26 1700-3244.1/538.9 = the male has the weight that is more extreme.

Winning the jackpot in a particular lottery requires that you select the correct four numbers between 1 and 26 and, in a separate drawing, you must also select the correct single number between 1 and 19. Find the probability of winning the jackpot. The probability of winning the jackpot is

26C4 19C1 1/(19*14950) = 1/284050

3/16

To the right are the outcomes that are possible when a couple has three children. Assume that boys and girls are equally likely, so that the eight simple events are equally likely. Find the probability that when a couple has three children, there is exactly 1 boy. 1st 2nd 3rd boy − boy − boy boy − boy − girl boy − girl − boy boy − girl − girl girl − boy − boy girl − boy − girl girl − girl − boy girl − girl − girl What is the probability of exactly 1 boy out of three children? three eighths38 (Type an integer or a simplified fraction.)

3/8

If radio station cal I letters must begin with either K or W andmust include either two or three additional letters, how many different possibilities are there?

36 504 possibilities

A thief steals an ATM card and must randomly guess the correct seven-digit pin code from a 5-key keypad. Repetition of digits is allowed. What is the probability of a correct guess on the first try?

5*5*5*5*5*5*5= Ans 1/Ans

A thief steals an ATM card and must randomly guess the correct four-digit pin code from a 5-key keypad. Repetition of digits is allowed. What is the probability of a correct guess on the first try? The number of possible codes is 625625. (Type an integer or fraction. Simplify your answer.) The probability that the correct code is given on the first try is StartFraction 1 Over 625 EndFraction1625

5^4 625 1/625

A thief steals an ATM card and must randomly guess the correct three-digit pin code from a 6-key keypad. Repetition of digits is allowed. What is the probability of a correct guess on the first try? The number of possible codes is 216 (Type an integer or fraction. Simplify your answer.) The probability that the correct code is given on the first try is StartFraction 1 Over 216 EndFraction1216.

6^3=

How many different ways can the letters of "barrette" be arranged?

9 letters in total, rr , tt , ee 9! / (2!)(2!)(2!)

In a genetics experiment on peas, one sample of offspring contained 391 green peas and 538 yellow peas. Based on those results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the value of 34 that was expected? The probability of getting a green pea is approximately 0.4210.421. (Type an integer or decimal rounded to three decimal places as needed.) Is this probability reasonably close to 34? Choose the correct answer below. A. No, it is not reasonably close. This is the correct answer. B. Yes, it is reasonably close.

A. No, it is not reasonably close.

Use the data in the following table, which lists drive-thru order accuracy at popular fast food chains. Assume that orders are randomly selected from those included in the table. Drive-thru Restaurant A B C D Order Accurate 331 263 249 149 Order Not Accurate 35 56 30 15 If one order is selected, find the probability of getting an order from Restaurant A or an order that is accurate. Are the events of selecting an order from Restaurant A and selecting an accurate order disjoint events? The probability of getting an order from Restaurant A or an order that is accurate is . 910.910. (Round to three decimal places as needed.) Are the events of selecting an order from Restaurant A and selecting an accurate order disjoint events? The events are not disjoint because it is possible to receive an accurate order from Restaurant A.

Add all numbers: 331 263 249 149 35 56 30 15 = E Add row for order accurate = X x+35 / (E)

In horse racing, a trifecta is a bet that the first three finishers in a race are selected, and they are selected in the correct order. Does a trifecta involve combinations or permutations? Explain.

Because the order of the first three finishers does make a difference, the trifecta involves permutations.

Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 51.1 miles per hour. Speed (miles per hour) 42−45 46−49 50−53 54−57 58−61 Frequency 25 16 6 3

C. The computed mean is not close to the actual mean because the difference between the means is more than 5%

Which of the following is NOT a requirement of the Combinations Rule, nCr=n!r!(n−r)!, for items that are all different?

That order is taken into account (consider rearrangements of the same items to be different sequences).

If radio station call letters must begin with either K or W and must include either two or three additional letters, how many different possibilities are there?

There are 36504 different possibilities.

The following are the ratings of males by females in an experiment involving speed dating. Use the given data to construct a boxplot and identify the 5-number summary.

The 5-number summary is 22, 4.54.5, 55, 5.55.5, and 88.

The conditional probability of B given A can be found by _______.

assuming that event A has occurred, and then calculating the probability that event B will occur

A presidential candidate plans to begin her campaign by visiting the capitals in 4 of 42 states. What is the probability that she selects the route of four specific capitals?

P(she selects the route of four specific capitals)=2686320

Use a simulation approach to find the probability that when five consecutive babies are born, there is a run of at least three babies of the same sex. Use the ten simulations shown below, where 0 represents a male and 1 represents a female, and determine whether such runs are unusual. Baby Number Sim. 1 2 3 4 5 1 0 0 1 1 0 2 1 0 1 1 1 3 1 0 0 0 0 4 1 0 0 0 1 5 0 1 1 0 0 6 0 1 1 0 0 7 1 1 1 0 0 8 1 0 0 1 1 9 0 1 1 1 0 10 1 1 0 0 0 Are runs of at least three babies of the same sex unusual? (Type an integer or a decimal.) A. Based on the simulations, such an event is unusual because its probability is nothing, which is less than 0.05. B. Based on the simulations, such an event is unusual because its probability is nothing, which is less than 0.5. C. Based on the simulations, such an event is not unusual because its probability is nothing, which is less than 0.5. D. Based on the simulations, such an event is not unusual because its probability is 0.60.6, which is greater than 0.05.

Based on the simulations, such an event is not unusual because its probability is 0.60.6, which is greater than 0.05.

In horse racing, a trifecta is a bet that the first three finishers in a race are selected, and they are selected in the correct order. Does a trifecta involve combinations or permutations? Explain. Choose the correct answer below. A. Because the order of the first three finishers does not make a difference, the trifecta involves permutations. B. Because the order of the first three finishers does make a difference, the trifecta involves permutations. Your answer is correct. C. Because the order of the first three finishers does not make a difference, the trifecta involves combinations. D. Because the order of the first three finishers does make a difference, the trifecta involves combinations.

Because the order of the first three finishers does make a difference, the trifecta involves permutations.

Use the following cell phone airport data speeds (Mbps) from a particular network. Find P40. 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.6 0.7 0.7 0.7 0.8 0.9 0.9 0.9 1.2 1.2 1.3 1.3 1.4 1.6 2.1 2.3 2.4 2.5 2.8 2.9 3.2 3.4 3.7 4.5 5.2 6.2 6.5 6.6 7.3 7.8 7.8 8.4 8.6 10.5 11.2 11.4 11.7 12.9 13.1 13.2 13.8 15.5 26.4

Compute L=k100•n, where n is the total number of values in the data set and k is the percentile being used. L is a locator that gives the position Pk, the kth percentile. First, determine the values for k and n. k=4040 and n=5050 (Type whole numbers.) Now compute L. L = k100•n = 40100•50 = 2020 (Type an integer or a decimal. Do not round.) If L is a whole number, the value of the kth percentile is midway between the Lth value and the next value in the sorted set of data. If L is not a whole number, change it by rounding up to the next largest whole number, and the value of the kth percentile is this value. Which value is P40? A. Midway between the 20th and 21st values Your answer is correct. B. The 20th value C. The 21st value D. Midway between the 19th and 20th values To determine P40, first identify the 20th and 21st values. The 20th value is 1.41.4 and the 21st value is 1.61.6. (Type integers or decimals. Do not round.) The value midway between these values is their sum divided by 2. P40 = 1.4+1.62 = 1.51.5 (Type an integer or a decimal. Do not round.) So, P40=1.5 Mbps.

Use the following cell phone airport data speeds (Mbps) from a particular network. Find P90. 0.1 0.1 0.1 0.2 0.2 0.3 0.4 0.4 0.6 0.6 0.6 0.6 0.7 0.7 1 1.1 1.3 1.5 1.8 1.8 2.2 2.5 2.7 2.7 2.8 3 3.8 4.5 5.4 9.6 11.8 12.8 13.3 13.9 17 17.1 17.2 17.5 17.7

Compute L=k100•n, where n is the total number of values in the data set and k is the percentile being used. L is a locator that gives the position Pk, the kth percentile. First, determine the values for k and n. k=90 and n=50 Now compute L. L = k100•n = 90100•50 = 45 If L is a whole number, the value of the kth percentile is midway between the Lth value and the next value in the sorted set of data. If L is not a whole number, change it by rounding up to the next largest whole number, and the value of the kth percentile is this value. Since L=45 is a whole number, the value P90 is midway between the 45th and 46th values. To determine P90, first identify the 45th and 46th values. The 45th value is 18.5 and the 46th value is 18.6. The value midway between these values is their sum divided by 2. P90 = 18.5+18.62 = 18.55 So, P90=18.55 Mbps.

Find the standard deviation, s, of sample data summarized in the frequency distribution table below by using the formula below, where x represents the class midpoint, f represents the class frequency, and n represents the total number of sample values. Also, compare the computed standard deviation to the standard deviation obtained from the original list of data values, 11.1.

Consider a difference of 20% between two values of a standard deviation to be significant. How does this computed value compare with the given standard deviation, Computed value is not sig. diff. from given value.

If A denotes some event, what does A denote? If P(A)=0.007, what is the value of P(A)? What does A denote? A. Event A denotes the complement of event A, meaning that A consists of all outcomes in which event A does not occur. Your answer is correct. B. Event A is always unusual. C. Event A denotes the complement of event A, meaning that A and A share some but not all outcomes. D. Events A and A share all outcomes. If P(A)=0.007, what is the value of P(A)? P(A)=0.9930.993 (Type an integer or a decimal. Do not round.)

Event A denotes the complement of event A, meaning that A consists of all outcomes in which event A does not occur.

Find the midpoints, Then f* midpoints Then sum the frequency and f*mdpts Then divide f*mdpts / frequency The computed mean is not close to the actual mean because the difference between the means is more than 5%. It's not close , if difference is more than 5% If difference is less than 5%, computed mean is close to actual mean

First arrange the data in order of lowest to highest. 120 125 130 130 131 131 131 134 135 140 143 144 145 149 Notice that the sorted data is now available in the top step. Now identify the 5-number summary. For a set of data, the 5-number summary consists of the five values listed below. 1. Minimum 2. First quartile, Q1 3. Second quartile, Q2 (same as the median) 4. Third quartile, Q3 5. Maximum Now that the data are in ascending order, the minimum is the first value and the maximum is the last value. The minimum value in the data set is 120 mm Hg. The maximum value in the data set is 149 mm Hg. Quartiles divide a set of data into four groups with about 25% of the values in each group. Each quartile corresponds to a particular percentile. First write the quartiles as percentiles. Q1=P25, Q2=P50, and Q3=P75 To locate a percentile, compute L=k100•n, where n is the total number of values in the data set and k is the percentile being used. L is a locator that gives the position Pk, the kth percentile. For P25, the value of k is 25. Determine the value of n. n=14 Now compute L for P25. L = k100•n = 25100•14 = 3.5 If L is a whole number, the value of the kth percentile is midway between the Lth value and the next value in the sorted set of data. If L is not a whole number, change it by rounding up to the next largest whole number, and the value of the kth percentile is this value. Since L is not a whole number, P25 is the 4th value. Identify the 4th value. P25=130 Now compute L for P50. L = k100•n = 50100•14 = 7 Since L is a whole number, P50 is midway between the 7th and 8th values. The 7th value is 131 and the 8th value is 134. The value midway between these values is their sum divided by 2. P50 = 131+1342 = 132.5 Finally, use this same process to determine that P75 is the 11th value. P75=143 The 5-number summary is 120, 130, 132.5, 143, and 149, all in mm Hg. To construct a boxplot using a 5-number summary, first construct a scale with values that include the minimum and maximum data values. Draw a line segment above the scale from the minimum value to the maximum value. Construct a box (rectangle) extending from Q1 to Q3, and draw a vertical line in the box at the value of Q2 (median). Recall that the 5-number summary is 120, 130, 132.5, 143, and 149, all in mm Hg. The boxplot below represents the data. 120130140150Blood Pressure (mm Hg)

Fourteen different second-year medical students at a hospital measured the blood pressure of the same person. The systolic readings (mm Hg) are listed below. Use the given data to construct a boxplot and identify the 5-number summary. 129 129 135 132 120 125 142 130 The 5-number summary is 120120, 129129, 133.5133.5, 142142, and 150150, all in mm Hg. (Use ascending order. Type integers or decimals. Do not round.) Which boxplot below represents the data?

First put in order Sorted Data Set: 120, 125, 129, 129, 130, 130, 132, 135, 139, 140, 142, 142, 146, 150 Then arrange in min, Q1 n=14, k=25 Q1 3.51 => 4th value Q2 7 => 7, 8 value Q3 P75 => 11th value

Which of the following is always true? Choose the correct answer below. A. In a symmetric and bell-shaped distribution, the mean, median, and mode are the same. B. The mean and median should be used to identify the shape of the distribution. C. Data skewed to the right have a longer left tail than right tail. D. For skewed data, the mode is farther out in the longer tail than the median.

In a symmetric and bell-shaped distribution, the mean, median, and mode are the same.

In this data set, grades for courses with more credits have a greater effect on the mean than grades for courses with less credits. Therefore, to find the mean of this data set, one should find the weighted mean. A weighted mean accounts for variations in the relative importance of data values. Each data value is assigned a weight and the weighted mean is calculated with the following formula. Weighted mean: x=∑(w•x)∑w The grades of A, B, C, and D represent data values of 4, 3, 2, and 1 respectively. The numbers of credits are the weights. Note that the sum of the weights is equal to the total number of class credits. Find the sum of all weights. ∑w=3+(2×3)+4+3=1616 Multiply each score by its weight and find the sum of these products. ∑(w•x)=4×3+3×(2×3)+2×4+1×3=4141 Find the weighted mean. x=∑(w•x)∑w=4116=2.62.6 (Round to the nearest tenth as needed.) Thus, the student's mean grade point score is 2.6.

In this data set, grades for courses with more credits have a greater effect on the mean than grades for courses with less credits. Therefore, to find the mean of this data set, one should find the weighted mean. A weighted mean accounts for variations in the relative importance of data values. Each data value is assigned a weight and the weighted mean is calculated with the following formula. Weighted mean: x=∑(w•x)∑w The grades of A, B, C, and D represent data values of 4, 3, 2, and 1 respectively. The numbers of credits are the weights. Note that the sum of the weights is equal to the total number of class credits. Find the sum of all weights. ∑w=4+(3×3)+3+2=18 Multiply each score by its weight and find the sum of these products. ∑(w•x)=4×4+3×(3×3)+2×3+1×2=51 Find the weighted mean. x=∑(w•x)∑w=5118≈2.8 Thus, the student's mean grade point score is 2.8.

Use software or a calculator to find the range, variance, and standard deviation of the F-scale measurements from the tornadoes listed the data set available below. Be careful to account for missing data. LOADING... Click the icon to view the tornado data. Since the data are missing at random, the tornadoes with missing values can be deleted from the data set. The range of the F-scale measurements is 4.04.0. (Round to one decimal place as needed.) The standard deviation, s, of the F-scale measurements is 1.21.2. (Round to one decimal place as needed.) The variance, s2, of the F-scale measurements is 1.51.5.

In general, it appears that males have lower pulse rates than females. The variation among the male pulse rates is similar to the variation among the female pulse rates. CH. 3

When randomly selecting an adult, let B represent the event of randomly selecting someone with type B blood. Write a sentence describing what the rule of complements below is telling us. PB or B=1

It is certain that the selected adult has type B blood or does not have type B blood.

When randomly selecting an adult, let B represent the event of randomly selecting someone with type B blood. Write a sentence describing what the rule of complements below is telling us. PB or B=1 Choose the correct answer below. A. It is impossible that the selected adult has type B blood or does not have type B blood. B. It is certain that the selected adult has type B blood or does not have type B blood. Your answer is correct. C. It is certain that the selected adult does not have type B blood. D. It is certain that the selected adult has type B blood.

It is certain that the selected adult has type B blood or does not have type B blood.

How could you use a random-digit generator or random-number table to simulate rain if you knew that 90% of the time with conditions as you have today, it will rain? Choose the correct answer below. A. Let the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 represent "rain," and let the digit 10 represent "no rain." Generate random digits and record the results. B. Let the digit 9 represent "rain," and let any other digit represent "no rain." Generate random digits and record the results. C. A random-digit generator or random-number table cannot be used to simulate rain. D. Let the digits 0, 1, 2, 3, 4, 5, 6, 7, and 8 represent "rain," and let the digit 9 represent "no rain." Generate random digits and record the results

Let the digits 0, 1, 2, 3, 4, 5, 6, 7, and 8 represent "rain," and let the digit 9 represent "no rain." Generate random digits and record the results.

How could you use a random-digit generator or random-number table to simulate rain if you knew that 70% of the time with conditions as you have today, it will rain? Choose the correct answer below. A. Let the digit 7 represent "rain," and let any other digit represent "no rain." Generate random digits and record the results. B. A random-digit generator or random-number table cannot be used to simulate rain. C. Let the digits 0, 1, 2, 3, 4, 5, and 6 represent "rain," and let the digits 7, 8, and 9 represent "no rain." Generate random digits and record the results. Your answer is correct. D. Let the digits 1, 2, 3, 4, 5, 6, and 7 represent "rain," and let the digits 8, 9, and 10 represent "no rain." Generate random digits and record the results.

Let the digits 0, 1, 2, 3, 4, 5, and 6 represent "rain," and let the digits 7, 8, and 9 represent "no rain." Generate random digits and record the results.

A student wants to simulate 31 birthdays, but she does not have a calculator or software program available, so she makes up 31 numbers between 1 and 365. Is it okay to conduct the simulation this way? Why or why not? Choose the correct answer below. A. Yes. People are better at picking birth dates than computers, since they can base the numbers on the birth dates of people they know. B. No. People generally favor some numbers over others so that they don't select numbers with a process that is truly random. This is the correct answer. C. No. Simulations must be performed with either a calculator or a statistical program. D. Yes. People are capable of randomly making up numbers between two values.

