Chapter 5 and 6

Ace your homework & exams now with Quizwiz!

Assume the readings on thermometers are normally distributed with a mean of 0degrees°C and a standard deviation of 1.00degrees°C. Find the probability that a randomly selected thermometer reads greater than 2.132.13 and draw a sketch of the region.

.0166

.0188

Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.\ 70

.0228

Assume the readings on thermometers are normally distributed with a mean of 0degrees°C and a standard deviation of 1.00degrees°C. Find the probability Upper P left parenthesis z less than minus 2.16 or z greater than 2.16 right parenthesisP(z<−2.16 or z>2.16), where z is the reading in degrees.

.0308

Assume the readings on thermometers are normally distributed with a mean of 0degrees°C and a standard deviation of 1.00degrees°C. Find the probability that a randomly selected thermometer reads between negative 2.12−2.12 and negative 1.61−1.61 and draw a sketch of the region.

.0367

Assume the readings on thermometers are normally distributed with a mean of 0degrees°C and a standard deviation of 1.00degrees°C. Find the probability Upper P left parenthesis negative 1.55 less than z less than 1.55 right parenthesisP(−1.55<z<1.55), where z is the reading in degrees.

.8788

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees0° and standard deviation of 1.00degrees°C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to Upper P 94P94, the 94 th94th percentile. This is the temperature reading separating the bottom 94 %94% from the top 6 %.

1.56

Assume that adults have IQ scores that are normally distributed with a mean of 105105 and a standard deviation of 15. Find the third quartile Upper Q 3Q3, which is the IQ score separating the top 25% from the others.

115.1

About _____% of the area is between zequals=negative 1.6−1.6 and zequals=1.61.6 (or within 1.61.6 standard deviationsdeviations of the mean).

About 89.0489.04% of the area is between zequals=negative 1.6−1.6 and zequals=1.61.6 (or within 1.61.6 standard deviationsdeviations of the mean).

A _______ random variable has infinitely many values associated with measurements.

Continuous

There is a 0.05310.0531 probability that a best-of-seven contest will last four games, a 0.11860.1186 probability that it will last five games, a 0.28090.2809 probability that it will last six games, and a 0.54740.5474 probability that it will last seven games. Verify that this is a probability distribution. Find its mean and standard deviation.

Mean = 6.32 SD = .88

Probability is .3409

A _______ variable is a variable that has a single numerical value, determined by chance, for each outcome of a procedure.

Random

Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. .85 = area

The indicated IQ score, x, is 115.6115.6

Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

The indicated IQ score, x, is 71.871.8.

What requirements are necessary for a normal probability distribution to be a standard normal probability distribution?

The mean and standard deviation have the values of mu equals 0μ=0 and sigma equals 1.

What requirements are necessary for a normal probability distribution to be a standard normal probability distribution? Choose the correct answer below.

The mean and standard deviation have the values of mu equals 0μ=0 and sigma equals 1.

What requirements are necessary for a normal probability distribution to be a standard normal probability distribution?

The mean and standard deviation have the values of mu equals μ=0 and sigma equals σ=1.

The probability is 0.97880.9788.

The probability is 0.98260.9826.

Conclusion of Central Limit Theorem

The mean of all sample means is the population mean muμ. The distribution of the sample means x overbarx will, as the sample size increases, approach a normal distribution. The standard deviation of all sample means is the population standard deviation divided by the square root of the sample size.

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 100μ=100 and a standard deviation sigma equals 20σ=20. Find the probability that a randomly selected adult has an IQ between 8686 and 114114.

The probability that a randomly selected adult has an IQ between 8686 and 114114 is 0.51610.5161.

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105μ=105 and a standard deviation sigma equals 20σ=20. Find the probability that a randomly selected adult has an IQ between 8787 and 123123.

The probability that a randomly selected adult has an IQ between 8787 and 123123 is 0.63190.6319.

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 100μ=100 and a standard deviation sigma equals 15σ=15. Find the probability that a randomly selected adult has an IQ between 9090 and 110110.

The probability that a randomly selected adult has an IQ between 9090 and 110110 is . 4972.4972.

The probability that a randomly selected adult has an IQ between 9090 and 110110 is 0.38290.3829.

Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Surveying 100 married individuals and recording if they have ever been divorcedSurveying 100 married individuals and recording if they have ever been divorced nothing

Yes because there are more than two possible outcomes

Determine whether the distribution is a discrete probability distribution. x P(x) 0 0.210.21 1 0.280.28 2 0.020.02 3 0.280.28 4

Yes, Because the probabilities sum to 1 and are all between 0 and 1 inclusive

Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. An experimental drug is administered to 120120 randomly selected individuals, with the number of individuals responding favorably recorded.

Yes, because the experiment satisfies all the criteria for a binomial experiment.

Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. area = .9

x = 80.8

Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. z = 0.75

x = 89.9

Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. z = .65

x = 94.2

Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. An experimental drug is administered to 200200 randomly selected individuals, with the number of individuals responding favorably recorded. Does the probability experiment represent a binomial experiment?

Yes, because the experiment satisfies all the criteria for a binomial experiment.

Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. An experimental drug is administered to 6060 randomly selected individuals, with the number of individuals responding favorably recorded.

Yes, because the experiment satisfies all the criteria for a binomial experiment.

If you are asked to find the 85th percentile, you are being asked to find _____.

a data value associated with an area of 0.85 to its left

f you are asked to find the 85th percentile, you are being asked to find _____.

a data value associated with an area of 0.85 to its left

Assume that adults have IQ scores that are normally distributed with a mean of 105105 and a standard deviation 1515. Find Upper P 15P15, which is the IQ score separating the bottom 1515% from the top 8585%.

he IQ score that separates the bottom 1515% from the top 8585% is Upper P 15P15equals= 89.4589.45.

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

he area of the shaded region is 0.5557

Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. An experimental drug is administered to 5050 randomly selected individuals, with the number of individuals responding favorably recorded.

Yes, because the experiment satisfies all the criteria for a binomial experiment.

Five males with a particular genetic disorder have one child each. The random variable x is the number of children among the five who inherit the genetic disorder. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. x 0 1 2 3 4 5 P(x) 0.34870.3487 0.40890.4089 0.19190.1919 0.04500.0450 0.00530.0053 0.00020.0002

mean = 0.9 SD = 0.9

In the accompanying table, the random variable x represents the number of televisions in a household in a certain country. Determine whether or not the table is a probability distribution. If it is a probability distribution, find its mean and standard deviation. x P(x) 0 0.040.04 1 0.110.11 2 0.260.26 3 0.320.32 4 0.160.16 5

mean = 2.8 SD 1.3

A candy company claims that 1212% of its plain candies are orange, and a sample of 200200 such candies is randomly selected.

mean = 24 SD = 4.6

Five males with a particular genetic disorder have one child each. The random variable x is the number of children among the five who inherit the genetic disorder. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. x 0 1 2 3 4 5 P(x) 0.00000.0000 0.00070.0007 0.01050.0105 0.08530.0853 0.34510.3451 0.55840.5584

mean = 4.5 SD = 0.7

A candy company claims that 2121% of its plain candies are orange, and a sample of 200200 such candies is randomly selected.

mu = 42 sigma = 5.8

There is a 0.10710.1071 probability that a best-of-seven contest will last four games, a 0.16990.1699 probability that it will last five games, a 0.29620.2962 probability that it will last six games, and a 0.42680.4268 probability that it will last seven games. Verify that this is a probability distribution. Find its mean and standard deviation.

mu = 6.04 Sigma = 1.01

Identify the expression for calculating the mean of a binomial distribution.

In the binomial probability formula, the variable x represents the _______.

number of successes

Assume that a procedure yields a binomial distribution with nequals=66 trials and a probability of success of pequals=0.600.60. Use a binomial probability table to find the probability that the number of successes x is exactly 11.

p (1) = .037

Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. nequals=9, xequals=4, pequals=0.45

p (4) = .26

p (4) = .263

ssume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. nequals=19, xequals=15, pequals=0.85

p(15) = .171

Ted is not particularly creative. He uses the pickup line "If I could rearrange the alphabet, I'd put U and I together." The random variable x is the number of girls Ted approaches before encountering one who reacts positively. Determine whether the table describes a probability distribution. If it does, find its mean and standard deviation. x P(x) 1 0.0010.001 2 0.0170.017 3 0.1050.105 4 0.2020.202 5 0.4500.450

- not a probability distribution

. 8962

.0132

.0150

Determine whether the distribution is a discrete probability distribution.

