Chapter 6 & 7 Statistics Questions
Assume that all grade-point averages are to be standardized on a scale between 0 and 4. How many grade-point averages must be obtained so that the sample mean is within 0.014 of the population mean? Assume that a 95% confidence level is desired. If using the range rule of thumb, σ can be estimated as range/4=4−0/4=1.
((1.96*1)/0.014)^2 = 19600 No, because fairly large.
A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 18 subjects had a mean wake time of 102.0min. After treatment, the 18 subjects had a mean wake time of 79.4min and a standard deviation of 21.2min. Assume that the 18 sample values appear to be from a normally distributed population and construct a 95% confidence interval estimate of the mean wake time for a population with drug treatments
(menu 6, 6, 2) 68.9, 89.9 The confidence interval does not include the mean wake time of 102.0min before the treatment, so the means before and after the treatment are different. This result suggests that the drug treatment has a significant effect.
A data set includes 106 body temperatures of healthy adult humans having a mean of 98.7°F and a standard deviation of 0.66°F. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body temperature?
(menu 6, 6, 2) 98.532, 98.868. This suggests that the mean body temperature could very possibly be 98.6°F.
Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find the bone density test scores that can be used as cutoff values separating the lowest 12% and highest 12%, indicating levels that are too low or too high, respectively.
-1.17, 1.17
A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 48.0 and 58.0 minutes. Find the probability that a given class period runs between 51.5 and 51.75 minutes.
0.025 because 0.25*0.1=0.025
A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 48.0 and 53.0 minutes. Find the probability that a given class period runs between 51.5 and 51.75 minutes.
0.05 because 0.25*0.2=0.05
The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 8 minutes. Find the probability that a randomly selected passenger has a waiting time less than 0.75 minutes.
0.094
Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the probability that a randomly selected thermometer reads between −1.99 and −0.88.
0.1661
A genetic experiment with peas resulted in one sample of offspring that consisted of 421 green peas and 159 yellow peas. a. Construct a 95% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations?
0.238<p<0.310 b. No
The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 6 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.
0.623
Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. z=0.54
0.7054 (calculator: menu, 5, 5, 2)
Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1. z= between -0.94 and 1.23
0.7171
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were born, and 272 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.
0.744<p<0.856 (Yes, the proportion of girls is significantly different from 0.5)
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 300 babies were born, and 270 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?
0.855<p<0.945 (yes)
Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1. z=-1.09
0.8621 because 1-0.1378=0.8621
Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than 3.65.
0.9999
Here are summary statistics for randomly selected weights of newborn girls: n=205, x=29.8hg, s=7.9hg. Construct a confidence interval estimate of the mean. Use a 90% confidence level. Are these results very different from the confidence interval 28.1hg<μ<31.7hg with only 12 sample values, x=29.9hg, and s=3.5hg?
28.9, 30.7 No, because the confidence interval limits are similar.
Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 98% confidence level. What does the confidence interval tell about the population of all college students in the state? 3.5, 2.9, 3.7, 4.8, 3.1, 4.1, 3.2, 4.4, 4.5, 4.4, 4.1, 3.9, 3.6, 3.7, 3.5
3.46, 4.21 No
Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 95% confidence level. What does the confidence interval tell about the population of all college students in the state? 4.0,2.9,3.9,4.6,3.1,4.4,3.7,4.5,4.1,4.1,4.2,3.6,3.2,4.0,3.8.
3.59, 4.15 No
Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 95% confidence level. What does the confidence interval tell about the population of all college students in the state? 3.7, 3.0, 3.9, 4.6, 3.1, 4.0, 3.8, 4.6, 4.5, 4.2, 4.6, 3.8, 3.5, 3.9, 4.0
3.67, 4.23 No
Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent." the ratings were obtained at one university in a state. construct a confidence interval using a 90% confidence level. what does the confidence interval tell about the population of all college students in the state? 3.93.13.74.43.24.13.34.44.3,4.44.5 3.93.24.2,4.