No. People generally favor some numbers over others so that they don't select numbers with a process that is truly random.

In a survey of consumers aged 12 and older, respondents were asked how many cell phones were in use by the household. (No two respondents were from the same household.) Among the respondents, 217 answered "none," 278 said "one," 375 said "two," 135 said "three," and 104 responded with four or more. A survey respondent is selected at random. Find the probability that his/her household has four or more cell phones in use. Is it unlikely for a household to have four or more cell phones in use? Consider an event to be unlikely if its probability is less than or equal to 0.05.

P(four or more cell phones)=0.094 No, because the probability of a respondent with four or more cell phones in use is greater than 0.05.

Use the following cell phone airport data speeds (Mbps) from a particular network. Find Q1. 0.1 0.1 0.1 0.2 0.2 0.3 0.4 0.4 0.6 0.6

Quartiles are measures of location, denoted Q1, Q2, and Q3, which divide a set of data into four groups with about 25% of the values in each group. First, write Q1 as a percentile. Q1=P25 Next, compute L=k100•n, where n is the total number of values in the data set and k is the percentile being used. L is a locator that gives the position Pk, the kth percentile. The value for k is 25. Determine the value for n. n=50 Now compute L. L = k100•n = 25100•50 = 12.5 If L is a whole number, the value of the kth percentile is midway between the Lth value and the next value in the sorted set of data. If L is not a whole number, change it by rounding up to the next largest whole number, and the value of the kth percentile is this value. Since L=12.5 is not a whole number, the value P25 is the 13th value. Identify the 13th value. P25=0.7 Since Q1 is the same value as P25, Q1=0.7 Mbps.

Use the following cell phone airport data speeds (Mbps) from a particular network. Find Q1. 0.1 0.1 0.1 0.2 0.2 0.3 0.4 0.4 0.6 0.6 0.6 0.6 0.7 0.7 1 1.1 1.3 1.5 1.8 1.8 2.2 2.5 2.7 2.7 2.8 3 3.8 4.5 5.4 9.6 11.8 12.8 13.3 13.9 17 17.1 17.2

Below are the jersey numbers of 11 players randomly selected from a football team. Find the range, variance, and standard deviation for the given sample data. What do the results tell us? 44 37 68 67 6 80 18 41 5 88 49

Range=83.083.0 (Round to one decimal place as needed.) Sample standard deviation=28.328.3 (Round to one decimal place as needed.) Sample variance=802.8802.8 (Round to one decimal place as needed.) What do the results tell us? A. Jersey numbers on a football team vary much more than expected. B. Jersey numbers are nominal data that are just replacements for names, so the resulting statistics are meaningless. This is the correct answer. C. The sample standard deviation is too large in comparison to the range. D. Jersey numbers on a football team do not vary as much as expected. Jersey numbers are nominal data that are just replacements for names, so the resulting statistics are meaningless.

The brain volumes (cm3) of 20 brains have a mean of 1221.4 cm3 and a standard deviation of 131.2 cm3. Use the given standard deviation and the range rule of thumb to identify the limits separating values that are significantly low or significantly high. For such data, would a brain volume of 1463.8 cm3 be significantly high?

Recall that if the standard deviation is known, the range rule of thumb for identifying significant values states that significantly high values are μ+2σ or higher and that significantly low values are μ−2σ or lower, where μ is the population mean and σ is the population standard deviation. For large and representative samples, x and s, the sample mean and sample standard deviation, respectively, can be used instead. As given in the problem statement, the mean is 1221.4 cm3 and the standard deviation is 131.2 cm3. Use this to calculate the limit separating value for significantly low values. limit separating value = (mean)−2×(standard deviation) = (1221.4)−2×(131.2) = 959.0 cm3 Next calculate the limit separating value for significantly high values. limit separating value = (mean)+2×(standard deviation) = (1221.4)+2×(131.2) = 1483.8 cm3 Based on the results above, a typical brain is between 959.0 cm3 and 1483.8 cm3. If 1463.8 cm3 falls between these values, it is not considered significant. If 1463.8 cm3 falls below the lower limit separating value, it is considered significantly low. If 1463.8 cm3 falls above the upper limit separating value, it is considered significantly high. Use this information to determine whether a brain volume of 1463.8 cm3 would be significantly high.

Consider a value to be significantly low if its z score less than or equal to −2 or consider a value to be significantly high if its z score is greater than or equal to 2. A test is used to assess readiness for college. In a recent year, the mean test score was 22.7 and the standard deviation was 4.6. Identify the test scores that are significantly low or significantly high. What test scores are significantly low? Select the correct answer below and fill in the answer box(es) to complete your choice. A. Test scores that are less than 13.513.5. (Round to one decimal place as needed.) Your answer is correct. B. Test scores that are greater than nothing. (Round to one decimal place as needed.) C. Test scores that are between nothing and nothing. (Round to one decimal place as needed. Use ascending order.) What test scores are significantly high? Select the correct answer below and fill in the answer box(es) to complete your choice. A. Test scores that are greater than 31.931.9. (Round to one decimal place as needed.) Your answer is correct. B. Test scores that are less than nothing. (Round to one decimal place as needed.) C. Test scores that are between nothing and nothing. (Round to one decimal place as needed. Use ascending order.)

Test scores that are less than 13.513.5. (Round to one decimal place as needed.) Test scores that are greater than 31.931.9. (Round to one decimal place as needed.) -2=x-22.7/4.6 +2=x+22.7/4.6

The brain volumes (cm3) of 50 brains vary from a low of 910 cm3 to a high of 1486 cm3. Use the range rule of thumb to estimate the standard deviation s and compare the result to the exact standard deviation of 162.6 cm3, assuming the estimate is accurate if it is within 15 cm3. The estimated standard deviation is 144144 cm3. (Type an integer or a decimal. Do not round.) Compare the result to the exact standard deviation. A. The approximation is accurate because the error of the range rule of thumb's approximation is greater than 15 cm3. B. The approximation is accurate because the error of the range rule of thumb's approximation is less than 15 cm3. C. The approximation is not accurate because the error of the range rule of thumb's approximation is less than 15 cm3. D. The approximation is not accurate because the error of the range rule of thumb's approximation is greater than 15 cm3.

The approximation is not accurate because the error of the range rule of thumb's approximation is greater than 15 cm3

A sample of blood pressure measurements is taken for a group of adults, and those values (mm Hg) are listed below. The values are matched so that 10 subjects each have a systolic and diastolic measurement. Find the coefficient of variation for each of the two samples; then compare the variation. Systolic 119 128 156 96 157 120 115 138 126 120 Diastolic 79 75 72 54 88 90 60 64 70 84

The coefficient of variation for the systolic measurements is 14.614.6%. (Type an integer or decimal rounded to one decimal place as needed.) The coefficient of variation for the diastolic measurements is 16.316.3%. (Type an integer or decimal rounded to one decimal place as needed.) Compare the variation. The coefficients of variation for each data set are within 5 percentage points of each other. Therefore, the systolic measurements vary about the same as the diastolic measurements.

In a state pick 4 lottery game, a bettor selects four numbers between 0 and 9 and any selected number can be used more than once. Winning the top prize requires that the selected numbers match those and are drawn in the same order. Do the calculations for this lottery involve the combinations rule or either of the two permutations rules? Why or why not? If not, what rule does apply? Choose the correct answer below. A. The permutation rule (with different items) applies to this problem because repetition is allowed. The permutation rule (with some identical items) and the combination rule cannot be used with repetition. B. The combination rule applies to this problem because the numbers are selected with replacement. Neither of the permutations rules allows replacement. C. The combination and permutations rules do not apply because repetition is allowed and numbers are selected with replacement. The multiplication counting rule applies to this problem. Your answer is correct. D. The combination and permutations rules do not apply because repetition is allowed and numbers are selected with replacement. The factorial rule applies to this problem. E. The permutation rule (with some identical items) applies to this problem because repetition is allowed. The permutation rule (with different items) and the combination rule cannot be used with repetition.

The combination and permutations rules do not apply because repetition is allowed and numbers are selected with replacement. The multiplication counting rule applies to this problem.

Listed below are pulse rates (beats per minute) from samples of adult males and females. Find the mean and median for each of the two samples and then compare the two sets of results. Does there appear to be a difference? Male: 83 58 59 56 69 78 59 52 63 88 63 84 57 60 72 Female: 74 71 91 91 76 62 94 86 87 80 68 85 83 84 82 Find the means. The mean for males is 66.766.7 beats per minute and the mean for females is 80.980.9 beats per minute. (Type integers or decimals rounded to one decimal place as needed.) Find the medians. The median for males is 6363 beats per minute and the median for females is 8383 beats per minute. (Type integers or decimals rounded to one decimal place as needed.) Compare the results. Choose the correct answer below. A. The mean and the median for males are both lower than the mean and the median for females. This is the correct answer. B. The mean is lower for males, but the median is lower for females. C. The mean and median appear to be roughly the same for both genders. D. The mean and the median for females are both lower than the mean and the median for males. E. The median is lower for males, but the mean is lower for females. Does there appear to be a difference? A. The pulse rates for females appear to be higher than the pulse rates for males. This is the correct answer. B. There does not appear to be any difference. C. The pulse rates for males appear to be higher than the pulse rates for females. D. Since the sample size is small, no meaningful information can be gained from analyzing the data.

The mean and the median for males are both lower than the mean and the median for females. The pulse rates for females appear to be higher than the pulse rates for males.

Compare the results. Choose the correct answer below. A. The median is lower for males, but the mean is lower for females. B. The mean and the median for males are both lower than the mean and the median for females. Your answer is correct. C. The mean and the median for females are both lower than the mean and the median for males. D. The mean is lower for males, but the median is lower for females. E. The mean and median appear to be roughly the same for both genders. Does there appear to be a difference? A. There does not appear to be any difference. B. The pulse rates for males appear to be higher than the pulse rates for females. C. Since the sample size is small, no meaningful information can be gained from analyzing the data. Your answer is not correct. D. The pulse rates for females appear to be higher than the pulse rates for males.

The mean and the median for males are both lower than the mean and the median for females. The pulse rates for females appear to be higher than the pulse rates for males.

Listed below are amounts (in millions of dollars) collected from parking meters by a security company in a certain city. A larger data set was used to convict 5 members of the company of grand larceny. Find the mean and median for each of the two samples and then compare the two sets of results. Do the limited data listed here show evidence of stealing by the security company's employees? Security Company: 1.5 1.4 1.5 1.4 1.6 1.3 1.7 1.3 1.3 1.2 Other Companies: 1.6 1.7 1.9 1.8 1.9 2.2 1.7 1.8 1.8 2.1 Find the means. The mean for the security company is $1.421.42 million and the mean for the other companies is $1.851.85 million. (Type integers or decimals rounded to two decimal places as needed.) Find the medians. The median for the security company is $1.41.4 million and the median for the other companies is $1.81.8 million. (Type integers or decimals rounded to two decimal places as needed.) Compare the results. Choose the correct answer below. A. The mean and the median for the collections performed by other companies are both lower than the mean and the median for the security company. B. The mean and the median for the security company are both lower than the mean and the median for the collections performed by other companies. This is the correct answer. C. The mean and median appear to be roughly the same for all collections. D. The median is lower for the collections performed by other companies, but the mean is lower for the security company. E. The mean is lower for the security company, but the median is lower for the collections performed by other companies. Do the limit data listed here show evidence of stealing by the security company's employees? A. Since the security company does not appear to have collected lower revenue than the other companies, there is no evidence of stealing by the security company's employees. B. Since the security company appears to have collected lower revenue than the other companies, there is some evidence of stealing by the security company's employees. This is the correct answer. C. The sample size is not large enough to show any meaningful results. D. Since the data is not matched, there is no evidence of stealing by the security company's employees.

The mean and the median for the security company are both lower than the mean and the median for the collections performed by other companies. Since the security company appears to have collected lower revenue than the other companies, there is some evidence of stealing by the security company's employees.

Listed below are the numbers of hurricanes that occurred in each year in a certain region. The data are listed in order by year. Find the range, variance, and standard deviation for the given sample data. Include appropriate units in the results. What important feature of the data is not revealed by any of the measures of variation? 19 6 6 The range of the sample data is 19.019.0 hurricanes.hurricanes. (Round to one decimal place as needed.) The standard deviation of the sample data is 5.95.9 hurricanes.hurricanes. (Round to one decimal place as needed.) The variance of the sample data is 34.234.2 hurricanes squared .hurricanes2. (Round to one decimal place as needed.) What important feature of the data is not revealed through the different measures of variation? A. The measures of variation do not reveal the difference between the largest number of hurricanes and the smallest number of hurricanes in the data. B. The measures of variation reveal nothing about how the numbers of hurricanes are spread. C. The measures of variation reveal nothing about the pattern over time. This is the correct answer. D. The measures of variation reveal no information about the scale of the data.

The measures of variation reveal nothing about the pattern over time.

Assume that 2200 births are randomly selected and 41 of the births are girls. Use subjective judgment to describe the number of girls as significantly high, significantly low, or neither significantly low nor significantly high. Choose the correct answer below. A. The number of girls is significantly low. This is the correct answer. B. The number of girls is significantly high. C. The number of girls is neither significantly low nor significantly high. D. It is impossible to make a judgment with the given information.

The number of girls is significantly low.

Refer to the sample data for pre-employment drug screening shown below. If one of the subjects is randomly selected, what is the probability that the test result is a false positive? Who would suffer from a false positive result? Why? Pre-Employment Drug Screening ResultsPositive test resultNegative test result Drug Use Is IndicatedDrug Use Is Not IndicatedSubject Uses Drugs3714Subject Is Not a Drug User230 The probability of a false positive test result is 0.0240.024. (Round to three decimal places as needed.) Who would suffer from a false positive result? Why? A. The employer would suffer because the person tested would be suspected of using drugs when in reality he or she does not use drugs. B. The person tested would suffer because he or she would be suspected of using drugs when in reality he or she does not use drugs. This is the correct answer. C. The person tested would suffer because he or she would not be suspected of using drugs when in reality he or she does use drugs. D. The employer would suffer because the person tested would not be suspected of using drugs when in reality he or she does use drugs.

The person tested would suffer because he or she would be suspected of using drugs when in reality he or she does not use drugs.

You want to obtain cash by using an ATM, but it's dark and you can't see your card when you insert it. The card must be inserted with the front side up and the printing configured so that the beginning of your name enters first. Complete parts (a) through (c). a. What is the probability of selecting a random position and inserting the card with the result that the card is inserted correctly? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The probability is one fourth14. (Type an integer or a simplified fraction.) B. This is a trick question. There is not enough information given to determine the answer. b. What is the probability of randomly selecting the card's position and finding that it is incorrectly inserted on the first attempt, but it is correctly inserted on the second attempt? (Assume that the same position used for the first attempt could also be used for the second attempt.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The probability is three sixteenths316. (Type an integer or a simplified fraction.) B. This is a trick question. There is not enough information given to determine the answer. c. How many random selections are required to be absolute sure that the card works because it is inserted correctly? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The number of random selections required is nothing. (Type an integer or a simplified fraction.) B. This is a trick question. There is no finite number of attempts, because it is possible to get the wrong position every time.

The probability is one fourth14. The probability is three sixteenths316. This is a trick question. There is no finite number of attempts, because it is possible to get the wrong position every time.

Testing for a disease can be made more efficient by combining samples. If the samples from two people are combined and the mixture tests negative, then both samples are negative. On the other hand, one positive sample will always test positive, no matter how many negative samples it is mixed with. Assuming the probability of a single sample testing positive is 0.1, find the probability of a positive result for two samples combined into one mixture. Is the probability low enough so that further testing of the individual samples is rarely necessary? The probability of a positive test result is 0.1900.190. (Round to three decimal places as needed.) Is the probability low enough so that further testing of the individual samples is rarely necessary? A. The probability is not low, so further testing of the individual samples will frequently be a necessary event. This is the correct answer. B. The probability is low, so further testing will be necessary for all of the combined mixtures. C. The probability is not low, so further testing will not be necessary for any of the mixtures. D. The probability is low, so further testing of the individual samples will be a rarely necessary event.

The probability is not low, so further testing of the individual samples will frequently be a necessary event.

When randomly selecting adults, let M denote the event of randomly selecting a male and let B denote the event of randomly selecting someone with blue eyes. What does P(M|B) represent? Is P(M|B) the same as P(B|M)? What does P(M|B) represent? A. The probability of getting a male or getting someone with blue eyes. B. The probability of getting a male and getting someone with blue eyes. C. The probability of getting someone with blue eyes, given that a male has been selected. D. The probability of getting a male, given that someone with blue eyes has been selected. This is the correct answer. Is P(M|B) the same as P(B|M)? A. No, because P(B|M) represents the probability of getting a male, given that someone with blue eyes has been selected. B. Yes, because P(B|M) represents the probability of getting someone with blue eyes, given that a male has been selected. C. Yes, because P(B|M) represents the probability of getting a male, given that someone with blue eyes has been selected. D. No, because P(B|M) represents the probability of getting someone with blue eyes, given that a male has been selected.

The probability of getting a male, given that someone with blue eyes has been selected. No, because P(B|M) represents the probability of getting someone with blue eyes, given that a male has been selected.

When randomly selecting adults, let M denote the event of randomly selecting a male and let B denote the event of randomly selecting someone with blue eyes. What does P(M|B) represent? Is P(M|B) the same as P(B|M)? What does P(M|B) represent? A. The probability of getting someone with blue eyes, given that a male has been selected. B. The probability of getting a male and getting someone with blue eyes. C. The probability of getting a male, given that someone with blue eyes has been selected. This is the correct answer. D. The probability of getting a male or getting someone with blue eyes. Is P(M|B) the same as P(B|M)? A. Yes, because P(B|M) represents the probability of getting someone with blue eyes, given that a male has been selected. B. Yes, because P(B|M) represents the probability of getting a male, given that someone with blue eyes has been selected. C. No, because P(B|M) represents the probability of getting someone with blue eyes, given that a male has been selected. This is the correct answer. D. No, because P(B|M) represents the probability of getting a male, given that someone with blue eyes has been selected.