Yes because the prob sum to 1 and are all between 0 and 1, inclusive

.0764

Assume the readings on thermometers are normally distributed with a mean of 0degrees°C and a standard deviation of 1.00degrees°C. Find the probability that a randomly selected thermometer reads between negative 2.29−2.29 and negative 0.75−0.75 and draw a sketch of the region.

.2156

.2843

.5478

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. z= .38

.6480

.9535

0.0192

Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. area = .9838

2.14 = z score

Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

70-105 =.6065

Assume that adults have IQ scores that are normally distributed with a mean of 100100 and a standard deviation 2020. Find Upper P 9P9, which is the IQ score separating the bottom 99% from the top 9191%.

73.2

About _____% of the area is between zequals=negative 2−2 and zequals=22 (or within 22 standard deviationsdeviations of the mean).

95.44

Determine whether the value is a discrete random variable, continuous random variable, or not a random variable. a. The time it takes for a light bulb to burn outtime it takes for a light bulb to burn out b. The number of points scored during a basketball gamenumber of points scored during a basketball game c. The gender of college studentsgender of college students d. The number of fish caught during a fishing tournamentnumber of fish caught during a fishing tournament e. The number of textbook authors now sitting at a computernumber of textbook authors now sitting at a computer f. The time required to download a file from the Internet

A- Contin B - Discrete C - not random D - Discrete E - Discrete F - Continuous

What requirements are necessary for a normal probability distribution to be a standard normal probability distribution? Choose the correct answer below.

A. The mean and standard deviation have the values of mu equals 0μ=0 and sigma equals 1.

Which of the following is not true?

A z-score is an area under the normal curve.

Which of the following is not true?

A z-score is an area under the normal curve. This is the correct answer.C.

The population of current statistics students has ages with mean muμ and standard deviation sigmaσ. Samples of statistics students are randomly selected so that there are exactly 4949 students in each sample. For each sample, the mean age is computed. What does the central limit theorem tell us about the distribution of those mean ages?

Because ngreater than>30, the sampling distribution of the mean ages can be approximated by a normal distribution with mean muμ and standard deviation StartFraction sigma Over StartRoot 49 EndRoot EndFraction σ 49.

The _______ tells us that for a population with any distribution, the distribution of the sample means approaches a normal distribution as the sample size increases.

Central Limit Theorem

The _______ tells us that for a population with any distribution, the distribution of the sample means approaches a normal distribution as the sample size increases.

Central limit Theorem

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. z= .72

The area of the shaded region is 0.76420.7642.

Which of the following is NOT a conclusion of the Central Limit Theorem? Choose the correct answer below.

The distribution of the sample data will approach a normal distribution as the sample size increases.

The capacity of an elevator is 1515 people or 23852385 pounds. The capacity will be exceeded if 1515 people have weights with a mean greater than 2385 divided by 15 equals 159 pounds.2385/15=159 pounds. Suppose the people have weights that are normally distributed with a mean of 165 lb165 lb and a standard deviation of 30 lb30 lb.

Find the probability that if a person is randomly selected, his weight will be greater than 159159 pounds. The probability is approximately 0.57930.5793. (Round to four decimal places as needed.) b. Find the probability that 1515 randomly selected people will have a mean that is greater than 159159 pounds. The probability is approximately 0.78070.7807. (Round to four decimal places as needed.) Does the elevator appear to have the correct weight limit? Why or why not? No, there is a good chance that 1515 randomly selected people will exceed the elevator capacity.

Five males with a particular genetic disorder have one child each. The random variable x is the number of children among the five who inherit the genetic disorder. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. x 0 1 2 3 4 5 P(x) 0.00140.0014 0.01940.0194 0.10490.1049 0.28360.2836 0.38340.3834 0.20730.2073

Find the mean of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. muμequals= 3.7 Find the standard deviation of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. equals= 1.01.0

The capacity of an elevator is 1010 people or 15701570 pounds. The capacity will be exceeded if 1010 people have weights with a mean greater than 1570 divided by 10 equals 157 pounds.1570/10=157 pounds. Suppose the people have weights that are normally distributed with a mean of 164 lb164 lb and a standard deviation of 26 lb26 lb.

Find the probability that if a person is randomly selected, his weight will be greater than 157157 pounds. The probability is approximately 0.60610.6061. Find the probability that 1010 randomly selected people will have a mean that is greater than 157157 pounds. The probability is approximately 0.80270.8027. No, there is a good chance that 1010 randomly selected people will exceed the elevator capacity.

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150150 lb and 191191 lb. The new population of pilots has normally distributed weights with a mean of 157 lb157 lb and a standard deviation of 34.7 lb34.7 lb.

If a pilot is randomly selected, find the probability that his weight is between 150150 lb and 191191 lb. The probability is approximately 0.41630.4163. If 4040 different pilots are randomly selected, find the probability that their mean weight is between 150150 lb and 191191 lb. The probability is approximately 0.89900.8990. Part (a) because the seat performance for a single pilot is more important.

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150150 lb and 211211 lb. The new population of pilots has normally distributed weights with a mean of 157 lb157 lb and a standard deviation of 29.2 lb29.2 lb.

If a pilot is randomly selected, find the probability that his weight is between 150150 lb and 211211 lb. The probability is approximately 0.56250.5625. (Round to four decimal places as needed.) b. If 3131 different pilots are randomly selected, find the probability that their mean weight is between 150150 lb and 211211 lb. The probability is approximately 0.90900.9090. (Round to four decimal places as needed.) c. When redesigning the ejection seat, which probability is more relevant? Part (a) because the seat performance for a single pilot is more important.

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150150 lb and 211211 lb. The new population of pilots has normally distributed weights with a mean of 159 lb159 lb and a standard deviation of 29.8 lb29.8 lb.

If a pilot is randomly selected, find the probability that his weight is between 150150 lb and 211211 lb. The probability is approximately 0.57820.5782. b. If 4040 different pilots are randomly selected, find the probability that their mean weight is between 150150 lb and 211211 lb. The probability is approximately 0.97190.9719. c. When redesigning the ejection seat, which probability is more relevant? Part (a) because the seat performance for a single pilot is more important.

Based on a survey, for women aged 18 to 24, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.8114.8 and a standard deviation of 13.313.3. Complete parts (a) through (c).

If a woman between the ages of 18 and 24 is randomly selected, find the probability that her systolic blood pressure is greater than 8080. 0.99560.9956 If 33 women in that age bracket are randomly selected, find the probability that their mean systolic blood pressure is greater than 8080. 0.00050.0005 Given that part (b) involves a sample size that is not larger than 30, why can the central limit theorem be used? ----Since the original population is normally distributed, the sampling distribution of sample means will be normally distributed for any sample size.

Mean 1.3 SD = 1.0

In the accompanying table, the random variable x represents the number of televisions in a household in a certain country. Determine whether or not the table is a probability distribution. If it is a probability distribution, find its mean and standard deviation. x P(x) 0 0.030.03 1 0.120.12 2 0.340.34 3 0.270.27 4 0.150.15 5 0.090.09 PrintDone

Mean 2.7 SD = 1.2

Five males with a particular genetic disorder have one child each. The random variable x is the number of children among the five who inherit the genetic disorder. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. x 0 1 2 3 4 5 P(x) 0.01650.0165 0.10490.1049 0.26710.2671 0.34000.3400 0.21640.2164 0.0551

Mean 2.8 SD 1.1

Groups of people aged 15-65 are randomly selected and arranged in groups of six. In the accompanying table, the random variable x is the number in the group who say that their family and/or partner contribute most to their happiness (based on a survey). x P(x) 0 0+ 1 0.0190.019 2 0.0110.011 3 0.1460.146 4 0.2870.287 5 0.3510.351 6 0.1860.186

Mean 4.5 SD 1.1

There is a 0.00780.0078 probability that a best-of-seven contest will last four games, a 0.23150.2315 probability that it will last five games, a 0.23790.2379 probability that it will last six games, and a 0.52280.5228 probability that it will last seven games. Verify that this is a probability distribution. Find its mean and standard deviation.