3.74, 4.13 No
Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 90% confidence level. What does the confidence interval tell about the population of all college students in the state?4.0,3.2,3.9,4.8,2.9,4.5,3.8,4.8,4.4,4.4,4.6,4.1,3.6,4.3,3.9
3.84, 4.32 The results tell nothing about the population of all college students in the state, since the sample is from only one university.
Here are summary statistics for randomly selected weights of newborn girls: n=203, x=32.1hg, s=6.7hg. Construct a confidence interval estimate of the mean. Use a 99% confidence level. Are these results very different from the confidence interval 30.6hg<μ<34.2hg with only 15 sample values, -x-=32.4hg, and s=2.3hg?
30.9, 33.3 No, because the confidence interval limits are similar.
Assume that all grade-point averages are to be standardized on a scale between 0 and 6. How many grade-point averages must be obtained so that the sample mean is within 0.005 of the population mean? Assume that a 95% confidence level is desired. If using the range rule of thumb, σ can be estimated as range/4=6−0/4=1.5. Does the sample size seem practical?
345,732 No, because fairly large.
The data given to the right includes data from 42 candies, and 6 of them are red. The company that makes the candy claims that 32% of its candies are red. Use the sample data to construct a 90% confidence interval estimate of the percentage of red candies.
5.4%<p<23.3% (no)
Assume that the sample is a simple random sample obtained from a normally distributed population of flight delays at an airport. Use the table below to find the minimum sample size needed to be 95% confident that the sample standard deviation is within 5% of the population standard deviation.
768. The computed minimum sample size is not likely correct.
Listed below are speeds (mi/h) measured from traffic on a busy highway. This simple random sample was obtained at 3:30 P.M. on a weekday. Use the sample data to construct a 95% confidence interval estimate of the population standard deviation.
CI estimate = 3.2, 7.6. No. The confidence interval is an estimate of the standard deviation of the population of speeds at 3:30 on a weekday, not other times.
Twelve different video games showing substance use were observed and the duration of times of game play (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. Use the sample data to construct an 80% confidence interval estimate of σ, the standard deviation of the duration times of game play.
CI estimate = 320.1, 563.3
The _____________ distribution is used to develop confidence interval estimates of variances or standard deviations.
Chi-square
Express the confidence interval 0.222 < p < 0.666 in the form ^p +- E.
Find point estimate (0.666+0.222)/2 = 0.444 AND the margin of error (0.666-0.222)/2 = 0.222 so it is 0.444 +- 0.222
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. #105
The area of the shaded region is 0.6305
Which of the following is not a commonly used practice?
If the distribution of the sample means is normally distributed, and n>30, then the population distribution is normally distributed.
A researcher collects a simple random sample of grade-point averages of statistics students, and she calculates the mean of this sample. Under what conditions can that sample mean be treated as a value from a population having a normal distribution?
If the population of grade-point averages has a normal distribution & the sample has more than 30 grade-point averages.
Weights of golden retriever dogs are normally distributed. Samples of weights of golden retriever dogs, each of size n=15, are randomly collected and the sample means are found. Is it correct to conclude that the sample means cannot be treated as being from a normal distribution because the sample size is too small?
No; the original population is normally distributed, so the sample means will be normally distributed for any sample size.
Annual incomes are known to have a distribution that is skewed to the right instead of being normally distributed. Assume that we collect a large (n>30) random sample of annual incomes. Can the distribution of incomes in that sample be approximated by a normal distribution because the sample is large?
No; the sample means will be normally distributed, but the sample of incomes will be skewed to the right.
A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the drug, 14 subjects had a mean wake time of 93.4min and a standard deviation of 42.7min. Assume that the 14 sample values appear to be from a normally distributed population and construct a 90% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments.
ONINE CALC: ignore the mean. CI estimate = 32.56, 63.43. No, the confidence interval does not indicate whether the treatment is effective.