Use the given probability value to determine whether the sample results could easily occur by chance, then form a conclusion. A study addressed the issue of whether pregnant women can correctly predict the gender of their baby. Among 104 pregnant women, 57 correctly predicted the gender of their baby. If pregnant women have no such ability, there is a 0.327 probability of getting such sample results by chance. What do you conclude? The probability shows that the sample results could have easily occurred by chance. It appears that there is not sufficient evidence to conclude that pregnant women can correctly predict the gender of their baby.

The probability shows that the sample results could have easily occurred by chance. It appears that there is not sufficient evidence to conclude that pregnant women can correctly predict the gender of their baby.

In a test of a gender-selection technique, results consisted of 210 baby girls and 8 baby boys. Based on this result, what is the probability of a girl born to a couple using this technique? Does it appear that the technique is effective in increasing the likelihood that a baby will be a girl?

The probability that a girl will be born using this technique is approximately 0.9630.963. (Type an integer or decimal rounded to three decimal places as needed.) Does the technique appear effective in improving the likelihood of having a girl baby? yes

A research center poll showed that 78% of people believe that it is morally wrong to not report all income on tax returns. What is the probability that someone does not have this belief?

The probability that someone does not believe that it is morally wrong to not report all income on tax returns is .22 1-.78

Let event A=subject is telling the truth and event B=polygraph test indicates that the subject is lying. Use your own words to translate the notation P(B|A) into a verbal statement.

The probability that the polygraph indicates lying given that the subject is actually telling the truth.

Let event A=subject is telling the truth and event B=polygraph test indicates that the subject is lying. Use your own words to translate the notation P(B|A) into a verbal statement. Choose the correct option below. A. The probability that the polygraph indicates lying given that the subject is actually lying. B. The probability that the polygraph indicates lying given that the subject is actually telling the truth. Your answer is correct. C. The probability that the polygraph indicates truthfulness given that the subject is actually telling the truth. D. The probability that the polygraph indicates truthfulness given that the subject is actually lying.

The probability that the polygraph indicates lying given that the subject is actually telling the truth.

Let event A=subject is telling the truth and event B=polygraph test indicates that the subject is lying. Use your own words to translate the notation P(B|A) into a verbal statement. Choose the correct option below. A. The probability that the polygraph indicates lying given that the subject is actually lying. B. The probability that the polygraph indicates truthfulness given that the subject is actually lying. C. The probability that the polygraph indicates lying given that the subject is actually telling the truth. This is the correct answer. D. The probability that the polygraph indicates truthfulness given that the subject is actually telling the truth. CH. 4

The probability that the polygraph indicates lying given that the subject is actually telling the truth.

Let event A=subject is telling the truth and event B=polygraph test indicates that the subject is lying. Use your own words to translate the notation P(B|A) into a verbal statement. Choose the correct option below. A. The probability that the polygraph indicates truthfulness given that the subject is actually telling the truth. B. The probability that the polygraph indicates truthfulness given that the subject is actually lying. C. The probability that the polygraph indicates lying given that the subject is actually lying. D. The probability that the polygraph indicates lying given that the subject is actually telling the truth.

The probability that the polygraph indicates lying given that the subject is actually telling the truth.

Find the range, variance, and standard deviation for the given sample data, if possible. If the measures of variation can be obtained for these values, do the results make sense? Biologists conducted experiments to determine whether a deficiency of carbon dioxide in the soil affects the phenotypes of peas. Listed below are the phenotype codes, where 1=smooth-yellow, 2=smooth-green, 3=wrinkled yellow, and 4=wrinkled-green. 1 2 2 2 4 1 4 1 2 2 2 3 2 2 2 2 2 2 2 Can the range of the sample data be obtained for these values? Choose the correct answer below and, if necessary, fill in the answer box within your choice. A. The range of the sample data is 33. (Type an integer or a decimal. Do not round.) B. The range of the sample data cannot be calculated. Can the standard deviation of the sample data be obtained for these values? Choose the correct answer below and, if necessary, fill in the answer box within your choice. A. The standard deviation of the sample data is 0.80.8. (Round to one decimal place as needed.) B. The standard deviation of the sample data cannot be calculated. Can the variance of the sample data be obtained for these values? Choose the correct answer below and, if necessary, fill in the answer box within your choice. A. The variance of the sample data is 0.60.6. (Round to one decimal place as needed.) B. The variance of the sample data cannot be calculated. Do the results make sense? A. The measures of variation do not make sense because the standard deviation cannot be greater than the variance. B. It makes sense that the measures of variation cannot be calculated because there is not a large enough sample size to calculate the measures of variation. C. While the measures of variation can be found, they do not make sense because the data are nominal; they don't measure or count anything. This is the correct answer. D. The measures of variation make sense because the data is numeric, so the spread between the values is meaningful.

The range of the sample data is 33. A. The standard deviation of the sample data is 0.80.8. The variance of the sample data is 0.60.6. While the measures of variation can be found, they do not make sense because the data are nominal; they don't measure or count anything.

Find the range, variance, and standard deviation for the given sample data. Include appropriate units in the results. Listed below are the measured radiation absorption rates (in W/kg) corresponding to various cell phone models. If one of each model is measured for radiation and the results are used to find the measures of variation, are the results typical of the population of cell phones that are in use?

The range of the sample data is 0.8100.810 Upper W divided by kg.W/kg. (Round to three decimal places as needed.) Sample standard deviation=0.2870.287 Upper W divided by kg.W/kg. (Round to three decimal places as needed.) Sample variance=0.0830.083 left parenthesis Upper W divided by kg right parenthesis squared .(W/kg)2. (Round to three decimal places as needed.) If one of each model is measured for radiation and the results are used to find the measures of variation, are the results typical of the population of cell phones that are in use? A. Yes, because the results from any sample of cell phones will be typical of the population. B. No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population. This is the correct answer. C. No, because it is necessary to have at least 5 of each cell phone in order to get a meaningful result. Only including one of each cell phone model is not representative of each cell phone model. D. Yes, because each model is being represented in the sample. Any sample that considers all possible cell phone models will produce results typical of the population of cell phones. No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population.

Listed below are prices in dollars for one night at different hotels in a certain region. Find the range, variance, and standard deviation for the given sample data. Include appropriate units in the results. How useful are the measures of variation for someone searching for a room? 191 194

The range of the sample data is 145.0145.0 dollars.dollars. (Round to one decimal place as needed.) The standard deviation of the sample data is 42.842.8 dollars.dollars. (Round to one decimal place as needed.) The variance of the sample data is 1828.11828.1 dollars squared .dollars2. (Round to one decimal place as needed.) How useful are the measures of variation for someone searching for a room? The measures of variation are not very useful because when searching for a room, low prices, location, and good accommodations are more important than the amount of variation in the area.

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below are the weights in pounds of 11 players randomly selected from the roster of a championship sports team. Are the results likely to be representative of all players in that sport's league? 185 258 298 201 246 219 203 239 185 186 186 . Find the mean. The mean is 218.7218.7 pound(s). (Type an integer or a decimal rounded to one decimal place as needed.) b. Find the median. The median is 203203 pound(s). (Type an integer or a decimal rounded to one decimal place as needed.) c. Find the mode. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The mode(s) is(are) 185 comma 186185,186 pound(s). (Type an integer or a decimal. Do not round. Use a comma to separate answers as needed.) B. There is no mode. d. Find the midrange. The midrange is 241.5241.5 pound(s). (Type an integer or a decimal rounded to one decimal place as needed.) e. Are the results likely to be representative of all players in that sport's league? A. The results are not likely to be representative because the championship team may not be representative of the entire league. This is the correct answer. B. The results are not likely to be representative because the median is not equal to the mean. C. The results are not likely to be representative because the median is not equal to the mode. D. The results are likely to be representative because a championship team is most likely representative of the entire league.

The results are not likely to be representative because the championship team may not be representative of the entire league.

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given questions. Listed below are selling prices (dollars) of TVs that are 60 inches or larger and rated as a "best buy" by a popular magazine. Are the resulting statistics representative of the population of all TVs that are 60 inches and larger? If you decide to buy one of these TVs, what statistic is most relevant, other than the measures of central tendency? 1400 1700 1650 1700 1800 1600 1700 1950 1600 1500 1700 1350 a. Find the mean. The mean is $1637.51637.5. (Type an integer or a decimal rounded to one decimal place as needed.) b. Find the median. The median is $16751675. (Type an integer or a decimal rounded to one decimal place as needed.) c. Find the mode. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The mode(s) is(are) $17001700. (Type an integer or a decimal. Do not round. Use a comma to separate answers as needed.) B. There is no mode. d. Find the midrange. The midrange is $16501650. (Type an integer or a decimal rounded to one decimal place as needed.) e. Are the resulting statistics representative of the population of all TVs that are 60 inches and larger? Choose the best answer below. A. Since the sample is random and the sample size is greater than 10, the sample can be considered to be representative of the population. B. The sample consists of the "best buy" TVs, so it is not a random sample and is not likely to be representative of the population. This is the correct answer. C. Since the sample is random and the sample size is greater than 10, the sample should not be considered to be representative of the population. D. The sample consists of the "best buy" TVs, so it is a random sample and is likely to be representative of the population. If you decide to buy one of these TVs, what statistic is most relevant, other than the measures of central tendency? Choose the best answer below. A. The price that occurs least frequently is a relevant statistic for someone planning to buy one of the TVs. B. The difference between the lowest price and the highest price is a relevant statistic for someone planning to buy one of the TVs. C. The highest price is a relevant statistic for someone planning to buy one of the TVs. D. The lowest price is a relevant statistic for someone planning to buy one of the TVs.

The sample consists of the "best buy" TVs, so it is not a random sample and is not likely to be representative of the population. The lowest price is a relevant statistic for someone planning to buy one of the TVs.

Let a population consist of the values 7 cigarettes, 8 cigarettes, and 22 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population. Sample Mean Absolute Deviation {7,7} 00 {7,8} 0.50.5 {7,22} 7.57.5 {8,7} 0.50.5 {8,8} 00 {8,22} 77 {22,7} 7.57.5 {22,8} 77 {22,22} 0 Calculate the mean of the sample mean absolute deviations. The mean of the sample mean absolute deviations is 3.33.3. (Type an integer or decimal rounded to one decimal place as needed.) Calculate the population mean absolute deviation. The population mean absolute deviation is 6.46.4. (Type an integer or decimal rounded to one decimal place as needed.) What do these values indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? A. The sample mean absolute deviation is a biased estimator of the population mean absolute deviation because the sample statistic centers around the mean value of the population. B. The sample mean absolute deviation is a biased estimator of the population mean absolute deviation because the sample statistic centers around a different value than the population parameter. This is the correct answer. C. The sample mean absolute deviation is an unbiased estimator of the population mean absolute deviation because the sample statistic centers around the mean value of the population. D. The sample mean absolute deviation is an unbiased estimator of the population mean absolute deviation because the sample statistic

The sample mean absolute deviation is a biased estimator of the population mean absolute deviation because the sample statistic centers around a different value than the population parameter.

Let a population consist of the values 11 cigarettes, 12 cigarettes, and 21 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population. Sample Mean Absolute Deviation {11,11} 00 {11,12} 0.50.5 {11,21} 55 {12,11} 0.50.5 {12,12} 00 {12,21} 4.54.5 {21,11} 55 {21,12} 4.54.5 {21,21} 0 Calculate the mean of the sample mean absolute deviations. The mean of the sample mean absolute deviations is 2.22.2. (Type an integer or decimal rounded to one decimal place as needed.) Calculate the population mean absolute deviation. The population mean absolute deviation is 4.24.2. (Type an integer or decimal rounded to one decimal place as needed.) What do these values indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? A. The sample mean absolute deviation is an unbiased estimator of the population mean absolute deviation because the sample statistic is the same value regardless of which sample is chosen. B. The sample mean absolute deviation is a biased estimator of the population mean absolute deviation because the sample statistic centers around the mean value of the population. C. The sample mean absolute deviation is a biased estimator of the population mean absolute deviation because the sample statistic centers around a different value than the population parameter. This is the correct answer. D. The sample mean absolute deviation is an unbiased estimator of the population mean absolute deviation because the sample statistic centers around the mean value of the population.

The sample mean absolute deviation is a biased estimator of the population mean absolute deviation because the sample statistic centers around a different value than the population parameter.

There are 15,958,866 adults in a region. If a polling organization randomly selects 1235 adults without replacement, are the selections independent or dependent? If the selections are dependent, can they be treated as independent for the purposes of calculations? Are the selections independent or dependent? A. The selections are independent, because the sample size is small relative to the population. B. The selections are independent, because the selection is done without replacement. C. The selections are dependent, because the sample size is not small relative to the population. D. The selections are dependent, because the selection is done without replacement. Your answer is correct. If the selections are dependent, can they be treated as independent for the purposes of calculations? A. No, because the sample size is greater than 5% of the population. B. Yes, because the sample size is less than 5% of the population. Your answer is correct. C. Yes, because the sample size is greater than 5% of the population. D. The selections are independent.

The selections are dependent, because the selection is done without replacement. Yes, because the sample size is less than 5% of the population.

Refer to the figure below in which surge protectors p and q are used to protect an expensive high-definition television. If there is a surge in the voltage, the surge protector reduces it to a safe level. Assume that each surge protector has a 0.95 probability of working correctly when a voltage surge occurs. Complete parts (a) through (c) below. pqpqTVTVSeries ConfigurationParallel Configuration a. If the two surge protectors are arranged in series, what is the probability that a voltage surge will not damage the television? 0.99750.9975 (Do not round.) b. If the two surge protectors are arranged in parallel, what is the probability that a voltage surge will not damage the television? 0.90250.9025 (Do not round.) c. Which arrangement should be used for better protection? The parallel arrangment provides better protection because it has a lower probability of protection. The series arrangment provides better protection because it has a higher probability of protection. This is the correct answer. The series arrangment provides better protection because it has a lower probability of protection. The parallel arrangment provides better protection because it has a higher probability of protection. CH. 4

The series arrangment provides better protection because it has a higher probability of protection.

A defunct website listed the "average" annual income for Florida as $35,031. What is the role of the term average in statistics? Should another term be used in place of average? Choose the correct answer below. A. The term average is not used in statistics. The term median should be used for the result obtained by adding all of the sample values and dividing by the total number of sample values. B. The term average is often used in statistics to represent the median. C. The term average is not used in statistics. The term mean should be used for the result obtained by adding all of the sample values and dividing by the total number of sample values. This is the correct answer. D. The term average is often used in statistics to represent the mean.

The term average is not used in statistics. The term mean should be used for the result obtained by adding all of the sample values and dividing by the total number of sample values.

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below are the amounts (dollars) it costs for marriage proposal packages at different sports venues. Are there any outliers? 49 50 50 55 55 55 65 75 80 85 150 150 200 209 225 225 325 400 425 500 500 500 500 1500 2000

The values $1500 and $2000 appear to be outliers.

If your score on your next statistics test is converted to a z score, which of these z scores would you prefer: −2.00, −1.00, 0, 1.00, 2.00? Why? A. The z score of 1.00 is most preferable because it is 1.00 standard deviation above the mean and would correspond to an above average test score. B. The z score of −1.00 is most preferable because it is 1.00 standard deviation below the mean and would correspond to an above average test score. C. The z score of −2.00 is most preferable because it is 2.00 standard deviations below the mean and would correspond to the highest of the five different possible test scores. D. The z score of 2.00 is most preferable because it is 2.00 standard deviations above the mean and would correspond to the highest of the five different possible test scores. This is the correct answer. E. The z score of 0 is most preferable because it corresponds to a test score equal to the mean.

The z score of 2.00 is most preferable because it is 2.00 standard deviations above the mean and would correspond to the highest of the five different possible test scores.

If your score on your next statistics test is converted to a z score, which of these z scores would you prefer: −2.00, −1.00, 0, 1.00, 2.00? Why? A. The z score of −2.00 is most preferable because it is 2.00 standard deviations below the mean and would correspond to the highest of the five different possible test scores. B. The z score of 0 is most preferable because it corresponds to a test score equal to the mean. C. The z score of 2.00 is most preferable because it is 2.00 standard deviations above the mean and would correspond to the highest of the five different possible test scores. Your answer is correct. D. The z score of −1.00 is most preferable because it is 1.00 standard deviation below the mean and would correspond to an above average test score. E. The z score of 1.00 is most preferable because it is 1.00 standard deviation above the mean and would correspond to an above average test score.

The z score of 2.00 is most preferable because it is 2.00 standard deviations above the mean and would correspond to the highest of the five different possible test scores.

A weather forecasting website indicated that there was a 90% chance of rain in a certain region. Based on that report, which of the following is the most reasonable interpretation? Choose the correct answer below. A. There is a 0.90 probability that it will rain somewhere in the region at some point during the day. This is the correct answer. B. 90% of the region will get rain today. C. In the region, it will rain for 90% of the day. D. None of the above interpretations are reasonable.

There is a 0.90 probability that it will rain somewhere in the region at some point during the day.

Listed below are amounts of strontium-90 (in millibecquerels, or mBq) in a simple random sample of baby teeth obtained from residents in a region born after 1979. Use the given data to construct a boxplot and identify the 5-number summary. 122 125 127 132 134 136 136 138 142 145 147 147 147 149 151 154 156 158 160 167

The 5-number summary is 122122, 135135, 146146, 152.5152.5, and 167167, all in mBq. (Use ascending order. Type integers or decimals. Do not round.) Which boxplot below represents the data?