Mean 6.38 SD = .84

Several psychology students are unprepared for a surprise true/false test with 1616 questions, and all of their answers are guesses. a. Find the mean and standard deviation for the number of correct answers for such students. b. Would it be unusual for a student to pass by guessing (which requires getting at least 1313 correct answers)? Why or why not?

Mean 8 SD = 2 Yes because 13 is greater than the maximum usual value

Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.350.35. The table on the right describes the probability of having x number of girls. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. Is it unusual for a family of three children to consist of three girls? x P(x) 0 0.2750.275 1 0.4440.444 2 0.2390.239 3 0.042

Mean = 1.05 SD = .83 Yes because the probability of having 3 girls is less than or equal to 0.05

In the accompanying table, the random variable x represents the number of televisions in a household in a certain country. Determine whether or not the table is a probability distribution. If it is a probability distribution, find its mean and standard deviation. x P(x) 0 0.020.02 1 0.110.11 2 0.240.24 3 0.340.34 4 0.160.16 5 0.13

Mean = 2.7 SD = 1.2

n the accompanying table, the random variable x represents the number of televisions in a household in a certain country. Determine whether or not the table is a probability distribution. If it is a probability distribution, find its mean and standard deviation. x P(x) 0 0.040.04 1 0.080.08 2 0.340.34 3 0.350.35 4 0.110.11 5 0.080.08

Mean = 2.7 SD = 1.2

In the accompanying table, the random variable x represents the number of televisions in a household in a certain country. Determine whether or not the table is a probability distribution. If it is a probability distribution, find its mean and standard deviation. x P(x) 0 0.050.05 1 0.090.09 2 0.270.27 3 0.340.34 4 0.150.15 5 0.100.10

Mean = 2.8 SD 1.3

A candy company claims that 1010% of its plain candies are orange, and a sample of 200200 such candies is randomly selected. Find the mean and standard deviation for the number of orange candies in such groups of 200200.

Mean = 20 SD = 4.2

A candy company claims that 1414% of its plain candies are orange, and a sample of 200200 such candies is randomly selected.

Mean = 28 SD = 4.9

Several psychology students are unprepared for a surprise true/false test with 99 questions, and all of their answers are guesses. a. Find the mean and standard deviation for the number of correct answers for such students. b. Would it be unusual for a student to pass by guessing (which requires getting at least 77 correct answers)? Why or why not?

Mean = 4.5 SD = 1.5 No because 7 is within the range of usual values

Several psychology students are unprepared for a surprise true/false test with 1010 questions, and all of their answers are guesses. a. Find the mean and standard deviation for the number of correct answers for such students. b. Would it be unusual for a student to pass by guessing (which requires getting at least 77 correct answers)? Why or why not?

Mean = 5 SD = 1.6 no because 7 is within the range of usual values

There is a 0.12480.1248 probability that a best-of-seven contest will last four games, a 0.13460.1346 probability that it will last five games, a 0.19080.1908 probability that it will last six games, and a 0.54980.5498 probability that it will last seven games. Verify that this is a probability distribution. Find its mean and standard deviation.

Mean = 6.17 SD = 1.08

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean muμ and standard deviation sigmaσ. Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigmaμ−2σ and the maximum usual value mu plus 2 sigmaμ+2σ. nequals=14601460, pequals=1 divided by 5

Muequals= 292292 sigmaσequals= 15.315.3 (Round to one decimal place as needed.) mu minus 2 sigmaμ−2σequals= 261.4261.4 (Round to one decimal place as needed.) mu plus 2 sigmaμ+2σequals= 322.6322.6 (Round to one decimal place as needed.)

Determine whether the distribution is a discrete probability distribution. x P(x) 0 0.250.25 1 0.300.30 2 negative 0.10−0.10 3 0.300.30 4

No Because some of the prob have values greater than 1 and or less than 0

Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Randomly selecting 50 citizens of a state and recording their nationalitiesRandomly selecting 50 citizens of a state and recording their nationalities nothing

No Because there are more than two possible outcomes

Determine whether the distribution is a discrete probability distribution. x P(x) 0 0.250.25 1 0.300.30 2 negative 0.10−0.10 3 0.300.30 4 0.25

No because some of the prob have values greater than 1 or less than 0

Determine whether the distribution is a discrete probability distribution. x P(x) 0 0.250.25 1 0.300.30 2 negative 0.10−0.10 3 0.300.30 4

No because some of the probabilities have values greater than 1 or less than 0

Determine whether the distribution is a discrete probability distribution. x P(x) 0 0.210.21 1 0.280.28 2 0.020.02 3 0.280.28 4 0.210.21

Yes because the probabilities sum to 1 and are all between 0 and 1 inclusive

The standard deviation of the distribution of sample means is _______.

SD/sqrt of N

For the binomial distribution, which formula finds the standard deviation?

Sort of npq

Assume that adults have IQ scores that are normally distributed with a mean of 105105 and a standard deviation 2020. Find Upper P 4P4, which is the IQ score separating the bottom 44% from the top 9696%.

The IQ score that separates the bottom 44% from the top 9696% is Upper P 4P4equals= 69.9969.99.

Assume that adults have IQ scores that are normally distributed with a mean of 105105 and a standard deviation 1515. Find Upper P 4P4, which is the IQ score separating the bottom 44% from the top 9696%.

The IQ score that separates the bottom 44% from the top 9696% is Upper P 4P4equals= 78.7478.74.

The IQ score that separates the bottom 44% from the top 9696% is Upper P 4P4equals= 78.7478.74. (Round to the nearest hundredth as needed.)

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. z = 0.18

The area of the shaded region is . 5714.5714.

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. z= .88

The area of the shaded region is . 8106.8106.

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

The area of the shaded region is 0.67360.6736.

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

The area of the shaded region is 0.74540.7454.

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees0° and standard deviation of 1.00degrees°C. Assume 3.33.3% of the thermometers are rejected because they have readings that are too high and another 3.33.3% are rejected because they have readings that are too low. Draw a sketch and find the two readings that are cutoff values separating the rejected thermometers from the others.

The cutoff values are negative 1.84 comma 1.84−1.84,1.84 degrees.

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees0° and standard deviation of 1.00degrees°C. Assume 3.23.2% of the thermometers are rejected because they have readings that are too high and another 3.23.2% are rejected because they have readings that are too low. Draw a sketch and find the two readings that are cutoff values separating the rejected thermometers from the others.

The cutoff values are negative 1.85 comma 1.85−1.85,1.85 degrees.

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees0° and standard deviation of 1.00degrees°C. Assume 2.52.5% of the thermometers are rejected because they have readings that are too high and another 2.52.5% are rejected because they have readings that are too low. Draw a sketch and find the two readings that are cutoff values separating the rejected thermometers from the others.

The cutoff values are negative 1.96 comma 1.96−1.96,1.96 degrees.

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees0° and standard deviation of 1.00degrees°C. Assume 2.42.4% of the thermometers are rejected because they have readings that are too high and another 2.42.4% are rejected because they have readings that are too low. Draw a sketch and find the two readings that are cutoff values separating the rejected thermometers from the others.

The cutoff values are negative 1.98 comma 1.98−1.98,1.98 degrees.

Which of the following is NOT a conclusion of the Central Limit Theorem?

The distribution of the sample data will approach a normal distribution as the sample size increases.

Which of the following does NOT describe the standard normal distribution?

The graph is uniform.

Men's heights are normally distributed with mean 68.668.6 in and standard deviation of 2.8 in.2.8 in. Women's heights are normally distributed with mean 63.763.7 in and standard deviation of 2.52.5 in. The standard doorway height is 8080 in. a. What percentage of men are too tall to fit through a standard doorway without bending, and what percentage of women are too tall to fit through a standard doorway without bending? b. If a statistician designs a house so that all of the doorways have heights that are sufficient for all men except the tallest 5%, what doorway height would be used?