Refer to the accompanying data set of mean drive-through service times at dinner in seconds at two fast food restaurants. Construct a 99% confidence interval estimate of the mean drive-through service time for Restaurant X at dinner; then do the same for Restaurant Y.
Restaurant X = 155.2, 203.1 Restaurant Y = 133.8, 171.4. The confidence interval estimates for the two restaurants overlap, so there does not appear to be a significant difference between the mean dinner times at the two restaurants.
Three randomly selected households are surveyed. The numbers of people in the households are 2, 4, and 9. Assume that samples of size n=2 are randomly selected with replacement from the population of 2, 4, and 9.
Sample proportion (number of even values over n)=0,0.5,1. Probabilities=1/9, 4/9, 4/9. The mean of the sample proportions=0.667. The population proportion=0.667.
Which of the following is NOT needed to determine the minimum sample size required to estimate a population proportion?
Standard Deviation
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. #102-#132
The area of the shaded region is 0.4305
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. #97
The area of the shaded region is 0.5793
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. #95-#125
The area of the shaded region is 0.5827
Which of the following is NOT a property of the sampling distribution of the sample mean?
The distribution of the sample mean tends to be skewed to the right or left.
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. x w/ 0.3949 area
The indicated IQ score is 104 (change the z score to positive because the known area is to the right)
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. x w/ 0.55 area to the left
The indicated IQ score, x, is 101.9 (calculator with area then use the formula x=meu+(z*sigma))
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. x w/ an area of 0.6 to the right
The indicated IQ score, x, is 96.2 (always change the z score sign)
Which of the following is NOT a property of the chi-square distribution?
The mean of the chi-square distribution is 0
Assume that the sample is a simple random sample obtained from a normally distributed population of IQ scores of statistics professors. Use the table below to find the minimum sample size needed to be 95% confident that the sample standard deviation s is within 20% of σ.
The minimum sample size needed is 48. Yes the sample size is practical, because the sample size is small enough for most applications.
Assume that military aircraft use ejection seats designed for men weighing between 133.7 lb and 202 lb. If women's weights are normally distributed with a mean of 171.6 lb and a standard deviation of 46.2 lb, what percentage of women have weights that are within those limits? Are many women excluded with those specifications?
The percentage of women that have weights between those limits is 53.86%. Yes, the percentage of women who are excluded, which is the complement of the probability found previously, shows that about half of women are excluded.
Assume a population of 40, 45, 47, and 53. Assume that samples of size n=2 are randomly selected with replacement from the population. Listed below are the sixteen different samples.
The population median is not equal to the mean of the sample medians (it is also not half or double the mean of the sample medians). The sample medians do not target the population median, so sample medians are biased estimators, because the mean of the sample medians does not equal the population median.
Which of the following is NOT a requirement of constructing a confidence interval estimate for a population variance?
The population must be skewed to the right.
An elevator has a placard stating that the maximum capacity is 1932 lb—12 passengers. So, 12 adult male passengers can have a mean weight of up to 1932/12=161 pounds. If the elevator is loaded with 12 adult male passengers, find the probability that it is overloaded because they have a mean weight greater than 161 lb.
The probability the elevator is overloaded is 0.7703 because z = (161-167)/(28/sqrt12) = -0.74 = calc552 >> 0.2297 >> 1-0.2297. It is indeed overloaded.
Which of the following is NOT a requirement for using the normal distribution as an approximation to the binomial distribution?
The sample is the result of conducting several dependent trials of an experiment in which the probability of success is p.
_____________ is the distribution of all values of the statistic when all possible samples of the same size n are taken from the same population.
The sampling distribution of a statistic
_____________ is the distribution of sample proportions, with all samples having the same sample size n taken from the same population.
The sampling distribution of the proportion
Express the confidence interval 0.222<p<0.444 in the form ^p±E.