To find the probability that event A occurs in one trial and event B occurs in another trial, multiply the probability of event A by the probability of event B, but be sure that the probability of event B is found by assuming that event A has already occurred. This approach is reflected in the formula below. P(A and B)=P(A)•P(B | A) In this situation, the simple event is selecting an order from a restaurant. When selecting an order from a restaurant, there are n=1152 different orders from which to select. Count the number of orders from Restaurant D (fourth column). There were 204 orders from Restaurant D. Divide the number of orders from Restaurant D by the total number of orders to calculate the probability that the first selection is from Restaurant D, rounding to four decimal places. P(A) = 2041152 = 0.1771 Recall that, in this case, the first selection is replaced before the second selection is made. Therefore, the events are independent and the probability of event B stays the same regardless of the outcome of event A. Therefore, P(B)=P(A)=0.1771. Multiply the probability of event A by the probability of event B to obtain the probability of A and B occurring, rounding to four decimal places. P(A and B) = 0.1771•0.1771 =

The sample space listing the eight simple events that are possible when a couple has three children is {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}. After identifying the sample space for a couple having four children, find the probability of getting two girls and two boys (in any order).

Use b to represent a boy and g to represent a girl and determine all possible four-letter combinations of b and g. Notice that each child could be either a boy or a girl. Begin with the situation(s) where the couple has four girls. Determine all possible four-letter combinations that use g four times and b zero times. gggggggg (Use a comma to separate answers as needed.) Continue with the situation(s) where the couple has three girls and one boy. Determine all possible four-letter combinations that use g three times and b one time. bggg comma gbgg comma ggbg comma gggbbggg,gbgg,ggbg,gggb (Use a comma to separate answers as needed.) Continue with the situation(s) where the couple has two girls and two boys. Determine all possible four-letter combinations that use g two times and b two times. bbgg comma bgbg comma bggb comma gbbg comma gbgb comma ggbbbbgg,bgbg,bggb,gbbg,gbgb,ggbb (Use a comma to separate answers as needed.) Continue with the situation(s) where the couple has one girl and three boys. Determine all possible four-letter combinations that use g one time and b three times. bbbg comma bbgb comma bgbb comma gbbbbbbg,bbgb,bgbb,gbbb (Use a comma to separate answers as needed.) Finish with the situation(s) where the couple has four boys. Determine all possible four-letter combinations that use g zero times and b four times. bbbbbbbb (Use a comma to separate answers as needed.) Thus, the sample space for a couple having four children is as shown below. {gggg, bggg, gbgg, ggbg, gggb, bbgg, bgbg, bggb, gbbg, gbgb, ggbb, bbbg, bbgb, bgbb, gbbb, bbbb}

Use b to represent a boy and g to represent a girl and determine all possible four-letter combinations of b and g. Notice that each child could be either a boy or a girl. Begin with the situation(s) where the couple has four girls. The only possible four-letter combination that uses g four times and b zero times is gggg. Continue with the situation(s) where the couple has three girls and one boy. The possible four-letter combinations that use g three times and b one time are bggg, gbgg, ggbg, and gggb. Continue with the situation(s) where the couple has two girls and two boys. The possible four-letter combinations that use g two times and b two times are bbgg, bgbg, bggb, gbbg, gbgb, and ggbb. Continue with the situation(s) where the couple has one girl and three boys. The possible four-letter combinations that use g one time and b three times are bbbg, bbgb, bgbb, and gbbb. Finish with the situation(s) where the couple has four boys. The only possible four-letter combination that uses g zero times and b four times is bbbb. Thus, the sample space for a couple having four children is as shown below. {gggg, bggg, gbgg, ggbg, gggb, bbgg, bgbg, bggb, gbbg, gbgb, ggbb, bbbg, bbgb, bgbb, gbbb, bbbb} If a procedure has n different simple events that are equally likely, and if event A can occur in s different ways, then the probability of event A is as given below. P(A)=number of ways A occursnumber of different simple events=sn The couple can have no girls and four boys 1 way (bbbb). The couple can have four children (in any order) 16 ways. Use the formula to calculate the probability of the event. number of ways A occursnumber of different simple events=0.0625 Therefore, the probability of getting no girls and four boys is 0.0625.

Use the body temperatures, in degrees Fahrenheit, listed in the accompanying table. The range of the data is 3.5°F. Use the range rule of thumb to estimate the value of the standard deviation. Compare the result to the actual standard deviation of the data rounded to two decimal places, 0.69°F, assuming the goal is to approximate the standard deviation within 0.2°F. LOADING... Click the icon to view the table of body temperatures. The estimated standard deviation is 0.880.88°F. (Round to two decimal places as needed.) Compare the result to the actual standard deviation. The estimated standard deviation is within 0.2 degrees ofwithin 0.2° of the actual standard deviation. Thus, the estimated standard deviation meets the goal.

Use the relative frequency approach. To approximate the probability of the event A, P(A), use the formula for the relative frequency approximation of the probability shown below. P(A)=number of times A occurrednumber of times the procedure was repeated For this problem, the formula can be written as shown below. P(false positive)=number of false positive test resultstotal number of test results A false positive test result occurs when a subject who is not a drug user gets a positive test result. The number of false positive test results is 16. Find the total number of test results. 42+6+16+31=95 Substitute these values into the formula to find the probability of a false positive, and then simplify, rounding to three decimal places. P(false positive) = number of false positive test resultstotal number of test results = 1695 = 0.168 Employers use drug screening to determine whether or not a potential employee uses drugs. However, drug screening results are not always accurate. In the false positive case, drug use is indicated by the test, but the subject is not a drug user. Use this information to determine who would suffer from a false positive result and why.

Use the F-scale measurements of tornadoes listed in the accompanying table. The range of the data is 5.0. Use the range rule of thumb to estimate the value of the standard deviation. Compare the result to the actual standard deviation of the data, 1.2.

Using the range rule of thumb, the standard deviation is approximately 1.31.3. (Type an integer or decimal rounded to one decimal place as needed.) Compare the result to the actual standard deviation. The estimated standard deviation is within 0.5 of the actual standard deviation. Thus, the estimated standard deviation is not substantially different from the actual standard deviation.

Using the accompanying table of data, blood platelet counts of women have a bell-shaped distribution with a mean of 255.2 and a standard deviation of 65.5. (All units are 1000 cells/μL.) Using Chebyshev's theorem, what is known about the percentage of women with platelet counts that are within 3 standard deviations of the mean? What are the minimum and maximum possible platelet counts that are within 3 standard deviations of the mean?

Using Chebyshev's theorem, what is known about the percentage of women with platelet counts that are within 3 standard deviations of the mean? At least 8989% of women have platelet counts within 3 standard deviations of the mean. (Round to the nearest integer as needed.) What are the minimum and maximum possible platelet counts that are within 3 standard deviations of the mean? The minimum possible platelet count within 3 standard deviations of the mean is 58.758.7. The maximum possible platelet count within 3 standard deviations of the mean is 451.7451.7.

Consider a value to be significantly low if its z score less than or equal to −2 or consider a value to be significantly high if its z score is greater than or equal to 2. A data set lists weights (grams) of a type of coin. Those weights have a mean of 5.17985 g and a standard deviation of 0.06931 g. Identify the weights that are significantly low or significantly high. What weights are significantly low? Select the correct answer below and fill in the answer box(es) to complete your choice. A. Weights that are greater than nothing. (Round to five decimal places as needed.) B. Weights that are less than 5.041235.04123. (Round to five decimal places as needed.) C. Weights that are between nothing and nothing. (Round to five decimal places as needed. Use ascending order.) What weights are significantly high? Select the correct answer below and fill in the answer box(es) to complete your choice. A. Weights that are greater than 5.318475.31847. (Round to five decimal places as needed.) B. Weights that are between nothing and nothing. (Round to five decimal places as needed. Use ascending order.) C. Weights that are less than nothing. (Round to five decimal places as needed.)

Weights that are less than 5.041235.04123. (Round to five decimal places as needed.) Weights that are greater than 5.318475.31847. (Round to five decimal places as needed.)

A classic counting problem is to determine the number of different ways that the letters of "incidentally" can be arranged. Find that number.

When some items are identical to others, the permutations rule states that the number of different permutations (order counts) when n items are available and all n are selected without replacement, but some of the items are identical to others: n1 are alike, n2 are alike, and nk are alike, is given by the expression below. The factorial symbol (!) denotes the product of decreasing positive whole numbers. For example, 4!=4•3•2•1=24. By special definition, 0!=1. n!n1!n2!•••nk! How many letters are in the word "incidentally"? n=1212 Since the word "incidentally" contains 12 letters with 2 i's, 2 n's, and 2 l's, use the permutations rule (when some items are identical) to count the number of different ways that the letters can be arranged. n!n1!n2!•••nk!=12!2!2!2! Evaluate the expression. 12!2!2!2!=5987520059875200 (Simplify your answer.) Therefore, the number of different ways that the letters of "incidentally" can be arranged is 59,875,200. COUNT letters 12!/(2!^3)

Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 46.8 miles per hour. Speed (miles per hour) 42−45 46−49 50−53 54−57 58−61 Frequency 29 16 7 3

When working with data summarized in a frequency distribution, the exact values falling in a particular class are unknown. To make calculations possible, pretend that in each class, all sample values are equal to the class midpoint. Then use the following formula, where f is the frequency and x is the midpoint of each class, to find the mean of the frequency distribution. Start by finding the class midpoints. The midpoint of two numbers is their mean. Speed Frequency f Class Midpoint x 42−45 29 42+452= 43.5 46−49 16 47.547.5 50−53 7 51.551.5 54−57 3 55.555.5 58−61 1 59.559.5 Now multiply the frequencies by their midpoints. Speed Frequency f Class Midpoint x f•x 42−45 29 43.5 1261.51261.5 46−49 16 47.5 760760 50−53 7 51.5 360.5360.5 54−57 3 55.5 166.5166.5 58−61 1 59.5 59.559.5 Sum the frequency column and the f•x column. Speed Frequency f Class Midpoint x f•x 42−45 29 43.5 1261.5 46−49 16 47.5 760 50−53 7 51.5 360.5 54−57 3 55.5 166.5 58−61 1 59.5 59.5 Totals: ∑f=5656 ∑(f•x)= 26082608 Substitute the sums found in the previous step into the formula to find the mean of the data from the frequency distribution. x = ∑(f•x)∑f = 260856 ≈ 46.646.6 miles per hour (Round to the nearest tenth as needed.) The computed mean and the actual mean are considered close if the difference is less than 5% of the actual mean. Otherwise the means are said to be substantially different.

Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 46.9 miles per hour. Speed (miles per hour) 42−45 46−49 50−53 54−57 58−61 Frequency 30 14 7 5

When working with data summarized in a frequency distribution, the exact values falling in a particular class are unknown. To make calculations possible, pretend that in each class, all sample values are equal to the class midpoint. Then use the following formula, where f is the frequency and x is the midpoint of each class, to find the mean of the frequency distribution. Start by finding the class midpoints. The midpoint of two numbers is their mean. Speed Frequency f Class Midpoint x 42−45 30 42+452= 43.5 46−49 14 47.5 50−53 7 51.5 54−57 5 55.5 58−61 2 59.5 Now multiply the frequencies by their midpoints. Speed Frequency f Class Midpoint x f•x 42−45 30 43.5 1305 46−49 14 47.5 665 50−53 7 51.5 360.5 54−57 5 55.5 277.5 58−61 2 59.5 119 Sum the frequency column and the f•x column. Speed Frequency f Class Midpoint x f•x 42−45 30 43.5 1305 46−49 14 47.5 665 50−53 7 51.5 360.5 54−57 5 55.5 277.5 58−61 2 59.5 119 Totals: ∑f=58 ∑(f•x)= 2727 Substitute the sums found in the previous step into the formula to find the mean of the data from the frequency distribution. x = ∑(f•x)∑f = 272758 ≈ 47.0 miles per hour The computed mean and the actual mean are considered close if the difference is less than 5% of the actual mean. Otherwise the means are said to be substantially different.

Use the magnitudes (Richter scale) of the 120 earthquakes listed in the accompanying data table. Use technology to find the range, variance, and standard deviation. If another value, 7.50, is added to those listed in the data set, do the measures of variation change much? LOADING... Click the icon to view the table of magnitudes.

Without the extra data value, the range is 3.5703.570. (Type an integer or decimal rounded to three decimal places as needed.) Without the extra data value, the standard deviation is 0.6530.653. (Type an integer or decimal rounded to three decimal places as needed.) Without the extra data value, the variance is 0.4260.426. (Type an integer or decimal rounded to three decimal places as needed.) With the extra data value, the range is 6.3506.350. (Type an integer or decimal rounded to three decimal places as needed.) With the extra data value, the standard deviation is 0.7920.792. (Type an integer or decimal rounded to three decimal places as needed.) With the extra data value, the variance is 0.6280.628. (Type an integer or decimal rounded to three decimal places as needed.) Do the measures of variation change much with the extra data value? Choose the correct answer below. The ranges are more than 5 percentage points apart, the variances are more than 5 percentage points apart, and the standard deviations are more than 5 percentage points apart, so all of them change significantly.

Is the magnitude of an earthquake measuring 7.0 on the Richter scale an outlier when considered in the context of the sample data given? A. No, because this value is not very far away from all of the other data values. B. Yes, because this value is the maximum data value. C. No, because this value is not the maximum data value. D. Yes, because this value is very far away from all of the other data values.

Yes, because this value is very far away from all of the other data values.

A corporation must appoint a president, chief executive officer (CEO), chief operating officer (COO), and chief financial officer (CFO). It must also appoint a planning committee with five different members. There are 10 qualified candidates, and officers can also serve on the committee. Complete parts (a) through (c) below.

a. How many different ways can the officers be appointed? Determine if the number of different ways the officers can be appointed can be found using combinations or permutations. Since the candidates are distinct, no one candidate can be chosen more than once, and the order in which the candidates are chosen matters, use permutations to determine how many different ways the officers can be appointed. A situation using permutations (when all the items are different) must satisfy three requirements. First, there must be n different items available. Also, one must select r of the n items without replacement. Finally, rearrangements of the same items are counted as being different. The number of permutations (or sequences) of r items selected from n different available items (without replacement) is given by the formula below. The factorial symbol (!) denotes the product of decreasing positive whole numbers. For example, 4!=4•3•2•1=24. By special definition, 0!=1. nPr=n!(n−r)! Substitute the appropriate values into the formula. 10P4=10!(10−4)! Now evaluate. 10!(10−4)!=5,040 Therefore, there are 5,040 different ways to appoint the officers. b. How many different ways can the committee be appointed? Since the candidates are distinct, no one candidate can be chosen more than once, and the order in which the candidates are chosen does not matter, use combinations to count the number of ways the committee can be appointed. A situation using combinations must satisfy three requirements. First, there must be n different items available. Also, one must select r of the n items without replacement. Finally, rearrangements of the same items are counted as being the same. The number of combinations of r items selected from n different items is given by the formula below. The factorial symbol (!) denotes the product of decreasing positive whole numbers. For example, 4!=4•3•2•1=24. By special definition, 0!=1. nCr=n!(n−r)!r! Substitute the appropriate values into the formula. 10C5=10!(10−5)!5! Now evaluate. 10!(10−5)!5!=252 Therefore, there are 252 different ways to appoint the committee. Assume that a given procedure has n different simple events and that each of those simple events has an equal chance of occurring. If event A can occur in s of these n ways, the P(A) is given by the formula below. Use the part (b) result to find the probability. P(A)=number of ways A occurnumber of different simple events=sn c. What is the probability of randomly selecting the committee members and getting the five youngest of the qualified candidates? Now find the probability of randomly selecting the committee members and getting the five youngest of the qualified candidates, where there are 252 different ways to appoint the committee. P(five youngest of the qualified candidates) = number of ways A occurnumber of different simple events = 1252 Therefore, the probability of randomly selecting the committee members and getting the five youngest of the qualified candidates is 1252.

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 79.2 Mbps. The complete list of 50 data speeds has a mean of x=14.70 Mbps and a standard deviation of s=22.46 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between −2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?

a. The difference can be calculated by subtracting the mean from the carrier's highest data speed or by subtracting the carrier's highest data speed from the mean. In this problem, subtract the mean from the carrier's highest data speed. First set up the difference. Then subtract. 79.2−14.70=64.5 The difference is 64.5 Mbps. b. Compare the difference 64.5 Mbps to the standard deviation 22.46 Mbps. Notice that to determine the number of standard deviations in the difference, divide to determine the number of groups that are the size of the standard deviation in the difference. Divide the difference by the standard deviation, rounding to two decimal places. 64.522.46=2.87 The difference is 2.87 standard deviations. c. A z score (or standard score or standardized value) is the number of standard deviations, s or σ, that a given value x is above or below the mean, x or μ. The z score is calculated by using one of the equations shown below. Sample Population z=x−xs or z=x−μσ In this case, the data are for a sample of 50 airports, not all airports, so use z=x−xs. Notice that the z score is the same as the value calculated in part (b). The z score is z=2.87. d. A value is significantly low if the z score less than or equal to −2. A value is significantly high if the z score is greater than or equal to 2. Otherwise, the value is not significant. The z score z=2.87 is greater than 2. So, the carrier's highest data speed is significantly high.

Five pulse rates are randomly selected from a set of measurements. The five pulse rates have a mean of 76.4 beats per minute. Four of the pulse rates are 77, 85, 63, and 76. a. Find the missing value. b. Suppose that you need to create a list of n values that have a specific known mean. Some of the n values can be freely selected. How many of the n values can be freely assigned before the remaining values are determined? (The result is referred to as the number of degrees of freedom.)

a. The mean of a data set is found by adding the data values and dividing the total by the number of data values. It is given by the formula Mean=∑xn, where ∑x denotes the sum of all data values and n denotes the number of data values. Let t represent the missing value. Substituting the known mean, the number of data values, and the known data values into this formula gives the following equation. 76.4 = 77+85+63+76+t5 Find the missing pulse rate by solving the equation for t. First, simplify the right side of this equation. 76.4 = 77+85+63+76+t5 76.4 = 301+t5 To solve this equation for t, multiply both sides of the equation by 5. 76.4 = 301+t5 5•76.4 = 301+t5•5 382 = 301+t Finally, subtract 301 from both sides of the equation. 382 = 301+t 382−301 = 301+t−301 81 = t Remember that t represents the missing data value. Therefore, the missing data value is 81 beats per minute. b. Using the approach from part (a), the known mean and known data values can be used to find a missing data value. Use this information and the formula for the mean to determine how many values out of n can be freely selected before the remaining values are determined.