The percentage of men who are too tall to fit through a standard door without bending is 0.000.00%. The percentage of women who are too tall to fit through a standard door without bending is 0.000.00%. The statistician would design a house with doorway height 73.273.2 in.

Men's heights are normally distributed with mean 7070 in and standard deviation of 2.8 in.2.8 in. Women's heights are normally distributed with mean 63.763.7 in and standard deviation of 2.52.5 in. The standard doorway height is 8080 in. a. What percentage of men are too tall to fit through a standard doorway without bending, and what percentage of women are too tall to fit through a standard doorway without bending? b. If a statistician designs a house so that all of the doorways have heights that are sufficient for all men except the tallest 5%, what doorway height would be used?

The percentage of men who are too tall to fit through a standard door without bending is 0.020.02%. (Round to two decimal places as needed.) The percentage of women who are too tall to fit through a standard door without bending is 0.000.00%. (Round to two decimal places as needed.) b. The statistician would design a house with doorway height 74.674.6 in. (Round to the nearest tenth as needed.)

A survey found that women's heights are normally distributed with mean 63.463.4 in and standard deviation 2.52.5 in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

The percentage of women who meet the height requirement is 98.4698.46%. (Round to two decimal places as needed.) Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? No, because only a small percentage of women are not allowed to join this branch of the military because of their height. For the new height requirements, this branch of the military requires women's heights to be at least 57.657.6 in and at most 68.568.5 in.

A survey found that women's heights are normally distributed with mean 63.963.9 in and standard deviation 2.32.3 in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

The percentage of women who meet the height requirement is 99.4899.48%. (Round to two decimal places as needed.) Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? A. No, because only a small percentage of women are not allowed to join this branch of the military because of their height. b. For the new height requirements, this branch of the military requires women's heights to be at least 58.558.5 in and at most 68.668.6 in.

The probability is 0.01430.0143.

Assume the readings on thermometers are normally distributed with a mean of 0degrees°C and a standard deviation of 1.00degrees°C. Find the probability that a randomly selected thermometer reads between negative 1.79−1.79 and negative 1.07−1.07 and draw a sketch of the region.

The probability is 0.10560.1056.

Assume the readings on thermometers are normally distributed with a mean of 0degrees°C and a standard deviation of 1.00degrees°C. Find the probability that a randomly selected thermometer reads between negative 1.19−1.19 and negative 0.46−0.46 and draw a sketch of the region.

The probability is 0.20580.2058.

The probability is 0.53590.5359.

Assume that women's heights are normally distributed with a mean given by mu equals 63.4 inμ=63.4 in, and a standard deviation given by sigma equals 2.3 inσ=2.3 in. (a) If 1 woman is randomly selected, find the probability that her height is less than 6464 in. (b) If 3636 women are randomly selected, find the probability that they have a mean height less than 6464 in.

The probability is approximately 0.60290.6029. (Round to four decimal places as needed.) (b) The probability is approximately 0.94120.9412. (Round to four decimal places as needed.)

Determine whether the distribution is a discrete probability distribution.

Yes because the probability sum to 1 and are all between 0 and 1, inclusive

he lengths of pregnancies are normally distributed with a mean of 268268 days and a standard deviation of 1515 days. a. Find the probability of a pregnancy lasting 308308 days or longer. b. If the length of pregnancy is in the lowest 22%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

The probability that a pregnancy will last 308308 days or longer is . 0038.0038. (Round to four decimal places as needed.) b. Babies who are born on or before 237237 days are considered premature.

The probability that a randomly selected adult has an IQ between 8585 and 115115 is 0.5467 .

peas are generated from parents having the green/yellow pair of genes, so there is a 0.750.75 probability that an individual pea will have a green pod. Find the probability that among the 1313 offspring peas, at leastat least 1212 havehave green pods.pods. Is it unusual to get at leastat least 1212 peaspeas with greengreen podspods when 1313 offspring peas are generated? Why or why not?

The probability that at leastat least 1212 of the 1313 offspring peas havehave green podspods is 0.127 Is it unusual to randomly select 1313 peas and find that at leastat least 1212 of them have a green pod? Note that a small probability is one that is less than 0.05. No, because the probability of this occurring is notis not small.

FourteenFourteen peas are generated from parents having the green/yellow pair of genes, so there is a 0.750.75 probability that an individual pea will have a green pod. Find the probability that among the 1414 offspring peas, at leastat least 1313 havehave green pods.pods. Is it unusual to get at leastat least 1313 peaspeas with greengreen podspods when 1414 offspring peas are generated? Why or why not?

The probability that at leastat least 1313 of the 1414 offspring peas havehave green podspods is 0.1010.101. (Round to three decimal places as needed.) Is it unusual to randomly select 1414 peas and find that at leastat least 1313 of them have a green pod? Note that a small probability is one that is less than 0.05. No, because the probability of this occurring is notis not small. This is the correct answer.C.

peas are generated from parents having the green/yellow pair of genes, so there is a 0.750.75 probability that an individual pea will have a green pod. Find the probability that among the 99 offspring peas, at leastat least 88 havehave green pods.pods. Is it unusual to get at leastat least 88 peaspeas with greengreen podspods when 99 offspring peas are generated? Why or why not?

The probability that at leastat least 88 of the 99 offspring peas havehave green podspods is . 3.3. (Round to three decimal places as needed.) Is it unusual to randomly select 99 peas and find that at leastat least 88 of them have a green pod? Note that a small probability is one that is less than 0.05. No because the probability of this occurring is notis not small.

A brand name has aa 4040% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 66 randomly selected consumers. Complete parts (a) through (d) below.

The probability that exactly 55 of the 66 consumers recognize the brand name is 0.0370.037. The probability that all of the selected consumers recognize the brand name is 0.0040.004. The probability that at least 55 of the selected consumers recognize the brand name is 0.0410.041. If 66 consumers are randomly selected, is 55 an unusually high number of consumers that recognize the brand name? ---Yes , because the probability that 55 or more of the selected consumers recognize the brand name is lessless than 0.05.

ThirteenThirteen peas are generated from parents having the green/yellow pair of genes, so there is a 0.750.75 probability that an individual pea will have a green pod. Find the probability that among the 1313 offspring peas, no more thanno more than 11 has ahas a green pod.pod. Is it unusual to get no more thanno more than 11 peapea with a greena green podpod when 1313 offspring peas are generated? Why or why not?

The probability that no more thanno more than 11 of the 1313 offspring peas has ahas a green podpod is 0.0000.000. (Round to three decimal places as needed.) Is it unusual to randomly select 1313 peas and find that no more thanno more than 11 of them have a green pod? Note that a small probability is one that is less than 0.05. yes because the prob of this occurring is very small

peas are generated from parents having the green/yellow pair of genes, so there is a 0.750.75 probability that an individual pea will have a green pod. Find the probability that among the 1515 offspring peas, no more thanno more than 11 has ahas a green pod.pod. Is it unusual to get no more thanno more than 11 peapea with a greena green podpod when 1515 offspring peas are generated? Why or why not?

The probability that no more thanno more than 11 of the 1515 offspring peas has ahas a green podpod is 0.0000.000. (Round to three decimal places as needed.) Is it unusual to randomly select 1515 peas and find that no more thanno more than 11 of them have a green pod? Note that a small probability is one that is less than 0.05. A. YesYes, because the probability of this occurring is veryis very small.

An airline has a policy of booking as many as 1212 persons on an airplane that can seat only 1111. (Past studies have revealed that only 86.086.0% of the booked passengers actually arrive for the flight.) Find the probability that if the airline books 1212 persons, not enough seats will be available. Is it unlikely for such an overbooking to occur?

The probability that not enough seats will be available is . 1637.1637. (Round to four decimal places as needed.) Is it unlikely for such an overbooking to occur? A. It is notis not unlikely for such an overbooking to occur, because the probability of the overbooking is greatergreater than 0.05.

An airline has a policy of booking as many as 1717 persons on an airplane that can seat only 1616. (Past studies have revealed that only 82.082.0% of the booked passengers actually arrive for the flight.) Find the probability that if the airline books 1717 persons, not enough seats will be available. Is it unlikely for such an overbooking to occur?

The probability that not enough seats will be available is 0.03430.0343. It isis unlikely for such an overbooking to occur, because the probability of the overbooking is less than or equal toless than or equal to than 0.05. This is the correct answer.B.

An airline has a policy of booking as many as 1717 persons on an airplane that can seat only 1616. (Past studies have revealed that only 87.087.0% of the booked passengers actually arrive for the flight.) Find the probability that if the airline books 1717 persons, not enough seats will be available. Is it unlikely for such an overbooking to occur?

The probability that not enough seats will be available is 0.09370.0937. It is notis not unlikely for such an overbooking to occur, because the probability of the overbooking is greatergreater than 0.05.

An airline has a policy of booking as many as 1515 persons on an airplane that can seat only 1414. (Past studies have revealed that only 86.086.0% of the booked passengers actually arrive for the flight.) Find the probability that if the airline books 1515 persons, not enough seats will be available. Is it unlikely for such an overbooking to occur?

The probability that not enough seats will be available is 0.10410.1041. Is it unlikely for such an overbooking to occur? It is notis not unlikely for such an overbooking to occur, because the probability of the overbooking is greatergreater than 0.05.

An airline has a policy of booking as many as 1010 persons on an airplane that can seat only 99. (Past studies have revealed that only 85.085.0% of the booked passengers actually arrive for the flight.) Find the probability that if the airline books 1010 persons, not enough seats will be available. Is it unlikely for such an overbooking to occur?

The probability that not enough seats will be available is 0.19690.1969. (Round to four decimal places as needed.) Is it unlikely for such an overbooking to occur? It is notis not unlikely for such an overbooking to occur, because the probability of the overbooking is greatergreater than 0.05.

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 1010 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If a particular shipment of thousands of aspirin tablets actually has a 22% rate of defects, what is the probability that this whole shipment will be accepted?

The probability that this whole shipment will be accepted is . 984.984.

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 1313 tablets. The entire shipment is accepted if at most 2 tablets do not meet the required specifications. If a particular shipment of thousands of aspirin tablets actually has a 3.03.0% rate of defects, what is the probability that this whole shipment will be accepted?

The probability that this whole shipment will be accepted is . 994.994.

Complete the following statement. If you are asked to find the 85th percentile, you are being asked to find _____.

a data value associated with an area of 0.85 to its left

What is the standard error of the mean?

The standard deviation of the sample means.

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees0° and standard deviation of 1.00degrees°C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to Upper P 92P92, the 92 nd92nd percentile. This is the temperature reading separating the bottom 92 %92% from the top 8 %.

The temperature for Upper P 92P92 is approximately 1.411.41degrees°.

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees0° and standard deviation of 1.00degrees°C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to Upper P 93P93, the 93 rd93rd percentile. This is the temperature reading separating the bottom 93 %93% from the top 7 %.

The temperature for Upper P 93P93 is approximately 1.481.48degrees°.

Determine whether the distribution is a discrete probability distribution. x P(x) 0 0.090.09 1 0.210.21 2 0.400.40 3 0.210.21 4 0.090.09

Yes because the probabilities sum to 1 and are all between 0 and 1, inclusive

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees0° and standard deviation of 1.00degrees°C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to Upper P 95P95, the 95 th95th percentile. This is the temperature reading separating the bottom 95 %95% from the top 5 %.

The temperature for Upper P 95P95 is approximately 1.641.64degrees°.

Assume that adults have IQ scores that are normally distributed with a mean of 100100 and a standard deviation of 15. Find the third quartile Upper Q 3Q3, which is the IQ score separating the top 25% from the others.

The third quartile, Upper Q 3Q3, is 110.1

Assume that adults have IQ scores that are normally distributed with a mean of 101101 and a standard deviation of 15. Find the third quartile Upper Q 3Q3, which is the IQ score separating the top 25% from the others.

The third quartile, Upper Q 3Q3, is 111.1

Assume that adults have IQ scores that are normally distributed with a mean of 106106 and a standard deviation of 15. Find the third quartile Upper Q 3Q3, which is the IQ score separating the top 25% from the others.

The third quartile, Upper Q 3Q3, is 116.1116.1.

Assume that adults have IQ scores that are normally distributed with a mean of 107107 and a standard deviation of 15. Find the third quartile Upper Q 3Q3, which is the IQ score separating the top 25% from the others.

The third quartile, Upper Q 3Q3, is 117.1

Which of the following is not a requirement of the binomial probability distribution?

The trials must be dependent.

Which of the following is not a requirement of the binomial probability distribution? Choose the correct answer below.

The trials must be dependent.

In a clinical trial of a cholesterol drug, 417417 subjects were given a placebo, and 1414% of them developed headaches. For such randomly selected groups of 417417 subjects given a placebo, identify the values of n, p, and q that would be used for finding the mean and standard deviation for the number of subjects who develop headaches.

The value of n is 417417. The value of p is . 14.14. (Type an integer or a decimal.) The value of q is . 86.86.

Assume that the given procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean muμ and standard deviation sigmaσ. Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigmaμ−2σ and the maximum usual value mu plus 2 sigmaμ+2σ. In an analysis of preliminary test results from a gender-selection method, 3737 babies are born and it is assumed that 50% of babies are girls, so nequals=3737 and pequals=0.50.5.

The value of the mean is muμequals= 18.518.5. (Simplify your answer.) The value of the standard deviation is sigmaσequals= 33. (Round to one decimal place as needed.) The minimum usual value is mu minus 2 sigmaμ−2σequals= 12.512.5. (Round to one decimal place as needed.) The maximum usual value is mu plus 2 sigmaμ+2σequals= 24.524.5. (Round to one decimal place as needed.)

A z-score is a conversion that standardizes any value from a normal distribution to a standard normal distribution.

True

If values are converted to standard z-scores, then procedures for working with all normal distributions are the same as those for the standard normal distribution.

True

The area in any normal distribution bounded by some score x is the same as the area bounded by the equivalent z-score in the standard normal distribution.

True

Which of the following is NOT one of the three methods for finding binomial probabilities that is found in the chapter on discrete probability distributions?

Use simulation

Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Treating 200 men with a special shampoo and recording Yes if they experience anyTreating 200 men with a special shampoo and recording Yes if they experience any burning or No otherwiseburning or No otherwise Choose the correct answer below.

Yea Because all 4 requirements are satisfied.

Determine whether the distribution is a discrete probability distribution. x P(x) 0 0.160.16 1 0.210.21 2 0.260.26 3 0.210.21 4 0.160.16

Yeas the probabilities sum to 1 and are all between 0 and 1 inclusive

Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Surveying 100 teenagers and recording if they have ever committed a crimeSurveying 100 teenagers and recording if they have ever committed a crime nothing

Yes because all 4 requirements are satisfied

Determine whether the value is a discrete random variable, continuous random variable, or not a random variable. a. The number of textbook authors now sitting at a computernumber of textbook authors now sitting at a computer b. The number of fish caught during a fishing tournamentnumber of fish caught during a fishing tournament c. The gender of college studentsgender of college students d. The time it takes for a light bulb to burn outtime it takes for a light bulb to burn out e. The amount of snowfall in December in City Upper Aamount of snowfall in December in City A f. The number of light bulbs that burn out in the next week in a room with 18 bulbs

a - Discrete b - Discrete c- it is not random d- continuous e - continuous f - discrete

Determine whether the value is a discrete random variable, continuous random variable, or not a random variable. a. The square footage of a housesquare footage of a house b. The number of bald eagles in a countrynumber of bald eagles in a country c. The eye color of people on commercial aircraft flightseye color of people on commercial aircraft flights d. The number of textbook authors now sitting at a computernumber of textbook authors now sitting at a computer e. The time required to download a file from the Internettime required to download a file from the Internet f. The number of statistics students now reading a book

a - continuous b - discrete c - not random d - discrete e - continuous F- discrete