Five pulse rates are randomly selected from a set of measurements. The five pulse rates have a mean of 61.4 beats per minute. Four of the pulse rates are 79, 53, 70, and 50. a. Find the missing value. b. Suppose that you need to create a list of n values that have a specific known mean. Some of the n values can be freely selected. How many of the n values can be freely assigned before the remaining values are determined? (The result is referred to as the number of degrees of freedom.)

a. The missing value is 55 beats per minute. 61.4=(79+53+70+50)+t/5 solve for t Select the correct choice below and fill in the answer box to complete your choice. Of the n values, n minus 1n−1 can be freely selected because the remaining value(s) can be expressed in terms of the assigned values and the known mean.

Identify the symbols used for each of the following: (a) sample standard deviation; (b) population standard deviation; (c) sample variance; (d) population variance. a. The symbol for sample standard deviation is ss. b. The symbol for population standard deviation is sigmaσ. c. The symbol for sample variance is s squareds2. d. The symbol for population variance is sigma squaredσ2.

a. The symbol for sample standard deviation is ss. b. The symbol for population standard deviation is sigmaσ. c. The symbol for sample variance is s squareds2. d. The symbol for population variance is sigma squaredσ2.

Assume that there is a 8% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive? b. If copies of all your computer data are stored on three independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

a. Use the rule of complements shown below to find the probability that catastrophe can be avoided, where A is the complement of A. Let A=at least 1 hard drive works correctly. P(A)=1−PA What is the complement of A, A? A. Both hard drives fail. This is the correct answer. B. Both hard drives work. C. At least one hard drive fails. D. At least one hard drive works. Your answer is not correct. Since the two hard drives operate separately, their failures are independent events. Use the multiplication rule for independent events to find the probability of the complement of event A, PA. The multiplication rule for independent events states that the probability of two independent events occurring is the product of their individual probabilities.The probability of any one of the hard drives failing to work correctly is 0.08. PA = P(hard drive 1 fails and hard drive 2 fails) = 0.08•0.08 = . 0064.0064 (Round to four decimal places as needed.) Now find P(A) by evaluating 1−PA. P(A) = 1−PA = 1−0.0064 = . 9936.9936 (Type an integer or a decimal.) Therefore, the probability that at least one of the two hard drives works correctly is 0.9936. b. Again let A=at least 1 hard drive works correctly. Using the same method as in part (a), find the probability of the complement of event A. PA = P(hard drive 1 fails and hard drive 2 fails and hard drive 3 fails) = 0.08•0.08•0.08 = 0.0005120.000512 (Round to six decimal places as needed.) Now find P(A) by evaluating 1−PA. P(A) = 1−PA = 1−0.000512 = . 999488.999488 (Type an integer or a decimal.) Therefore, the probability that at least one hard drive out of three works is 0.999488

A modified roulette wheel has 44 slots. One slot is 0, another is 00, and the others are numbered 1 through 42, respectively. You are placing a bet that the outcome is an odd number. (In roulette, 0 and 00 are neither odd nor even.)

a. What is your probability of winning? Assume that a given procedure has n different simple events and that each of those simple events has an equal chance of occurring. If event A can occur in s of these n ways, then the probability of event A occurring is given below. P(A)=number of ways A can occurnumber of different simple events=sn A win can occur 21 ways because there are 21 odd slots on the wheel. There are 44 simple events, because there are 44 slots on the wheel, and "slots on the wheel" cannot be broken down into smaller events. What is your probability of winning? The probability of winning is the number of winning slots divided by the total number of slots, or 2144. b. What are the actual odds against winning? The actual odds against event A occurring are the ratio PAP(A), usually expressed in the form a:b, where a and b are integers having no common factors. Recall that P(odd)=2144. Now, to determine P(not odd), recall that the complement of an event A consists of all outcomes in which event A does not occur. P(not odd) = 1−P(odd) = 1−2144 = 2344 Thus, the actual odds against winning are 23442144. Simplify the fraction and write in the form a:b to find that the actual odds against winning are 23:21. c. When you bet that the outcome is an odd number, the payoff odds are 1:1. How much profit do you make if you bet $20 and win? The payoff odds against event A represent the ratio of net profit (if you win) to the amount bet. payoff odds against A=(net profit):(amount bet) When you bet that the outcome is an odd number, the payoff odds are 1:1. How much profit do you make if you bet $20 and win? When the payoff odds are 1:1, you win exactly as much as you bet. So if you win, the payoff is $20. d. How much profit should you make on the $20 bet if you could somehow convince the casino to change its payoff odds so that they are the same as the actual odds against winning? The actual odds against winning are 23:21, so if the payoff odds were the same as the actual odds, then a winning $20 bet would be paid 20•2321=$21.90.

In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it on gum. The results are summarized in the table. Complete parts (a) through (c) below. Purchased Gum Kept the Money Students Given Four Quarters 26 15 Students Given a $1 Bill 19 35 a. Find the probability of randomly selecting a student who spent the money, given that the student was given four quarters. The probability is . 634.634. (Round to three decimal places as needed.) b. Find the probability of randomly selecting a student who kept the money, given that the student was given four quarters. The probability is . 366.366. (Round to three decimal places as needed.) c. What do the preceding results suggest? A. A student given four quarters is more likely to have kept the money. B. A student given four quarters is more likely to have spent the money than a student given a $1 bill. C. A student given four quarters is more likely to have kept the money than a student given a $1 bill. D. A student given four quarters is more likely to have spent the money.

add columns

Each of two parents has the genotype blue/brown, which consists of the pair of alleles that determine eye color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one blue allele, that color will dominate and the child's eye color will be blue. a. List the different possible outcomes. Assume that these outcomes are equally likely. b. What is the probability that a child of these parents will have the brown/brown genotype? c. What is the probability that the child will have blue eye color?

blue/blue, blue/brown, brown/blue and brown/brown b. The probability that a child of these parents will have the brown/brown genotype is 0.250.25. (Round to two decimal places as needed.) c. The probability that the child will have blue eye color is 0.750.75.

The principle of redundancy is used when system reliability is improved through redundant or backup components. Assume that a student's alarm clock has a 15.8% daily failure rate. Complete parts (a) through (d) below. a. What is the probability that the student's alarm clock will not work on the morning of an important final exam? . 158.158 (Round to three decimal places as needed.) b. If the student has two such alarm clocks, what is the probability that they both fail on the morning of an important final exam? . 02496.02496 (Round to five decimal places as needed.) c. What is the probability of not being awakened if the student uses three independent alarm clocks? 0.003940.00394 (Round to five decimal places as needed.) d. Do the second and third alarm clocks result in greatly improved reliability? A. Yes, because you can always be certain that at least one alarm clock will work. B. No, because the malfunction of both is equally or more likely than the malfunction of one. C. Yes, because total malfunction would not be impossible, but it would be unlikely. Your answer is correct. D. No, because total malfunction would still not be unlikely.

convert 15.8%=> decimal b) multiply twice c) ans^3 Yes, because total malfunction would not be impossible, but it would be unlikely.

When a man observed a sobriety checkpoint conducted by a police department, he saw 656 drivers were screened and 4 were arrested for driving while intoxicated. Based on those results, we can estimate that P(W)=0.00610, where W denotes the event of screening a driver and getting someone who is intoxicated. What does PW denote, and what is its value? What does PW represent? A. PW denotes the probability of a driver passing through the sobriety checkpoint. B. PW denotes the probability of screening a driver and finding that he or she is intoxicated. C. PW denotes the probability of screening a driver and finding that he or she is not intoxicated. Your answer is correct. D. PW denotes the probability of driver being intoxicated. PW=. 9939.9939 (Round to five decimal places as needed.)

denotes the probability of screening a driver and finding that he or she is not intoxicated. 1-.00610

Use the following cell phone airport data speeds (Mbps) from a particular network. Find P90. 0.2 0.3 0.3 0.3 0.3 0.3 0.4 0.5 0.5 0.5 0.5 0.7 0.7 0.9 0.9 1.1 1.3 1.3 1.4 1.7 1.8 2.2 2.4 2.7 2.9 2.9 3.4 3.4 3.5 4.7 5.4 6.6 7.7 8.4 8.4 8.8 9.3 9.3 9.7 11.8 12.5 12.8 12.9 14.7 14.7 15.1 15.2 15.2 15.8 29.7 P90=14.914.9 Mbps

k=90 n=50 L=(K/100)*n (40/100)*50

The brain volumes (cm3) of 20 brains have a mean of 1090.6 cm3 and a standard deviation of 128.1 cm3. Use the given standard deviation and the range rule of thumb to identify the limits separating values that are significantly low or significantly high. For such data, would a brain volume of 1316.8 cm3 be significantly high? Significantly low values are 834.4834.4 cm3 or lower. (Type an integer or a decimal. Do not round.) Significantly high values are 1346.81346.8 cm3 or higher. (Type an integer or a decimal. Do not round.) Is 1316.8 cm3 significantly high? A. No, because it is above the upper limit separating value. B. Yes, because it is between the limits separating values. C. Yes, because it is below the lower limit separating value. D. No, because it is between the limits separating values. This is the correct answer. E. Yes, because it is above the upper limit separating value. F. No, because it is below the lower limit separating value.

no b/w limit separating values.

First, find the range of the sample. The range of a set of data values is the difference between the maximum data value and the minimum data value. range = (maximum data value)−(minimum data value) = (3,250−41) = 3,209 Recall that the units of the range are the same as the units of the original data values. Therefore, the range of the sample is 3,209 dollars. While technology could be used to find the standard deviation and variance, in this case find the sample standard deviation, s, of the data using one of the formulas below, where n is the sample size, x are the individual data values, and x is the sample mean. Note that ∑ denotes the sum of a set of values. ∑x−x2n−1 or s=n∑x2−∑x2n(n−1) To find the sample standard deviation with the formula ∑x−x2n−1, first find the sample mean, where ∑x is the sum of all data values and n is the number of data values in the sample, rounding to one decimal place. x = ∑xn = 41+45+55+...+2250+325015 = 7,63115 = 508.7 Calculate (x−x) for the first value in the sample data. (x−x) = (41−508.7) = −467.7 Calculate (x−x) for the remaining data values. Recall that x=508.7. x (x−x) x (x−x) x (x−x) 41 −467.7 80 −428.7 240 −268.7 45 −463.7 100 −408.7 275 −233.7 55 −453.7 175 −333.7 550 41.3 70 −438.7 200 −308.7 2250 1741.3 75 −433.7 225 −283.7 3250 2741.3 Calculate x−x2 for the first value in the sample data, rounding to the nearest integer. x−x2 = (41−508.7)2 = (−467.7)2 =

A combination lock uses three numbers between 1 and 76 with repetition, and they must be selected in the correct sequence. Is the name of "combination lock" appropriate? Why or why not? Choose the correct answer below. A. No, because factorials would be used to determine the total number of combinations. B. Yes, because the combinations rule would be used to determine the total number of combinations. C. No, because the multiplication counting rule would be used to determine the total number of combinations. Your answer is correct. D. No, because the permutations rule would be used to determine the total number of combinations.

No, because the multiplication counting rule would be used to determine the total number of combinations.

The principle of redundancy is used when system reliability is improved through redundant or backup components. Assume that a student's alarm clock has a 17.9% daily failure rate. Complete parts (a) through (d) below. a. What is the probability that the student's alarm clock will not work on the morning of an important final exam? 0.1790.179 (Round to three decimal places as needed.) b. If the student has two such alarm clocks, what is the probability that they both fail on the morning of an important final exam? 0.032040.03204 (Round to five decimal places as needed.) c. What is the probability of not being awakened if the student uses three independent alarm clocks? 0.005740.00574 (Round to five decimal places as needed.) d. Do the second and third alarm clocks result in greatly improved reliability? A. Yes, because you can always be certain that at least one alarm clock will work. B. No, because the malfunction of both is equally or more likely than the malfunction of one. C. No, because total malfunction would still not be unlikely. D. Yes, because total malfunction would not be impossible, but it would be unlikely.

Yes, because total malfunction would not be impossible, but it would be unlikely.

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below are the jersey numbers of 11 players randomly selected from the roster of a championship sports team. What do the results tell us? 46 61 83 39 14 38 48 7 53 56 69

. Find the mean. The mean is 46.746.7. (Type an integer or a decimal rounded to one decimal place as needed.) b. Find the median. The median is 4848. (Type an integer or a decimal rounded to one decimal place as needed.) c. Find the mode. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The mode(s) is(are) nothing. (Type an integer or a decimal. Do not round. Use a comma to separate answers as needed.) B. There is no mode. This is the correct answer. d. Find the midrange. The midrange is 4545. (Type an integer or a decimal rounded to one decimal place as needed.) e. What do the results tell us? A. The midrange gives the average (or typical) jersey number, while the mean and median give two different interpretations of the spread of possible jersey numbers. B. The jersey numbers are nominal data and they do not measure or count anything, so the resulting statistics are meaningless. This is the correct answer. C. The mean and median give two different interpretations of the average (or typical) jersey number, while the midrange shows the spread of possible jersey numbers. D. Since only 11 of the jersey numbers were in the sample, the statistics cannot give any meaningful results. There is no mode. The jersey numbers are nominal data and they do not measure or count anything, so the resulting statistics are meaningless.

Assume that there is a 10% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive? b. If copies of all your computer data are stored on three independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive? a. With two hard disk drives, the probability that catastrophe can be avoided is . 99.99. (Round to four decimal places as needed.) b. With three hard disk drives, the probability that catastrophe can be avoided is . 999.999.

.1*.1=.01 1-Ans=0.99 .1*.1*.1 = 0.001 1-Ans .999

Use the same scales to construct modified boxplots for the pulse rates of males and females from the accompanying data sets. Identify any outliers. Use the boxplots to compare the two data sets.

102 outliers 99, 99, 102, 106 In general, it appears that males have lower pulse rates than females. The variation among the male pulse rates is similar to the variation among the female pulse rates. Compare the two boxplots. Choose the correct answer below.

The brain volumes (cm3) of 20 brains have a mean of 1088.1 cm3 and a standard deviation of 127.4 cm3. Use the given standard deviation and the range rule of thumb to identify the limits separating values that are significantly low or significantly high. For such data, would a brain volume of 1292.9 cm3 be significantly high? Significantly low values are 833.3833.3 cm3 or lower. (Type an integer or a decimal. Do not round.) Significantly high values are 1342.91342.9 cm3 or higher. (Type an integer or a decimal. Do not round.) Is 1292.9 cm3 significantly high? A. Yes, because it is between the limits separating values. B. Yes, because it is above the upper limit separating value. C. Yes, because it is below the lower limit separating value. D. No, because it is above the upper limit separating value. E. No, because it is between the limits separating values. Your answer is correct. F. No, because it is below the lower limit separating value.

1088.1-2*127.4 lower higher 1088.1+2*127.4 hIGHER No, between

In a genetics experiment on peas, one sample of offspring contained 378 green peas and 124 yellow peas. Based on those results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the value of 34 that was expected? The probability of getting a green pea is approximately 0.7530.753. (Type an integer or decimal rounded to three decimal places as needed.) Is this probability reasonably close to 34? Choose the correct answer below. A. Yes, it is reasonably close. Your answer is correct. B. No, it is not reasonably close.

124+376=502 378/502 Yes, it is reasonably close.

A moving company has a truck filled for deliveries to three different sites. If the order of the deliveries is randomly selected, what is the probability that it is the shortest route? P(randomly selecting the shortest route)=

6! 1/6

In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it on gum. The results are summarized in the table. Complete parts (a) through (c) below. Purchased Gum Kept the Money Students Given Four Quarters 28 16 Students Given a $1 Bill 14 35 a. Find the probability of randomly selecting a student who spent the money, given that the student was given four quarters. The probability is 0.6360.636. (Round to three decimal places as needed.) b. Find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill. The probability is 0.2860.286. (Round to three decimal places as needed.) c. What do the preceding results suggest? A. A student given four quarters is more likely to have spent the money than a student given a $1 bill. This is the correct answer. B. A student given a $1 bill is more likely to have spent the money than a student given four quarters. C. A student was more likely to be given four quarters than a $1 bill. D. A student was more likely to have spent the money than to have kept the money.

A student given four quarters is more likely to have spent the money than a student given a $1 bill.

In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it on gum. The results are summarized in the table. Complete parts (a) through (c) below. Purchased Gum Kept the Money Students Given Four Quarters 33 13 Students Given a $1 Bill 18 34 a. Find the probability of randomly selecting a student who spent the money, given that the student was given four quarters. The probability is 0.7170.717. (Round to three decimal places as needed.) b. Find the probability of randomly selecting a student who kept the money, given that the student was given four quarters. The probability is 0.2830.283. (Round to three decimal places as needed.) c. What do the preceding results suggest? A. A student given four quarters is more likely to have kept the money. B. A student given four quarters is more likely to have spent the money than a student given a $1 bill. C. A student given four quarters is more likely to have kept the money than a student given a $1 bill. D. A student given four quarters is more likely to have spent the money.

A student given four quarters is more likely to have spent the money.

In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it on gum. The results are summarized in the table. Complete parts (a) through (c) below. Purchased Gum Kept the Money Students Given Four Quarters 35 12 Students Given a $1 Bill 17 33 a. Find the probability of randomly selecting a student who spent the money, given that the student was given four quarters. The probability is 0.7450.745. (Round to three decimal places as needed.) b. Find the probability of randomly selecting a student who kept the money, given that the student was given four quarters. The probability is 0.2550.255. (Round to three decimal places as needed.) c. What do the preceding results suggest? A. A student given four quarters is more likely to have spent the money than a student given a $1 bill. B. A student given four quarters is more likely to have spent the money. This is the correct answer. C. A student given four quarters is more likely to have kept the money than a student given a $1 bill. D. A student given four quarters is more likely to have kept the money.