Determine whether the value is a discrete random variable, continuous random variable, or not a random variable. a. The number of textbook authors now sitting at a computernumber of textbook authors now sitting at a computer b. The square footage of a housesquare footage of a house c. The political party affiliation of adults in the United Statespolitical party affiliation of adults in the United States d. The time it takes for a light bulb to burn outtime it takes for a light bulb to burn out e. The number of people in a restaurant that has a capacity of 300number of people in a restaurant that has a capacity of 300 f. The distance a baseball travels in the air after being hitdistance a baseball travels in the air after being hit a. Is the number of textbook authors now sitting at a computernumber of textbook authors now sitting at a computer a discrete random variable, a continuous random variable, or not a random variable? A. It is a continuouscontinuous random variable. B. It is a discretediscrete random variable. Your answer is correct.C. It is not a random variable. b. Is the square footage of a housesquare footage of a house a discrete random variable, a continuous random variable, or not a random variable? A. It is a continuouscontinuous random variable. Your answer is correct.B. It is a discretediscrete random variable. C. It is not a random variable c. Is the political party affiliation of adults in the United Statespolitical party affiliation of adults in the United States a discrete random variable, a continuous random variable, or not a random variable? A. It is a discrete random variable. B. It is a continuous random variable. C. It is not a random variable. Your answer is correct. d. Is the time it takes for a light bulb to burn outtime it takes for a light bulb to burn out a discrete random variable, a continuous random variable, or not a random variable? A. It is a discretediscrete random variable. B. It is a continuouscontinuous random variable. Your answer is correct.C. It is not a random variable. e. Is the number of people in a restaurant that has a capacity of 300number of people in a restaurant that has a capacity of 300 a discrete random variable, a continuous random variable, or not a random variable?

a - discrete b - continuous c - not random d - Continuous e -Discrete f - Continuous

Determine whether the value is a discrete random variable, continuous random variable, or not a random variable. a. The number of bald eagles in a countrynumber of bald eagles in a country b. The height of a randomly selected giraffeheight of a randomly selected giraffe c. The political party affiliation of adults in the United Statespolitical party affiliation of adults in the United States d. The number of textbook authors now sitting at a computernumber of textbook authors now sitting at a computer e. The number of free dash throw attempts before the first shot is madenumber of free-throw attempts before the first shot is made f. The number of people in a restaurant that has a capacity of 150

a- Discrete b- Continuous C- Not a random D- Discrete E- Discrete F- Discrete

In a clinical trial of a drug used to help subjects stop smoking, 818818 subjects were treated with 1 mg1 mg doses of the drug. That group consisted of 3232 subjects who experienced nausea. The probability of nausea for subjects not receiving the treatment was 0.01120.0112. Complete parts (a) through (c).

a. Assuming that the drug has no effect, so that the probability of nausea was 0.01120.0112, find the mean and standard deviation for the numbers of people in groups of 818818 that can be expected to experience nausea. The mean is 9.29.2 people. (Round to one decimal place as needed.) The standard deviation is 33 people. (Round to one decimal place as needed.) b. Based on the result from part (a), is it unusual to find that among 818818 people, there are 3232 who experience nausea? Why or why not? - It is unusual because 3232 is outside the range of usual values. Based on the preceding results, does nausea appear to be an adverse reaction that should be of concern to those who use the drug? - The drug does appear to be the cause of some nausea. Since the nausea rate is still quite low (about 44%), it appears to be an adverse reaction that does not occur very often.

A candy company claims that 1010% of its plain candies are orange, and a sample of 100100 such candies is randomly selected.

a. Find the mean and standard deviation for the number of orange candies in such groups of 100100. muμequals= 1010 sigmaσequals= 33 (Round to one decimal place as needed.) b. A random sample of 100100 candies contains 99 orange candies. Is this result unusual? Does it seem that the claimed rate of 1010% is wrong? No because 9 is within the range of usual values. Thus, the claimed rate of 10 %10% is not necessarilynot necessarily wrong.

A government agency has specialists who analyze the frequencies of letters of the alphabet in an attempt to decipher intercepted messages. In standard English text, a particular letter is used at a rate of 7.97.9%.

a. Find the mean and standard deviation for the number of times this letter will be found on a typical page of 35003500 characters. muμequals= 276.5276.5 sigmaσequals= 1616 (Round to one decimal place as needed.) b. In an intercepted message, a page of 35003500 characters is found to have the letter occurring 321321 times. Is this unusual? Yes because 321 is greater than the max usual value

In a region, there is a 0.80.8 probability chance that a randomly selected person of the population has brown eyes. Assume 1414 people are randomly selected. Complete parts (a) through (d) below.

a. Find the probability that all of the selected people have brown eyes. The probability that all of the 1414 selected people have brown eyes is 0.0440.044. (Round to three decimal places as needed.) b. Find the probability that exactly 1313 of the selected people have brown eyes. The probability that exactly 1313 of the selected people have brown eyes is 0.1540.154. (Round to three decimal places as needed.) c. Find the probability that the number of selected people that have brown eyes is 1212 or more. The probability that the number of selected people that have brown eyes is 1212 or more is 0.4480.448. (Round to three decimal places as needed.) d. If 1414 people are randomly selected, is 1212 an unusually high number for those with brown eyes? - No, because the probability that 1212 or more of the selected people have brown eyes is greatergreater than 0.05.

The capacity of an elevator is 1212 people or 19321932 pounds. The capacity will be exceeded if 1212 people have weights with a mean greater than 1932 divided by 12 equals 161 pounds.1932/12=161 pounds. Suppose the people have weights that are normally distributed with a mean of 169 lb169 lb and a standard deviation of 31 lb31 lb.

a. Find the probability that if a person is randomly selected, his weight will be greater than 161161 pounds. The probability is approximately 0.60180.6018. (Round to four decimal places as needed.) b. Find the probability that 1212 randomly selected people will have a mean that is greater than 161161 pounds. The probability is approximately 0.81430.8143. (Round to four decimal places as needed.) c. Does the elevator appear to have the correct weight limit? Why or why not? No, there is a good chance that 1212 randomly selected people will exceed the elevator capacity.

Which of the following groups of terms can be used interchangeably when working with normal distributions?

areas, probability, and relative frequencies

The capacity of an elevator is 1515 people or 24602460 pounds. The capacity will be exceeded if 1515 people have weights with a mean greater than 2460 divided by 15 equals 164 pounds.2460/15=164 pounds. Suppose the people have weights that are normally distributed with a mean of 173 lb173 lb and a standard deviation of 26 lb26 lb.

a. Find the probability that if a person is randomly selected, his weight will be greater than 164164 pounds. The probability is approximately 0.63540.6354. (Round to four decimal places as needed.) b. Find the probability that 1515 randomly selected people will have a mean that is greater than 164164 pounds. The probability is approximately 0.91000.9100. (Round to four decimal places as needed.) c. Does the elevator appear to have the correct weight limit? Why or why not? No, there is a good chance that 1515 randomly selected people will exceed the elevator capacity.

A TV show, Lindsay and Tobias, recently had a share of 2020, meaning that among the TV sets in use, 2020% were tuned to that show. Assume that an advertiser wants to verify that 2020% share value by conducting its own survey, and a pilot survey begins with 1414 households having TV sets in use at the time of a Lindsay and Tobias broadcast.

a. Find the probability that none of the households are tuned to Lindsay and Tobias. . 044.044 (Round to three decimal places as needed.) b. Find the probability that at least one household is tuned to Lindsay and Tobias. . 956.956 (Round to three decimal places as needed.) c. Find the probability that at most one household is tuned to Lindsay and Tobias. . 198.198 (Round to three decimal places as needed.) d. If at most one household is tuned to Lindsay and Tobias, does it appear that the 2020% share value is wrong? Why or why not? - NoNo, because with a 2020% rate, the probability of at most one household is greatergreater than 0.05.