A student given four quarters is more likely to have spent the money.

Use z scores to compare the given values. Based on sample data, newborn males have weights with a mean of 3196.7 g and a standard deviation of 590.6 g. Newborn females have weights with a mean of 3101.4 g and a standard deviation of 742.7 g. Who has the weight that is more extreme relative to the group from which they came: a male who weighs 1500 g or a female who weighs 1500 g?

A z score (or standard score or standardized value) is the number of standard deviations, s or σ, that a given value x is above or below the mean, x or μ. The z score is calculated by using one of the equations shown below. Sample Population z=x−xs or z=x−μσ In this case, the given values are from a sample, so use z=x−xs. To calculate the z score for the male, first identify the values of x and s. x=3196.7 g and s=590.6 g Next, identify the value of x for the male. x=1500 g Now calculate the z score for the male, rounding to two decimal places. z = x−xs = 1500−3196.7590.6 = −2.87 So, the weight of the male is 2.87 standard deviations less than the mean. Next, identify the values of x and s for females. x=3101.4 g and s=742.7 g Next, identify the value of x for the female. x=1500 g Now calculate the z score for the female, rounding to two decimal places. z = x−xs = 1500−3101.4742.7 = −2.16 So, the weight of the female is 2.16 standard deviations less than the mean. Since the z score for the male is z=−2.87 and the z score for the female is z=−2.16, the male has the weight that is more extreme.

Use z scores to compare the given values. Based on sample data, newborn males have weights with a mean of 3216.5 g and a standard deviation of 597.5 g. Newborn females have weights with a mean of 3028.7 g and a standard deviation of 587.5 g. Who has the weight that is more extreme relative to the group from which they came: a male who weighs 1700 g or a female who weighs 1700 g?

A z score (or standard score or standardized value) is the number of standard deviations, s or σ, that a given value x is above or below the mean, x or μ. The z score is calculated by using one of the equations shown below. Sample Population z=x−xs or z=x−μσ Which equation is appropriate for this case? z=x−xs Your answer is correct. z=x−μσ To calculate the z score for the male, first identify the values of x and s. x=3216.53216.5 g and s=597.5597.5 g (Type integers or decimals. Do not round.) Next, identify the value of x for the male. x=17001700 g (Type an integer or a decimal. Do not round.) Now calculate the z score for the male. z = x−xs = 1700−3216.5597.5 = negative 2.54−2.54 (Round to two decimal places as needed.) So, the weight of the male is 2.54 standard deviations less than the mean. Next, identify the values of x and s for females. x=3028.73028.7 g and s=587.5587.5 g (Type integers or decimals. Do not round.) Next, identify the value of x for the female. x=17001700 g (Type an integer or a decimal. Do not round.) Now calculate the z score for the female. z = x−xs = 1700−3028.7587.5 = negative 2.26−2.26 (Round to two decimal places as needed.) So, the weight of the female is 2.26 standard deviations less than the mean.

Refer to the data set of body temperatures in degrees Fahrenheit given in the accompanying table and use software or a calculator to find the mean and median. Do the results support or contradict the common belief that the mean body temperature is 98.6°F? LOADING... Click the icon for the body temperature data. The mean of the data set is 98.2498.24°F. (Round to two decimal places as needed.) The median of the data set is 98.2598.25°F. (Type an integer or a decimal. Do not round.) Do the results support or contradict the common belief that the mean body temperature is 98.6°F? A. The results contradict the belief that the mean body temperature is 98.6°F because both the mean and the median are less than 98.6°F. This is the correct answer. B. The results are inconclusive because the median body temperature is not equal to 98.6°F while the mean is approximately equal to 98.6°F. C. The results are inconclusive because the mean body temperature is not equal to 98.6°F while the median is approximately equal to 98.6°F. D. The results support the belief that the mean body temperature is 98.6°F because both the mean and the median are equal to 98.6°F.

A. The results contradict the belief that the mean body temperature is 98.6°F because both the mean and the median are less than 98.6°F.

In a computer instant messaging survey, respondents were asked to choose the most fun way to flirt, and it found that P(D)=0.650, where D is directly in person. If someone is randomly selected, what does PD represent, and what is its value? What does PD represent? A. PD is the probability of randomly selecting someone who chooses a direct in-person encounter as the most fun way to flirt. B. PD is the probability of randomly selecting someone who does not choose a direct in-person encounter as the most fun way to flirt. This is the correct answer. C. PD is the probability of randomly selecting someone who did not participate in the survey. D. PD is the probability of randomly selecting someone who did not have a preference in regards to the most fun way to flirt. PD=0.3500.350 (Simplify your answer.)

B. PD is the probability of randomly selecting someone who does not choose a direct in-person encounter as the most fun way to flirt.

Use the same scales to construct boxplots for the pulse rates of males and females from the accompanying data sets. Use the boxplots to compare the two data sets.

First identify the 5-number summary. For a set of data, the 5-number summary consists of the five values listed below. 1. Minimum 2. First quartile, Q1 3. Second quartile, Q2 (same as the median) 4. Third quartile, Q3 5. Maximum Remember that each quartile corresponds to a particular percentile. To locate a percentile, use the formula for the locator L, shown below, where k is the percentile and n is the number of values. If L is a whole number, the location of the value for the percentile is the mean of the value at L and the value at L+1. If L is not a whole number, round up to the nearest integer to find the location of the value for the percentile. L=k100n Place the men's data in ascending order as shown below. 46 52 53 55 55 56 57 58 58 59 61 62 62 63 63 64 65 66 66 67 68 68 69 70 70 71 71 71 73 74 74 75 76 78 79 81 85 87 87 88 The minimum number in the data set is 46, and the maximum number in the data set is 88. Next find the first quartile, Q1, which is the 25th percentile. Use the formula for the locator L, shown below. L=k100n To calculate the locator L for Q1, substitute k=25 and n=40 in the formula and simplify. L = 25100(40) = 10 Since L is a whole number, Q1 is the mean of the value at L and the value at L+1. Calculate Q1. The ordered set is listed below for reference. 46 52 53 55 55 56 57 58 58 59 61 62 62 63 63 64 65 66 66 67 68 68 69 70 70 71 71 71 73 74 74 75 76 78 79 81 85 87 87 88 Q1 = 59+612 = 60 Recall that the median of a data set is the measure of center that is the middle value when the original data values are arranged in order of increasing (or decreasing) magnitude. Find the median. Median = lesser 'middle' value+greater 'middle' value2 = 67+682 = 67.5 Next, find the third quartile, Q3, which is the 75th percentile. To calculate the locator L for Q3, substitute k=75 and n=40 in the formula for the locator, L=k100n, and simplify. L = 75100(40) = 30 Since L is a whole number, Q3 is the mean of the value at L and the value at L+1. Calculate Q3. The ordered set is listed below for reference. 46 52 53 55 55 56 57 58 58 59 61 62 62 63 63 64 65 66 66 67 68 68 69 70 70 71 71 71 73 74 74 75 76 78 79 81 85 87 87 88 Q3 = 74+742 = 74 Thus, the 5-number summary is 46, 60, 67.5, 74, and 88. To construct a modified boxplot, recall that a data value is an outlier if it is above Q3 by an amount greater than 1.5×IQR or below Q1 by an amount greater than 1.5×IQR, where IQR= Q3−Q1. Calculate the IQR. 74−60=14 Calculate 1.5×IQR. 1.5×14=21 In a modified boxplot, a data value is an outlier if it is above Q3 by an amount greater than 1.5×IQR or below Q1 by an amount greater than 1.5×IQR. Subtract 1.5×IQR from Q1 and add 1.5×IQR to Q3. Q1−1.5×IQR =60−21 Q3+1.5×IQR =74+21 =39 =95 Any data points below 39 and above 95 are considered outliers. Since the minimum of the data set is 46 and the maximum of the data set is 88, none of the values in the data set are outliers. To construct a modified boxplot using a 5-number summary, first construct a scale with values that include the minimum and maximum data values. Construct a box (rectangle) extending from Q1 to Q3, and draw a line in the box at the value of Q2 (median). Draw lines extending outward from the box to the minimum and maximum data values that are not outliers, and plot points at the outliers. Using the 5-number summary and the steps given above, the boxplot of the men's data is constructed, as shown below. 405060708090100110 A boxplot is plotted above a horizontal scale from 40 to 110 in increments of 5. The boxplot consists of a box that extends from 60 to 74, a vertical line segment drawn through the box at 68, and horizontal line segments that extend outward from the left and right edges of the box to 46 and 88, respectively. All values are approximate. Find the 5-number summary for the women's data by following the previous steps for the men's 5-number summary. The women's data is placed in ascending order below for reference. 55 61 62 63 64 66 66 69 69 70 71 71 72 72 73 73 74 76 76 76 76 76 77 77 79 79 80 80 80 81 81 87 90 90 91 96 96 97 99 103 The 5-number summary for the women's data is shown below. Minimum = 55 Q1 = 70.5 Median = 76 Q3 = 81 Maximum = 103 To construct a modified boxplot using a 5-number summary, first construct a scale with values that include the minimum and maximum data values. Construct a box (rectangle) extending from Q1 to Q3, and draw a line in the box at the value of Q2 (median). Draw lines extending outward from the box to the minimum and maximum data values that are not outliers, and plot points at the outliers. Use the same process used to find the outliers in the men's data to find the outliers in the women's data. Since the points below 54.75 and above 96.75 are outliers, the following data values are outliers. 97, 99, 103 Using the 5-number summary and the steps given above, the boxplot of the women's data is constructed, as shown below. 405060708090100110 A boxplot is plotted above a horizontal scale from 40 to 110 in increments of 5. The boxplot consists of a box that extends from 71 to 81, a vertical line segment drawn through the box at 76, and horizontal line segments that extend outward from the left and right edges of the box to 55 and 103, respectively. Points are plotted at 97 comma 99 comma and 103 to the right of the box. All values are approximate. Compare the two boxplots, shown below, to decide whether men or women in general have lower or higher pulse rates. Locate the positions of the two boxplots on their respective scales to determine where the majority of the data lies. Men's Pulse Rates Women's Pulse Rates 405060708090100110 A boxplot is plotted above a horizontal scale from 40 to 110 in increments of 5. The boxplot consists of a box that extends from 60 to 74, a vertical line segment drawn through the box at 68, and horizontal line segments that extend outward from the left and right edges of the box to 46 and 88, respectively. All values are approximate. 405060708090100110

A research center poll showed that 88% of Americans believe that it is morally wrong to not report all income on tax returns. What is the probability that an American does not have this belief?

Let P(A) be the probability that someone believes it is morally wrong to not report all income on tax returns. To find the complement of A, use the following formula. P(A)=1−P(A) Substitute the probability P(A) into the equation. P(A)=1−0.88 Calculate the complement of A. P(A)=0.12 Therefore, the probability that someone does not believe it is morally wrong to not report all income on tax returns is 0.12.

Addition Rule

P(A or B) = P(A) - P(A and B)

4+3*3+2+3 =18 (4*4)+3(3*3)+(2*2)+3=50 50/18 2.8

Which of the following values cannot be probabilities? 0, 5/3, 1, 0.06, 3/5, 2, 1.32, −0.54 S

5/3 -0.54 1.32 sqrt 2

In a study of speed dating, female subjects were asked to rate the attractiveness of their male dates, and a sample of the results is listed below (1=not attractive; 10=extremely attractive). Find the range, variance, and standard deviation for the given sample data. Can the results be used to describe the variation among attractiveness ratings for the population of adult males? 5 10 8 4 2 8 8 2 6 8 4 10 1 5 10 3 4 7 8 9 9 9 1 8 2 9 The range of the sample data is 9.09.0. (Round to one decimal place as needed.) The standard deviation of the sample data is 3.03.0. (Round to one decimal place as needed.) The variance of the sample data is 9.29.2. (Round to one decimal place as needed.) Can the results be used to describe the variation among attractiveness ratings for the population of adult males? A. The results can be used to describe the population because the sample is random. B. Since it is likely that the male subjects volunteered to participate in speed dating, they may not be representative of all adult males. Therefore, the results cannot be used to describe the population.

Since it is likely that the male subjects volunteered to participate in speed dating, they may not be representative of all adult males. Therefore, the results cannot be used to describe the population.

In a certain country, the true probability of a baby being a boy is 0.534. Among the next four randomly selected births in the country, what is the probability that at least one of them is a girl? The probability is . 919.919.

.535^4 =Ans 1-Ans

In a small private school, 6 students are randomly selected from 12 available students. What is the probability that they are the six youngest students?

1/924

Use the same scales to construct modified boxplots for the pulse rates of males and females from the accompanying data sets. Identify any outliers. Use the boxplots to compare the two data sets.

100, 104 C. In general, it appears that males have lower pulse rates than females. The variation among the male pulse rates is similar to the variation among the female pulse rates.

How many ways can you make change for a quarter? (Different arrangements of the same coins are not counted separately.) Using only pennies, nickels, and dimes, there are 12 ways to make change for a quarter.

The brain volumes (cm3) of 50 brains vary from a low of 908 cm3 to a high of 1496 cm3. Use the range rule of thumb to estimate the standard deviation s and compare the result to the exact standard deviation of 187.1 cm3, assuming the estimate is accurate if it is within 15 cm3. The estimated standard deviation is 147147 cm3. (Type an integer or a decimal. Do not round.) Compare the result to the exact standard deviation. A. The approximation is accurate because the error of the range rule of thumb's approximation is less than 15 cm3. B. The approximation is not accurate because the error of the range rule of thumb's approximation is greater than 15 cm3. Your answer is correct. C. The approximation is accurate because the error of the range rule of thumb's approximation is greater than 15 cm3. D. The approximation is not accurate because the error of the range rule of thumb's approximation is less than 15 cm3.

1496-908 =10588 Ans/4 The approximation is not accurate because the error of the range rule of thumb's approximation is greater than 15 cm3.

If you know the names of the remaining two students in the spelling bee, what is the probability of randomly selecting an order and getting the order that is used in the spelling bee?

2! 1/2

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 77.7 Mbps. The complete list of 50 data speeds has a mean of x=18.25 Mbps and a standard deviation of s=20.96 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between −2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant? a. The difference is 59.4559.45 Mbps. (Type an integer or a decimal. Do not round.) b. The difference is 2.842.84 standard deviations. (Round to two decimal places as needed.) c. The z score is z=2.842.84. (Round to two decimal places as needed.) d. The carrier's highest data speed is significantly high.

77.7-18.25=59.45 part A b and C Ans/20.06 D The carrier's highest data speed is significantly high.

When eight basketball players are about to have a free-throw competition, they often draw names out of a hat to randomly select the order in which they shoot. What is the probability that they shoot free throws in alphabetical order? Assume each player has a different name. P(shoot free throws in alphabetical order)=

8! 1/40320

A classic counting problem is to determine the number of different ways that the letters of "dissipate" can be arranged. Find that number. The number of different ways that the letters of "dissipate" can be arranged is 90720.

9!/(2!^2)

In horse racing, a trifecta is a bet that the first three finishers in a race are selected, and they are selected in the correct order. Does a trifecta involve combinations or permutations? Explain. Choose the correct answer below. A. Because the order of the first three finishers does make a difference, the trifecta involves permutations. This is the correct answer. B. Because the order of the first three finishers does not make a difference, the trifecta involves combinations. C. Because the order of the first three finishers does not make a difference, the trifecta involves permutations. D. Because the order of the first three finishers does make a difference, the trifecta involves combinations.

Because the order of the first three finishers does make a difference, the trifecta involves permutations.

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given questions. Listed below are the highest amounts of net worth (in millions of dollars) of all celebrities. What do the results tell us about the population of all celebrities? Based on the nature of the amounts, what can be inferred about their precision? 280 200 195 185 175 175 170 170 170 170 a. Find the mean. The mean is $189189 million. (Type an integer or a decimal rounded to one decimal place as needed.) b. Find the median. The median is $175175 million. (Type an integer or a decimal rounded to one decimal place as needed.) c. Find the mode. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The mode(s) is(are) $170170 million. (Type an integer or a decimal. Do not round. Use a comma to separate answers as needed. . Find the midrange. The midrange is $225225 million. The results tell us that the most common celebrity net worth is the mode, but all other celebrities are expected to have net worths approximately equal to the mean, median, or midrange. B. The results tell us that all celebrities are expected to have amounts of net worth approximately equal to one of the measures of center found in parts (a) through (d). C. Apart from the fact that all other celebrities have amounts of net worth lower than those given, nothing meaningful can be known about the population. This is the correct answer. D. Apart from the fact that all other celebrities have amounts of net worth lower than those given, the results in parts (a), (b), and (d) do not given meaningful results. However, the result from part (c) shows that the most common celebrity net worth is equal to the mode. Based on the nature of the amounts, what can be inferred about their precision? A. Since no information is given, nothing can be said about the precision of the given values. B. Since celebrity information is public, these values can be assumed to be unrounded. C. The values are all whole numbers, so they appear to be accurate to the nearest whole number. D. The values all end in 0 or 5, so they appear to be rounded estimates.

Apart from the fact that all other celebrities have amounts of net worth lower than those given, nothing meaningful can be known about the population. The values all end in 0 or 5, so they appear to be rounded estimates.

Which of the following values cannot be probabilities? 0.05, −0.41, 1, 0, 3/5, 1.22, 2,

For any event A, the probability of A is between 0 and 1 inclusive. That is, 0≤P(A)≤1. First analyze 0.05. 0.05 could be the probability of an event because 0≤P(A)≤1, where P(A)=0.05. Now analyze −0.41. −0.41 could not be the probability of an event since −0.41<0. Using the same concepts as above, determine all the numbers that could be a probability of an event. If P(A)=1, 0, 0.05 or 3/5 the statement 0≤P(A)≤1 is true, so all these values could be a probability. However, the statement is not true if P(A)=1.22 as 1.22>1, so 1.22 could not be a probability. Similarly, 2>1 and 5/3>1, so they also could not be probabilities. Therefore, the numbers that could not be probabilities of an event are 1.22, −0.41, 2, and 5/3.