Researchers conducted a study to determine whether there were significant differences in graduation rates between medical students admitted through special programs and medical students admitted through the regular admissions criteria. It was found that the graduation rate was 9191% for the medical students admitted through special programs in all medical schools. Complete parts (a) and (b) below.

a. If 1010 of the students admitted through special programs are randomly selected, find the probability that at least 99 of them graduated. The probability that at least 99 of the 1010 students graduated is . 775.775. (Round to three decimal places as needed.) b. Suppose a medical school has 1010 students in one of the special programs of its medical program. Does the probability calculated in part (a) apply to these students? -No, because the students admitted through a single special program at a specific medical school are not a random sample.

n airliner carries 250250 passengers and has doors with a height of 7575 in. Heights of men are normally distributed with a mean of 69.069.0 in and a standard deviation of 2.82.8 in. Complete parts (a) through (d). a. If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending. The probability is . 9838.9838. (Round to four decimal places as needed.) b. If half of the 250250 passengers are men, find the probability that the mean height of the 125125 men is less than 7575 in. The probability is 11. (Round to four decimal places as needed.)

a. If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending. The probability is . 9838.9838. (Round to four decimal places as needed.) b. If half of the 250250 passengers are men, find the probability that the mean height of the 125125 men is less than 7575 in. The probability is 11. c. When considering the comfort and safety of passengers, which result is more relevant: the probability from part (a) or the probability from part (b)? Why? The probability from part (a) is more relevant because it shows the proportion of male passengers that will not need to bend. When considering the comfort and safety of passengers, why are women ignored in this case? A. Since men are generally taller than women, a design that accommodates a suitable proportion of men will necessarily accommodate a greater proportion of women.

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 120120 lb and 181181 lb. The new population of pilots has normally distributed weights with a mean of 129 lb129 lb and a standard deviation of 30.5 lb30.5 lb.

a. If a pilot is randomly selected, find the probability that his weight is between 120120 lb and 181181 lb. The probability is approximately 0.57190.5719. (Round to four decimal places as needed.) b. If 3131 different pilots are randomly selected, find the probability that their mean weight is between 120120 lb and 181181 lb. The probability is approximately 0.94980.9498. (Round to four decimal places as needed.) c. When redesigning the ejection seat, which probability is more relevant? A. Part (a) because the seat performance for a single pilot is more important.

Based on a survey, for women aged 18 to 24, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.9114.9 and a standard deviation of 13.113.1. Complete parts (a) through (c).

a. If a woman between the ages of 18 and 24 is randomly selected, find the probability that her systolic blood pressure is greater than 110110. 0.64580.6458 (Round to four decimal places as needed.) b. If 55 women in that age bracket are randomly selected, find the probability that their mean systolic blood pressure is greater than 110110. 0.79850.7985 (Round to four decimal places as needed.) c. Given that part (b) involves a sample size that is not larger than 30, why can the central limit theorem be used? Since the original population is normally distributed, the sampling distribution of sample means will be normally distributed for any sample size.

Based on a survey, for women aged 18 to 24, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.8114.8 and a standard deviation of 13.113.1. Complete parts (a) through (c).

a. If a woman between the ages of 18 and 24 is randomly selected, find the probability that her systolic blood pressure is greater than 120120. 0.34570.3457 (Round to four decimal places as needed.) b. If 33 women in that age bracket are randomly selected, find the probability that their mean systolic blood pressure is greater than 120120. 0.24590.2459 (Round to four decimal places as needed.) c. Given that part (b) involves a sample size that is not larger than 30, why can the central limit theorem be used? Since the original population is normally distributed, the sampling distribution of sample means will be normally distributed for any sample size.

What conditions would produce a negative z-score?

a z-score corresponding to an area located entirely in the left side of the curve

a. If a woman between the ages of 18 and 24 is randomly selected, find the probability that her systolic blood pressure is greater than 120120. 0.34790.3479 (Round to four decimal places as needed.) b. If 55 women in that age bracket are randomly selected, find the probability that their mean systolic blood pressure is greater than 120120. 0.19100.1910 (Round to four decimal places as needed.) c. Given that part (b) involves a sample size that is not larger than 30, why can the central limit theorem be used? Since the original population is normally distributed, the sampling distribution of sample means will be normally distributed for any sample size.

Men's heights are normally distributed with mean 68.768.7 in and standard deviation of 2.8 in.2.8 in. Women's heights are normally distributed with mean 63.463.4 in and standard deviation of 2.52.5 in. The standard doorway height is 8080 in. a. What percentage of men are too tall to fit through a standard doorway without bending, and what percentage of women are too tall to fit through a standard doorway without bending? b. If a statistician designs a house so that all of the doorways have heights that are sufficient for all men except the tallest 5%, what doorway height would be used?

a. The percentage of men who are too tall to fit through a standard door without bending is 0.000.00%. (Round to two decimal places as needed.) The percentage of women who are too tall to fit through a standard door without bending is 0.000.00%. (Round to two decimal places as needed.) b. The statistician would design a house with doorway height 73.373.3 in.

a. The percentage of men who are too tall to fit through a standard door without bending is 0.010.01%. (Round to two decimal places as needed.) The percentage of women who are too tall to fit through a standard door without bending is 00%. (Round to two decimal places as needed.) b. The statistician would design a house with doorway height 73.373.3 in.

Finding probabilities associated with distributions that are standard normal distributions is equivalent to _______.

finding the area of the shaded region representing that probability.

Men's heights are normally distributed with mean 7070 in and standard deviation of 2.8 in.2.8 in. Women's heights are normally distributed with mean 6464 in and standard deviation of 2.52.5 in. The standard doorway height is 8080 in. a. What percentage of men are too tall to fit through a standard doorway without bending, and what percentage of women are too tall to fit through a standard doorway without bending? b. If a statistician designs a house so that all of the doorways have heights that are sufficient for all men except the tallest 5%, what doorway height would be used?

a. The percentage of men who are too tall to fit through a standard door without bending is 0.020.02%. (Round to two decimal places as needed.) The percentage of women who are too tall to fit through a standard door without bending is 0.000.00%. (Round to two decimal places as needed.) b. The statistician would design a house with doorway height 74.674.6 in. (Round to the nearest tenth as needed.)

A survey found that women's heights are normally distributed with mean 63.463.4 in and standard deviation 2.42.4 in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

a. The percentage of women who meet the height requirement is 98.7898.78%. (Round to two decimal places as needed.) Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? A. No, because only a small percentage of women are not allowed to join this branch of the military because of their height. b. For the new height requirements, this branch of the military requires women's heights to be at least 57.857.8 in and at most 68.368.3 in.

A survey found that women's heights are normally distributed with mean 63.663.6 in and standard deviation 2.22.2 in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

a. The percentage of women who meet the height requirement is 99.45% No, because only a small percentage of women are not allowed to join this branch of the military because of their height. b. For the new height requirements, this branch of the military requires women's heights to be at least 58.558.5 in and at most 68.1 in.

If a gambler places a bet on the number 7 in roulette, he or she has a 1/38 probability of winning. a. Find the mean and standard deviation for the number of wins of gamblers who bet on the number 7 three hundredthree hundred times. b. Would 0 wins in three hundredthree hundred bets be an unusually low number of wins?

a. The value of the mean is muμequals= 7.97.9. (Round to one decimal place as needed.) The value of the standard deviation is sigmaσequals= 2.82.8. (Round to one decimal place as needed.) b. Would 0 wins in 300300 bets be an unusually low number of wins? Yes because 0 is below the minimum usual value

A brand name has aa 5050% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 44 randomly selected consumers. Complete parts (a) through (d) below.

a. What is the probability that exactly 33 of the selected consumers recognize the brand name? The probability that exactly 33 of the 44 consumers recognize the brand name is 0.2500.250. (Round to three decimal places as needed.) b. What is the probability that all of the selected consumers recognize the brand name? The probability that all of the selected consumers recognize the brand name is 0.0630.063. (Round to three decimal places as needed.) c. What is the probability that at least 33 of the selected consumers recognize the brand name? The probability that at least 33 of the selected consumers recognize the brand name is 0.3130.313. (Round to three decimal places as needed.) d. If 44 consumers are randomly selected, is 33 an unusually high number of consumers that recognize the brand name? No, because the probability that 33 or more of the selected consumers recognize the brand name is greatergreater than 0.05.