Which of the following is NOT a requirement of the Permutations Rule, nPr=n!(n−r)!, for items that are all different?

Order is not taken into account (rearrangements of the same items are considered to be the same).

Confusion of the inverse occurs when we incorrectly believe Upper P left parenthesis B|A right parenthesis equals Upper P left parenthesis A|B right parenthesis .

P(B|A)=P(A|B)

Listed below are foot lengths in inches for 11 randomly selected people taken in 1988. Find the range, variance, and standard deviation for the given sample data. Include appropriate units in the results. Are the statistics representative of the current population of all people? 8.6 10.1 9.9 9.9 9.5 8.6 8.7 9.8 9.6 8.8 9.7 The range of the sample data is 1.51.5 inches.inches. (Type an integer or a decimal. Do not round.) The standard deviation of the sample data is 0.580.58 inches.inches. (Round to two decimal places as needed.) The variance of the sample data is 0.340.34 inches squared .inches2. (Round to two decimal places as needed.) Are the statistics representative of the current population of all people? A. Since the measurements were made in 1988, it is not necessarily representative of the population today. This is the correct answer. B. The statistics are representative because they are taken from a random sample. C. The statistics are not representative because a smaller sample is needed to represent the population. D. The statistics are representative because the standard deviation of the sample data is less than 1.

Since the measurements were made in 1988, it is not necessarily representative of the population today.

A group of adult males has foot lengths with a mean of 27.16 cm and a standard deviation of 1.45 cm. Use the range rule of thumb to identify the limits separating values that are significantly low or significantly high. Is the adult male foot length of 24.0 cm significantly low or significantly high? Explain. Significantly low values are 24.2624.26 cm or lower. (Type an integer or a decimal. Do not round.) Significantly high values are 30.0630.06 cm or higher. (Type an integer or a decimal. Do not round.) Select the correct choice below and fill in the answer box(es) to complete your choice. A. The adult male foot length of 24.0 cm is not significant because it is between nothing cm and nothing cm. (Type integers or decimals. Do not round.) B. The adult male foot length of 24.0 cm is significantly high because it is greater than nothing cm. (Type an integer or a decimal. Do not round.) C. The adult male foot length of 24.0 cm is significantly low because it is less than 24.2624.26 cm. (Type an integer or a decimal. Do not round.)

The adult male foot length of 24.0 cm is significantly low because it is less than 24.2624.26 cm.

Listed below are pulse rates (beats per minute) from samples of adult males and females. Does there appear to be a difference? Find the coefficient of variation for each of the two samples; then compare the variation. Male 89 72 63 68 72 53 63 52 82 71 62 61 98 54 65 Female 67 83 81 69 76 84 88 84 88 88 90 68 90 79 77

The coefficient of variation for the male pulse rates is 19.119.1%. (Type an integer or decimal rounded to one decimal place as needed.) The coefficient of variation for the female pulse rates is 9.89.8%. (Type an integer or decimal rounded to one decimal place as needed.) Compare the variation. The coefficients of variation for each data set are more than 5 percentage points apart. Therefore, the male pulse rates vary significantly more than the female pulse rates.

he boxplot shown below results from the heights (cm) of males listed in a data set. What do the numbers in that boxplot tell us?

The minimum height is 152152 cm, the first quartile Q1 is 168.5168.5 cm, the second quartile Q2 (or the median) is 175.8175.8 cm, the third quartile Q3 is 182.2182.2 cm, and the maximum height is 192192 cm.

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the given sample data. An experiment was conducted to determine whether a deficiency of carbon dioxide in the soil affects the phenotype of peas. Listed below are the phenotype codes where 1=smooth-yellow, 2=smooth-green, 3=wrinkled-yellow, and 4=wrinkled-green. Do the results make sense? 2 2 1 1 1 2 3 4 3 2 3 3 2 2 (a) The mean phenotype code is 2.22.2. (Round to the nearest tenth as needed.) (b) The median phenotype code is 22. (Type an integer or a decimal.) (c) Select the correct choice below and fill in any answer boxes within your choice. A. The mode phenotype code is 22. (Use a comma to separate answers as needed.) B. There is no mode. (d) The midrange of the phenotype codes is 2.52.5. (Type an integer or a decimal.) Do the measures of center make sense? A. All the measures of center make sense since the data is numerical. B. Only the mean, median, and mode make sense since the data is numerical. C. Only the mode makes sense since the data is nominal. This is the correct answer. D. Only the mean, median, and midrange make sense since the data is nominal.

The mode phenotype code is 22. Only the mode makes sense since the data is nominal.

Assume that 1300 births are randomly selected and 6 of the births are girls. Use subjective judgment to describe the number of girls as significantly high, significantly low, or neither significantly low nor significantly high. Choose the correct answer below. A. The number of girls is significantly high. B. The number of girls is neither significantly low nor significantly high. C. The number of girls is significantly low. This is the correct answer. D. It is impossible to make a judgment with the given information.

The number of girls is significantly low.

Which of the following is an important business application related to counting?

Traveling Salesman Problem

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below are foot lengths in inches of randomly selected women in a study of a country's military in 1988. Are the statistics representative of the current population of all women in that country's military? 10.4 8.5 8.7 8.5 8.6 9.1 9.7 10.5 9.2 8.7 8.7

a. Find the mean. The mean is 9.159.15 inch(es). (Type an integer or a decimal rounded to two decimal places as needed.) b. Find the median. The median is 8.78.7 inch(es). (Type an integer or a decimal. Do not round.) c. Find the mode. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. . The mode(s) is(are) 8.78.7 inch(es). Since the measurements were made in 1988, they are not necessarily representative of the current population of all women in the country's military.

Five pulse rates are randomly selected from a set of measurements. The five pulse rates have a mean of 84.8 beats per minute. Four of the pulse rates are 83, 92, 95, and 65. a. Find the missing value. b. Suppose that you need to create a list of n values that have a specific known mean. Some of the n values can be freely selected. How many of the n values can be freely assigned before the remaining values are determined? (The result is referred to as the number of degrees of freedom.)

a. The missing value is 8989 beats per minute. Of the n values, n minus 1n−1 can be freely selected because the remaining value(s) can be expressed in terms of the assigned values and the known mean.

The principle of redundancy is used when system reliability is improved through redundant or backup components. Assume that a student's alarm clock has a 16.0% daily failure rate. Complete parts (a) through (d) below.

a. What is the probability that the student's alarm clock will not work on the morning of an important final exam? To convert a percentage to a decimal number, remove the % symbol and divide by 100. For the stated failure rate of 16.0% remove the percent symbol and divide by 100. 16.0% = 16.0100 = 0.160 Thus, the probability that the student's alarm clock will not work on the morning of an important final exam is 0.160. b. If the student has two such alarm clocks, what is the probability that they both fail on the morning of an important final exam? Use the formal multiplication rule that states that if P(A) is the probability of event A occurring and P(B|A) is the probability of B occurring given that A has occurred, the probability of both A and B occurring is given by the formula below. P(A and B)=P(A)•P(B|A) Let B be the event that the second alarm clock functions properly, B be the event that it fails, and B|A be the event that the second alarm clock fails given that the first alarm clock has failed. Then, the formal multiplication rule becomes the equation below. PA and B=PA•PA|B The functioning of the second alarm clock is not affected by the failure of the first alarm clock, so by definition A and B are independent events. Since the failure of the second alarm clock is independent of the failure of the first, PB|A=PB. The formula from the formal multiplication rule becomes the formula shown below. PA and B = PA•PB Substitute the appropriate probabilities into the formula, and then simplify to find the probability of both alarm clocks not functioning correctly. PA and B = PA•PB = 0.160•0.160 = 0.0256 Thus, the probability that both alarm clocks will fail on the morning of an important final exam is 0.0256. c. What is the probability of not being awakened if the student uses three independent alarm clocks? Let C be the event that the third alarm clock functions properly, C be the event that it fails. Recall from part (b) that the failure of one alarm clock is independent of the failure of another alarm clock, so use the formula for PA and B and C given below. PA and B and C=PA•PB•PC Substitute PA=PB=PC=0.160 to find the probability of not being awakened if the student uses three alarm clocks. Round to five decimal places. P(all three fail) = PA and B and C = PA•PB•PC = 0.160•0.160•0.160 = 0.00410 Thus, the probability of not being awakened if the student uses three alarm clocks is 0.00410. d. Do the second and third alarm clocks result in greatly improved reliability? Compare the probability of one alarm clock not working to the probabilities of 2 or 3 alarm clocks not working. In general, when an event will occur with probability 1, it is called certain. An event occurring with probability less than or equal to 0.05 is called unlikely. An event occurring with probability 0 is called impossible. Use this information to determine if the second and third alarm clocks result in greatly improved reliability.

One common system for computing a grade point average (GPA) assigns 4 points to an A, 3 points to a B, 2 points to a C, 1 point to a D, and 0 points to an F. What is the GPA of a student who gets an A in a 3-credit course, a B in each of two 2-credit courses, a C in a 3-credit course, and a D in a 2-credit course? In this data set, grades for courses with more credits have a greater effect on the mean than grades for courses with less credits. Therefore, to find the mean of this data set, one should find the weighted mean. A weighted mean accounts for variations in the relative importance of data values. Each data value is assigned a weight and the weighted mean is calculated with the following formula. Weighted mean: x=∑(w•x)∑w The grades of A, B, C, and D represent data values of 4, 3, 2, and 1 respectively. The numbers of credits are the weights. Note that the sum of the weights is equal to the total number of class credits. Find the sum of all weights. ∑w=3+(2×2)+3+2=12 Multiply each score by its weight and find the sum of these products. ∑(w•x)=4×3+3×(2×2)+2×3+1×2=3232 Find the weighted mean. x=∑(w•x)∑w=3212=2.72.7 (Round to the nearest tenth as needed.) Thus, the student's mean grade point score is 2.7.

first, add all of the class weights then multiply each class weight by the units then divide by 12

Listed below are amounts (in millions of dollars) collected from parking meters by a security service company and other companies during similar time periods. Do the limited data listed here show evidence of stealing by the security service company's employees? Security Service Company: 1.4 1.7 1.6 1.3 1.6 1.4 1.7 1.8 1.5 1.6 Other Companies: 1.6 1.9 1.5 1.8 1.6 1.9 1.7 1.6 1.7 1.8 Find the coefficient of variation for each of the two samples, then compare the variation. The coefficient of variation for the amount collected by the security service company is 10.110.1%. (Round to one decimal place as needed.) The coefficient of variation for the amount collected by the other companies is 8.08.0%. (Round to one decimal place as needed.) Do the limited data listed here show evidence of stealing by the security service company's employees? Consider a difference of greater than 1% to be significant. A. Yes. There is not a significant difference in the variation. B. Yes. There is a significant difference in the variation. This is the correct answer. C. No. There is a significant difference in the variation.

yes, there is sig. difference in variation

A common computer programming rule is that names of variables must be between one and eight characters long. The first character can be any of the 26 letters, while successive characters can be any of the 26 letters or any of the 10 digits. For example, allowable variable names include A, BB, and M3477K. How many different variable names are possible? (Ignore the difference between uppercase and lowercase letters.)

2,095,681,645,538

A combination lock uses three numbers between 1 and 85 with repetition, and they must be selected in the correct sequence. Is the name of "combination lock" appropriate? Why or why not?

No, because the multiplication counting rule would be used to determine the total number of combinations.

Listed below are the amounts (dollars) it costs for marriage proposal packages at different baseball stadiums. Find the range, variance, and standard deviation for the given sample data. Include appropriate units in the results. Are there any outliers, and are they likely to have much of an effect on the measures of variation? 35 50 50 65 70 75 95 140 185 216 270 300 500 1500 3000 The range of the sample data is 29652965 dollars.dollars. (Type an integer or a decimal. Do not round.) The standard deviation of the sample data is 798.8798.8 dollars.dollars. (Round to one decimal place as needed.) The variance of the sample data is 638,024.4638,024.4 dollars squared .dollars2. (Round to one decimal place as needed.) Are there any outliers and, if so, are they likely to have much of an effect on the measures of variation?

Yes, the largest amounts are much higher than the rest of the data, and appear to be outliers. It is likely that these are having a large effect on the measures of variation.

1/4

A classic counting problem is to determine the number of different ways that the letters of "embarrass" can be arranged. Find that number. The number of different ways that the letters of "embarrass" can be arranged is 45360

9!/((2!)(2!)(2!)) = 45360

A classic counting problem is to determine the number of different ways that the letters of "necessary" can be arranged. Find that number. When some items are identical to others, the permutations rule states that the number of different permutations (order counts) when n items are available and all n are selected without replacement, but some of the items are identical to others: n1 are alike, n2 are alike, and nk are alike, is given by the expression below. The factorial symbol (!) denotes the product of decreasing positive whole numbers. For example, 4!=4•3•2•1=24. By special definition, 0!=1. n!n1!n2!•••nk! How many letters are in the word "necessary"? n=99 Since the word "necessary" contains 9 letters with 2 e's and 2 s's, use the permutations rule (when some items are identical) to count the number of different ways that the letters can be arranged. n!n1!n2!•••nk!=9!2!2! Evaluate the expression. 9!2!2!=9072090720 (Simplify your answer.) Therefore, the number of different ways that the letters of "necessary" can be arranged is 90,720.

9!/((2!)(2!))

With a short time remaining in the day, a delivery driver has time to make deliveries at 3 locations among the 9 locations remaining. How many different routes are possible? There are 504504 possible different routes.

9P3

What do the preceding results suggest? A. A student given a $1 bill is more likely to have spent the money. B. A student given a $1 bill is more likely to have kept the money. This is the correct answer. C. A student given a $1 bill is more likely to have kept the money than a student given four quarters. D. A student given a $1 bill is more likely to have spent the money than a student given four quarters.

A student given a $1 bill is more likely to have kept the money.

What do the preceding results suggest? A. A student given four quarters is more likely to have kept the money than a student given a $1 bill. B. A student given four quarters is more likely to have spent the money. Your answer is correct. C. A student given four quarters is more likely to have kept the money. D. A student given four quarters is more likely to have spent the money than a student given a $1 bill.

A student given four quarters is more likely to have spent the money.

A presidential candidate plans to begin her campaign by visiting the capitals in 4 of 39 states. What is the probability that she selects the route of four specific capitals?

Determine if the possible arrangements of visiting the capitals are combinations or permutations of events. Since the capitals are distinct, no one capital can be visited more than once, and the order in which capitals are visited matters, use permutations to count the number of possible routes. A situation using permutations (when all the items are different) must satisfy three requirements. First, there must be n different items available. Also, one must select r of the n items without replacement. Finally, rearrangements of the same items are counted as being different. The number of permutations (or sequences) of r items selected from n different available items (without replacement) is given by the formula below. The factorial symbol (!) denotes the product of decreasing positive whole numbers. For example, 4!=4•3•2•1=24. By special definition, 0!=1. nPr=n!(n−r)! Substitute the appropriate values into the formula. 39P4=39!(39−4)! Now evaluate. 39!(39−4)!=1,974,024 Therefore, there are 1,974,024 different possible ways to visit the capitals in 4 of the 39 states. The probability that she selects the route of four specific capitals is equal to the proportion of the number of ways to visit the four specific capitals to the number of possible routes. The number of ways to visit the four specific capitals is 1, and the number of possible routes is 1,974,024. Assume that a given procedure has n different simple events and that each of those simple events has an equal chance of occurring. If event A can occur in s of these n ways, the P(A) is given by the formula below. P(A)=number of ways A occurnumber of different simple events=sn Now find the probability that she selects the route of four specific capitals. P(selects the route of four specific capitals) = number of ways A occur number of different simple events = 11,974,024 Therefore, the probability that she selects the route of four specific capitals is 11,974,024.

A modified roulette wheel has 36 slots. One slot is 0, another is 00, and the others are numbered 1 through 34, respectively. You are placing a bet that the outcome is an odd number. (In roulette, 0 and 00 are neither odd nor even.) a. What is your probability of winning? The probability of winning is StartFraction 17 Over 36 EndFraction1736. (Type an integer or a simplified fraction.) b. What are the actual odds against winning? The actual odds against winning are 1919:1717. c. When you bet that the outcome is an odd number, the payoff odds are 1:1. How much profit do you make if you bet $13 and win? If you win, the payoff is $13 d. How much profit should you make on the $13 bet if you could somehow convince the casino to change its payoff odds so that they are the same as the actual odds against winning? $14.53 (Round to the nearest cent as needed.)

First, find out how many odd numbers are between 0 and 36 Part a) 17/36 Part b) 1-Ans Ans>Frac Part C) $13 Part D) 13*(19/17) = 14.53

In a computer instant messaging survey, respondents were asked to choose the most fun way to flirt, and it found that P(D)=0.880, where D is directly in person. If someone is randomly selected, what does PD represent, and what is its value?

If A represents an event, A is used to indicate that A does not occur. Based on the given information, P(D) is the probability of randomly selecting someone who chooses a direct in-person encounter as the most fun way to flirt. To determine what PD represents, think about what must be true in the case that event D does not occur. Based on the rule of complement, PD is given by the following formula. PD=1−P(D) Substitute probability P(D) into the equation. PD=1−0.880 Calculate the complement of D. PD=0.120 Therefore, the probability of randomly selecting someone who does not choose a direct in-person encounter as the most fun way to flirt, PD, is 0.120.

Based on a poll, 72% of Internet users are more careful about personal information when using a public Wi-Fi hotspot. What is the probability that among three randomly selected Internet users, at least one is more careful about personal information when using a public Wi-Fi hotspot? How is the result affected by the additional information that the survey subjects volunteered to respond?

Let A=at least 1 of the 3 randomly selected Internet users being more careful with personal information when using a public Wi-Fi hotspot. To find the probability that at least one of the three randomly selected Interner users is more careful with personal information, first find the probability that none of the three randomly selected Internet users are more careful. The probability of "at least one" can be computed using the rule of complements. The rule of complements states that the following expression is true for events A and A, where A indicates that event A did not take place. P(A)=1−PA Identify the event that is the complement of A. A = none of the Internet users are more careful To find the probability of the complement, first find the probability that an Internet user is not more careful with personal information while using a public Wi-Fi hotspot. P(is not more careful) = 1−P(is more careful) = 1−0.72 = 0.28 Find the probability of the complement using the multiplication rule for independent events, rounding to three decimal places. The multiplication rule for independent events states that the probability of two independent events occurring is the product of their individual probabilities. This can be extended to 3 independent events. PA = P(user 1 is not more careful and user 2 is not more careful and user 3 is not more careful) = 0.28•0.28•0.28 = 0.022 Now find P(A) using the rule of complements, P(A)=1−PA. P(A) = 1−PA = 1−0.022 = 0.978 Therefore, if three Internet users are randomly selected, there is a 0.978 probability that at least one of them is more careful with personal information when using a public Wi-Fi hotspot. A probability result can be invalid if the sample is not randomly selected or if the sample is not representative of the population or if one outcome has an effect on later trials. Use this information and the information given in the problem statement to determine if there is any reason why the probability result might be invalid.

PD is the probability of randomly selecting someone who chooses a directin-person encounter as the most fun way to flirt. P(D_)=0.45 1-.550=

When a man observed a sobriety checkpoint conducted by a police department, he saw 692 drivers were screened and 4 were arrested for driving while intoxicated. Based on those results, we can estimate that P(W)=0.00578, where W denotes the event of screening a driver and getting someone who is intoxicated. What does PW denote, and what is its value?

PW denotes the probability of screening a driver and finding that he or she is not intoxicated 1-0.00578=0.99422

For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of (m)(n) ways. This is called _______. For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of (m)(n) ways. This is called the fundamental counting rule

The Fundamental Counting Rule says that for a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of (m)(n) ways. Next Question

The combination and permutations rules do not apply because repetition is allowed and numbers are selected with replacement. The multiplication counting rule applies to this problem.

When a man observed a sobriety checkpoint conducted by a police department, he saw 767 drivers were screened and 13 were arrested for driving while intoxicated. Based on those results, we can estimate that P(W)=0.01695, where W denotes the event of screening a driver and getting someone who is intoxicated. What does PW denote, and what is its value?

The complement of event A, denoted by A, consists of all outcomes in which event A does not occur. In this situation, it is given that W denotes the event of screening a driver and getting someone who is intoxicated. In this situation, W is the event of screening a driver and getting someone who is not intoxicated. Therefore, PW denotes the probability of screening a driver and finding that he or she is not intoxicated. If A represents an event and A represents the complement of that event, then the probability of A can be found by subtracting P(A) from 1, where P(A) is the probability of A occurring. Subtract P(W) from 1 to find PW. PW = 1−0.01695 Calculate PW. PW = 0.98305 Therefore, the probability of screening a driver and finding that he or she is not intoxicated, PW, is 0.98305.

In a clinical trial of 2195 subjects treated with a certain drug, 21 reported headaches. In a control group of 1600 subjects given a placebo, 21 reported headaches. Denoting the proportion of headaches in the treatment group by pt and denoting the proportion of headaches in the control (placebo) group by pc, the relative risk is pt/pc. The relative risk is a measure of the strength of the effect of the drug treatment. Another such measure is the odds ratio, which is the ratio of the odds in favor of a headache for the treatment group to the odds in favor of a headache for the control (placebo) group, found by evaluating pt/1−ptpc/1−pc. The relative risk and odds ratios are commonly used in medicine and epidemiological studies. Find the relative risk and odds ratio for the headache data. What do the results suggest about the risk of a headache from the drug treatment? Find the relative risk for the headache data. The relative risk=0.7290.729 (Round to three decimal places as needed.) Find the odds ratio for the headache data. The odds ratio=0.7290.729 (Round to three decimal places as needed.) What do the results suggest about the risk of a headache from the drug treatment? A. The drug does not appear to pose a risk of headaches because pt is slightly less than pc. Your answer is correct. B. The drug has no risk because the relative risk and odds ratio are almost equal. C. The drug appears to pose a risk of headaches because the odds ratio is greater than 1.0. D. The drug appears to pose a risk of headaches because pt is greater than pc.

The drug does not appear to pose a risk of headaches because pt is slightly less than pc. Find the relative risk for the headache data. The relative risk=0.7290.729 (Round to three decimal places as needed.) Calculate this, first 21/2195=A 21/1600=B A/B = C C/(1-C) Ans/B = Ans-0.005

If you decide to buy one of these TVs, what statistic is most relevant, other than the measures of central tendency? Choose the best answer below. A. The price that occurs least frequently is a relevant statistic for someone planning to buy one of the TVs. Your answer is not correct. B. The difference between the lowest price and the highest price is a relevant statistic for someone planning to buy one of the TVs. C. The lowest price is a relevant statistic for someone planning to buy one of the TVs. This is the correct answer. D. The highest price is a relevant statistic for someone planning to buy one of the TVs.

The lowest price is a relevant statistic for someone planning to buy one of the TVs.

The latest results from a method of gender selection shows that among 152 sets of parents using the method for increasing the likelihood of a boy, 129 actually had boys and the other 23 had girls. Assuming that the method has no effect and that boys and girls are equally likely, a simulation of 152 births was conducted. The simulated number of boys and girls are shown on the right for 10 cases. Is it unusual to get 129 boys in 152 births? What does the result suggest about the method? Is it unusual to get 129 boys in 152 births? Yes Your answer is correct. No What does the result suggest about the method? A. The method is effective because it is not unusual to get 129 boys in 152 births when boys and girls are equally likely. B. The method is not effective because it is not unusual to get 129 boys in 152 births when boys and girls are equally likely. C. The method is not effective because it is unusual to get 129 boys in 152 births when boys and girls are equally likely. D. The method is effective because it is unusual to get 129 boys in 152 births when boys and girls are equally likely.

The method is effective because it is unusual to get 129 boys in 152 births when boys and girls are equally likely.

The latest results from a method of gender selection shows that among 152 sets of parents using the method for increasing the likelihood of a boy, 124 actually had boys and the other 28 had girls. Assuming that the method has no effect and that boys and girls are equally likely, a simulation of 152 births was conducted. The simulated number of boys and girls are shown on the right for 10 cases. Is it unusual to get 124 boys in 152 births? What does the result suggest about the method? Is it unusual to get 124 boys in 152 births? Yes What does the result suggest about the method? A. The method is not effective because it is not unusual to get 124 boys in 152 births when boys and girls are equally likely. B. The method is not effective because it is unusual to get 124 boys in 152 births when boys and girls are equally likely. Your answer is not correct. C. The method is effective because it is not unusual to get 124 boys in 152 births when boys and girls are equally likely. D. The method is effective because it is unusual to get 124 boys in 152 births when boys and girls are equally likely.

The method is effective because it is unusual to get 124 boys in 152 births when boys and girls are equally likely.

The probability is StartFraction 1 Over 100,000 1/100,000

Use the data in the following table, which lists drive-thru order accuracy at popular fast food chains. Assume that orders are randomly selected from those included in the table. If one order is selected, find the probability of getting an order that is not accurate.A B C D Order Accurate 328 268 249 128 Order Not Accurate 35 57 38 13

The probability of getting an order that is not accurate is . 128 First add all of the columns with "Order not accurate" and then divide by total number which is 1116 143/116 = 0.128

The probability of getting food that is not from Restaurant A is . 676 First add all of the numbers 365/1128 = 0.32 1-Ans = 0.676

In a certain country, the true probability of a baby being a girl is 0.484. Among the next nine randomly selected births in the country, what is the probability that at least one of them is a boy?

The probability of "at least one" can be computed using the rule of complements. Let A represent the event that at least one of the next nine births is a boy. Use the rule of complements below to find the probability of event A, P(A), where A is the complement of A. P(A)=1−PA What is the complement of A, A? The next nine births are all girls Since each birth has no effect on any of the other births, the births are all independent events. The probability that the next nine births are all girls can be found using the multiplication rule for independent events. The probability of the event can be written as shown below. P(all 9 are girls) = P1st is a girl and 2nd is a girl...and last is a girl What is the probability of a birth being a girl? 0.484 Use the multiplication rule for independent events to find the probability that the next nine births are all girls. The multiplication rule for independent events states that the probability of two independent events occurring is the product of their individual probabilities. This can be extended to 9 independent events. PA = P1st is a girl and 2nd is a girl...and last is a girl = 0.484•0.484•0.484•0.484•0.484•0.484•0.484•0.484•0.484 = 0.001 (Round to three decimal places as needed.) Then use the rule of complements to find the probability that at least one of the next nine randomly selected births is a boy. P(A) = 1−PA = 1−0.001 = . 999

The conditional probability P(B|A) is the probability of event B occurring, given that event A has already occurred. In this problem, event A=subject is telling the truth, so it is assumed that the subject is telling the truth when calculating P(B|A). Since event B= polygraph test indicates that the subject is lying, P(B|A) is the probability that the polygraph indicates lying, given that the subject is actually telling the truth. Let event A=subject is telling the truth and event B=polygraph test indicates that the subject is lying. Use your own words to translate the notation P(B|A) into a verbal statement.

The probability that the polygraph indicates lying given that the subject is actually telling the truth.

Are the resulting statistics representative of the population of all TVs that are 60 inches and larger? Choose the best answer below. A. The sample consists of the "best buy" TVs, so it is not a random sample and is not likely to be representative of the population. This is the correct answer. B. Since the sample is random and the sample size is greater than 10, the sample can be considered to be representative of the population. Your answer is not correct. C. Since the sample is random and the sample size is greater than 10, the sample should not be considered to be representative of the population. D. The sample consists of the "best buy" TVs, so it is a random sample and is likely to be representative of the population.

The sample consists of the "best buy" TVs, so it is not a random sample and is not likely to be representative of the population.

The term average is not used in statistics. The term mean should be used for the result obtained by adding all of the sample values and dividing by the total number of sample values. B. The term average is not used in statistics. The term median should be used for the result obtained by adding all of the sample values and dividing by the total number of sample values. C. The term average is often used in statistics to represent the median. D. The term average is often used in statistics to represent the mean.

The term average is not used in statistics. The term mean should be used for the result obtained by adding all of the sample values and dividing by the total number of sample values.

In a clinical trial of 2079 subjects treated with a certain drug, 23 reported headaches. In a control group of 1651 subjects given a placebo, 20 reported headaches. Denoting the proportion of headaches in the treatment group by pt and denoting the proportion of headaches in the control (placebo) group by pc, the relative risk is pt/pc. The relative risk is a measure of the strength of the effect of the drug treatment. Another such measure is the odds ratio, which is the ratio of the odds in favor of a headache for the treatment group to the odds in favor of a headache for the control (placebo) group, found by evaluating pt/1−ptpc/1−pc. The relative risk and odds ratios are commonly used in medicine and epidemiological studies. Find the relative risk and odds ratio for the headache data. What do the results suggest about the risk of a headache from the drug treatment?

To find the relative risk for the headache data, first calculate the proportion of headaches in the treatment group, rounding to four decimal places. pt = number of subjects who reported headaches in the treatment grouptotal number of subjects in the treatment group = 232079 = 0.0111 Then calculate the proportion of headaches in the control group, rounding to four decimal places. pc = number of subjects who reported headaches in the control group total number of subjects in the control group = 201651 = 0.0121 Now calculate the relative risk using the formula given in the problem statement, rounding to three decimal places. The relative risk = ptpc = 0.01110.0121 = 0.917 To find the odds ratio, first calculate the odds in favor of a headache for the treatment group. Recall that pt=0.0111, rounding to four decimal places. pt1−pt = 0.01111−0.0111 0.0112 Then calculate the odds in favor of a headache for the control group. Recall that pc=0.0121, rounding to four decimal places. pc1−pc = 0.01211−0.0121 = 0.0122 Now calculate the odds ratio using the formula given in the problem statement, rounding to three decimal places. The odds ratio = pt/1−ptpc/1−pc 0.01120.0122 = 0.918 What do the results suggest about the risk of a headache from the drug treatment? Compare the proportion pt of a headache with the drug to the proportion pc of a headache with the placebo. The drug does not pose a risk of headaches as long as pt is less than than pc. Since pt=0.0111 is less than pc=0.0121, the drug does not pose a risk of headaches.

Find the indicated complement. A certain group of women has a 0.49% rate of red/green color blindness. If a woman is randomly selected, what is the probability that she does not have red/green color blindness?

What is the probability that the woman selected does not have red/green color blindness? . 9951 1-0.0049

Find the indicated complement. A certain group of women have a 0.39% rate of red/green color blindness. If a woman is randomly selected, what is the probability that she does not have red/green color blindness?

What is the probability that the woman selected does not have red/green color blindness? To find this probability, first determine the probability that the woman selected does have red/green color blindness. P(woman has red/green color blindness)=0.0039 The question asks for the probability that the woman selected does not have red/green color blindness. This is the complement of the event where the woman does have red/green color blindness. Use the following formula to determine the complement. P(not having red/green color blindness) = 1−P(having red/green color blindness) = 1−0.0039 = 0.9961

a. If the two surge protectors are arranged in series, what is the probability that a voltage surge will not damage the television? . 9991 (Do not round.) 1-.97=0.03 Ans^2 1-Ans If the two surge protectors are arranged in parallel, what is the probability that a voltage surge will not damage the television? . 9409 (Do not round.) .97^2 = 0.9409

Assume that there is a 12% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive? b. If copies of all your computer data are stored on three independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

a. Use the rule of complements shown below to find the probability that you can avoid catastrophe, where A is the complement of A. Let A=at least 1 hard drive works correctly. P(A)=1−PA Identify the event that is the complement of A. A = both hard drives fail Since the two hard drives operate separately, their failures are independent events. Use the multiplication rule for independent events to find the probability of the complement of event A, A. The multiplication rule for independent events states that the probability of two independent events occurring is the product of their individual probabilities.The probability of any one of the hard drives failing to work correctly is 0.12. PA = P(hard drive 1 fails and hard drive 2 fails) = 0.12•0.12 = 0.0144 Now find P(A) by evaluating 1−PA. P(A) = 1−PA = 1−0.0144 = 0.9856 Therefore, the probability that at least one of the two hard drives works correctly is 0.9856. b. Again let A=at least 1 hard drive works correctly. Using the same method as in part (a), find the probability of the complement of event A. PA = P(hard drive 1 fails and hard drive 2 fails and hard drive 3 fails) = 0.12•0.12•0.12 = 0.001728 Now find P(A) by evaluating 1−PA. P(A) = 1−PA = 1−0.001728 = 0.998272 Therefore, the probability that at least one hard drive out of three works is 0.998272.

Assume that there is a 11% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive? b. If copies of all your computer data are stored on four independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

a. With two hard disk drives, the probability that catastrophe can be avoided is .9879 (Round to four decimal places as needed.) b. With four hard disk drives, the probability that catastrophe can be avoided is . 999854 (Round to six decimal places as needed.)

In the game of blackjack played with one deck, a player is initially dealt 2 different cards from the 52 different cards in the deck. A winning "blackjack" hand is won by getting 1 of the 4 aces and 1 of 16 other cards worth 10 points. The two cards can be in any order. Find the probability of being dealt a blackjack hand. What approximate percentage of hands are winning blackjack hands? The probability is StartFraction 32 Over 663 EndFraction32663. (Type an integer or a simplified fraction.) The percentage of hands that are winning blackjack hands is 4.83%.

nCr in selecting 2 of the 52 cards in the deck, n = 52 and r = 2 52C2= (52!)/(2!(52-2)! A blackjack consists of 1 of 4 aces and 1 of 16 other cards that are worth 10 points. Ace: 4 ways Other card: 16 ways 4*16 = 64 64 of the 1326 possible hands result in a blackjack. The probably is then number of favorable outcomes divided by number of possible outcomes P(Blackjack) = 64/1326 = 32/663 = 0.04827 = 4.827% The probability is StartFraction 32 Over 663 EndFraction32663 32/663 The percentage of hands that are winning blackjack hands is 4.83%

A combination lock uses three numbers between 1 and 47 with repetition, and they must be selected in the correct sequence. Is the name of "combination lock" appropriate? Why or why not? Use the permutations rule (when all of the items are different) to find the number of different permutations (order counts) when n different items are available, but only r of them are selected without replacement. Use the permutations rule (when some items are identical to others) to find the number of different permutations (order counts) when n items are available and all n are selected without replacement, but some of the items are identical to others. Use the combinations rule to find the number of different combinations (order does not count) when n different items are available, but only r of them are selected without replacement. The multiplication counting rule states that the number of ways that two events can occur, given that the first event can occur m ways and the second event can occur n ways is given by m•n. Use the factorial rule to find the number of different permutations (order counts) of n different items when all n of them are selected.

No, because the multiplication counting rule would be used to determine the total number of combinations.

A magazine published a list consisting of the state tax on each gallon of gas. If we add the 50 state tax amounts and then divide by 50, we get 27.3 cents. Is the value of 27.3 cents the mean amount of state sales tax paid by all U.S. drivers? Why or why not? Choose the correct answer below. A. Yes, the value of 27.3 cents is the mean because a mean is found by summing all sample values and then dividing the by total number of values. B. No, the value of 27.3 cents is not the mean because the 50 amounts are all weighted equally in the calculation, but some states consume more gas than others, so the mean amount of state sales tax should be calculated using a weighted mean. C. Yes, the value of 27.3 cents is the mean because it is a measure of center and all measures of centers can be refered to as either an average or a mean. D. No, the value of 27.3 cents is not the mean because it is the median that is found by summing all sample values and then dividing the by total number of values.

No, the value of 27.3 cents is not the mean because the 50 amounts are all weighted equally in the calculation, but some states consume more gas than others, so the mean amount of state sales tax should be calculated using a weighted mean.

If you know the names of the remaining two students in the spelling bee, what is the probability of randomly selecting an order and getting the order that is used in the spelling bee?

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