A brand name has aa 5050% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 55 randomly selected consumers. Complete parts (a) through (d) below.

a. What is the probability that exactly 44 of the selected consumers recognize the brand name? The probability that exactly 44 of the 55 consumers recognize the brand name is 0.1560.156. (Round to three decimal places as needed.) b. What is the probability that all of the selected consumers recognize the brand name? The probability that all of the selected consumers recognize the brand name is 0.0310.031. (Round to three decimal places as needed.) c. What is the probability that at least 44 of the selected consumers recognize the brand name? The probability that at least 44 of the selected consumers recognize the brand name is 0.1880.188. (Round to three decimal places as needed.) d. If 55 consumers are randomly selected, is 44 an unusually high number of consumers that recognize the brand name? - No, because the probability that 44 or more of the selected consumers recognize the brand name is greatergreater than 0.05.

A brand name has aa 6060% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 66 randomly selected consumers. Complete parts (a) through (d) below.

a. What is the probability that exactly 55 of the selected consumers recognize the brand name? The probability that exactly 55 of the 66 consumers recognize the brand name is 0.1870.187. (Round to three decimal places as needed.) b. What is the probability that all of the selected consumers recognize the brand name? The probability that all of the selected consumers recognize the brand name is 0.0470.047. (Round to three decimal places as needed.) c. What is the probability that at least 55 of the selected consumers recognize the brand name? The probability that at least 55 of the selected consumers recognize the brand name is 0.2330.233. (Round to three decimal places as needed.) d. If 66 consumers are randomly selected, is 55 an unusually high number of consumers that recognize the brand name? No, because the probability that 55 or more of the selected consumers recognize the brand name is greatergreater than 0.05.

A brand name has aa 6060% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 77 randomly selected consumers. Complete parts (a) through (d) below.

a. What is the probability that exactly 66 of the selected consumers recognize the brand name? The probability that exactly 66 of the 77 consumers recognize the brand name is 0.1310.131. (Round to three decimal places as needed.) b. What is the probability that all of the selected consumers recognize the brand name? The probability that all of the selected consumers recognize the brand name is 0.0280.028. (Round to three decimal places as needed.) c. What is the probability that at least 66 of the selected consumers recognize the brand name? The probability that at least 66 of the selected consumers recognize the brand name is 0.1590.159. (Round to three decimal places as needed.) d. If 77 consumers are randomly selected, is 66 an unusually high number of consumers that recognize the brand name? NoNo, because the probability that 66 or more of the selected consumers recognize the brand name is greatergreater than 0.05.

Several psychology students are unprepared for a surprise true/false test with 1313 questions, and all of their answers are guesses. a. Find the mean and standard deviation for the number of correct answers for such students. b. Would it be unusual for a student to pass by guessing (which requires getting at least 88 correct answers)? Why or why not?

a. muμequals= 6.56.5 sigmaσequals= 1.81.8 b. No because 8 is within the range of usual values

Several psychology students are unprepared for a surprise true/false test with 1414 questions, and all of their answers are guesses. a. Find the mean and standard deviation for the number of correct answers for such students. b. Would it be unusual for a student to pass by guessing (which requires getting at least 1010 correct answers)? Why or why not?

a. muμequals= 77 sigmaσequals= 1.91.9 (Round to one decimal place as needed.) b. Choose the correct answer below. - No because 10 is within the range of usual values

A survey found that women's heights are normally distributed with mean 63.663.6 in and standard deviation 2.32.3 in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

b. For the new height requirements, this branch of the military requires women's heights to be at least 58.258.2 in and at most 68.368.3 in. (Round to one decimal place as needed.) No, because only a small percentage of women are not allowed to join this branch of the military because of their height.

A _______ random variable has either a finite or a countable number of values.

discrete

equals= 105105 sigmaσequals= 5.65.6 (Round to one decimal place as needed.) mu minus 2 sigmaμ−2σequals= 93.893.8 (Round to one decimal place as needed.) mu plus 2 sigmaμ+2σequals= 116.2116.2 (Round to one decimal place as needed.)

The _______ of a discrete random variable represents the mean value of the outcomes

expected value

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees0° and standard deviation of 1.00degrees°C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to Upper P 81P81, the 81 st81st percentile. This is the temperature reading separating the bottom 81 %81% from the top 19 %.

he temperature for Upper P 81P81 is approximately 0.880.88degrees°.

Assume that a procedure yields a binomial distribution with 66 trials and a probability of success of 0.700.70. Use a binomial probability table to find the probability that the number of successes is exactly 00.

successes = 0 = .001

When someone buys a ticket for an airline flight, there is a 0.09870.0987 probability that the person will not show up for the flight. A certain jet can seat 2121 passengers. Is it wise to book 2323 passengers for a flight on the jet? Explain.

t is not a wise decision because the probability that there are not enough seats on the jet is . 3224.3224. So, overbooking is not an unlikely event.

Assume that women's heights are normally distributed with a mean given by mu equals 62.6 inμ=62.6 in, and a standard deviation given by sigma equals 2.1 inσ=2.1 in. (a) If 1 woman is randomly selected, find the probability that her height is less than 6363 in. (b) If 4646 women are randomly selected, find the probability that they have a mean height less than 6363 in.

the probability is approximately 0.57550.5755. The probability is approximately 0.90180.9018.

the notation P(zless than<a) denotes _______.

the probability that the z-score is less than a.

Where would a value separating the top 15% from the other values on the graph of a normal distribution be found?

the right side of the horizontal scale of the graph

The area in any normal distribution bounded by some score x is the same as the area bounded by the equivalent z-score in the standard normal distribution.

true

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 inμ=62.5 in, and a standard deviation given by sigma equals 2.9 inσ=2.9 in. (a) If 1 woman is randomly selected, find the probability that her height is less than 6363 in. (b) If 4444 women are randomly selected, find the probability that they have a mean height less than 6363 in.

(a) The probability is approximately . 5675.5675. (Round to four decimal places as needed.) (b) The probability is approximately . 8729.8729.

Assume that women's heights are normally distributed with a mean given by mu equals 62.4 inμ=62.4 in, and a standard deviation given by sigma equals 2.4 inσ=2.4 in. (a) If 1 woman is randomly selected, find the probability that her height is less than 6363 in. (b) If 4343 women are randomly selected, find the probability that they have a mean height less than 6363 in.

(a) The probability is approximately 0.59870.5987. (Round to four decimal places as needed.) (b) The probability is approximately 0.94940.9494.

Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. FiveFive cards are selected from a standard 52-card deck without replacement. The number of fivesfives selected is recorded.

No, because the trials of the experiment are not independent and the probability of success differs from trial to trial.

The capacity of an elevator is 1010 people or 16301630 pounds. The capacity will be exceeded if 1010 people have weights with a mean greater than 1630 divided by 10 equals 163 pounds.1630/10=163 pounds. Suppose the people have weights that are normally distributed with a mean of 172 lb172 lb and a standard deviation of 34 lb34 lb.

No, there is a good chance that 1010 randomly selected people will exceed the elevator capacity.

Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. An experimental drug is administered to 140140 randomly selected individuals, with the number of individuals responding favorably recorded.

Yes, because the experiment satisfies all the criteria for a binomial experiment.

Assume that women's heights are normally distributed with a mean given by mu equals 62.4 inμ=62.4 in, and a standard deviation given by sigma equals 1.7 inσ=1.7 in. (a) If 1 woman is randomly selected, find the probability that her height is less than 6363 in. (b) If 3131 women are randomly selected, find the probability that they have a mean height less than 6363 in.

a) The probability is approximately 0.63790.6379. (Round to four decimal places as needed.) (b) The probability is approximately 0.97530.9753. (Round to four decimal places as needed.)

Chapter 5 and 6

Related study sets

Space Program Project

Ch. 10 Study guide

Part 4 General Anesthesia

Test

Nursing 335 Week 5

MKT Exam3

Real Estate Exam

Sociology Final Review (Tests 1-3)

Chapter 4 - Reproductive system concerns (EAQ's)

Management Chapter 4- Dr. Loes (Belmont University)

Math Unit 3 Test

Chapter 12

Chap 5

Geriatrics Chapter 33 and 38 Foundation

Ch. 19 & 20 (Lungs and Cardiac)

Microsoft Word

Chapter 40 Biology

Neuro Passpoint Questions

Exam 3 Practice of Quizzes

DIRECTING AND SUPERVISING THE LABORATORY