^p+-E = 0.333 +- 0.111
A drug is used to help prevent blood clots in certain patients. In clinical trials, among 4394 patients treated with the drug, 147 developed the adverse reaction of nausea. Construct a 90% confidence interval for the proportion of adverse reactions.
a) 0.033 b) 0.004 c) 0.029 < p < 0.037 d) One has 90% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
A genetic experiment with peas resulted in one sample of offspring that consisted of 438 green peas and 161 yellow peas. a. Construct a 90% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations?
a) 0.239<p<0.299b) No, the confidence interval includes 0.25, so the true percentage could easily equal 25%
In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2501 subjects randomly selected from an online group involved with ears. 1082 surveys were returned. Construct a 99% confidence interval for the proportion of returned surveys.
a) 0.433 b) 0.026 c) 0.407 < p < 0.459 d) One has 99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n=910 and x=512 who said "yes." Use a 90% confidence level.
a) 0.563 b) 0.027 c) 0.536 < p < 0.590 d) One has 90% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n=1023 and x=531 who said "yes." Use a 90% confidence level.
a) ^p = x/n = 0.519 b)E = 1.645 sqrt(0.519(1-0.519))/1023 = 0.026 c) All requirements are satisfied. The confidence interval is 0.493 < p < 0.545. d) would actually contain
Assume that females have pulse rates that are normally distributed with a mean of μ=74.0 beats per minute and a standard deviation of σ=12.5 beats per minute.
a. (convert pulse rate to z-score using z=(x-μ)/σ ) then use that in the calculator (menu, 5, 5, 2). Therefore, the probability that a randomly selected adult female's pulse rate is less than 81 beats per minute is 0.7123.
A genetics experiment involves a population of fruit flies consisting of 2 males named Barry and Carlos and 2 females named Diana and Erin.
a. 0 -> 1/4 0.5 -> 1/2 1 -> 1/4 b. The mean of the sampling distribution is 0.5. c. Yes, the sample mean is equal to the population proportion of males. These values are always equal, because proportion is an unbiased estimator.
Assume that 26.7% of people have sleepwalked. Assume that in a random sample of 1542 adults, 456 have sleepwalked.
a. 0.0059 b. Yes, because less, 0.05. c. Since the result of 456 adults that have sleepwalked is significantly high, it is strong evidence against the assumed rate of 26.7%.
Based on a smartphone survey, assume that 42% of adults with smartphones use them in theaters. In a separate survey of 284 adults with smartphones, it is found that 98 use them in theaters.
a. 0.0062 b. Yes, because the probability of this event is less than the probability cutoff that corresponds to a significant event, which is 0.05.
In a survey of 1428 people, 968 people said they voted in a recent presidential election. Voting records show that 65% of eligible voters actually did vote.
a. 0.0140 b. Some people are being less than honest because P(x≥968) is less than 5%
In a survey of 1370 people, 913 people said they voted in a recent presidential election. Voting records show that 64% of eligible voters actually did vote.
a. 0.0222 because subtract from 1 b. Some people are being less than honest because P(x>913) is less than 5%
A gender-selection technique is designed to increase the likelihood that a baby will be a girl. In the results of the gender-selection technique, 880 births consisted of 442 baby girls and 438 baby boys. In analyzing these results, assume that boys and girls are equally likely.
a. 0.0266 b. 1-0.5396=0.4597 No, because it is not far. c. The result from part (b) is more relevant, because one wants the probability of a result that is at least as extreme as the one obtained. d. No, because the probability of having 505 or more girls in 969 births is not unlikely, andthus, is attributable to random chance.
Based on a smartphone survey, assume that 51% of adults with smartphones use them in theaters. In a separate survey of 296 adults with smartphones, it is found that 150 use them in theaters.
a. 0.4787 b. Not significantly low.
TInterval (13.046,22.15) -x-=17.598 Sx=16.01712719 n=50
a. 13.05, 22.15 (menu 6, 6, 2) b. same as -x- so 17.60 Mbps & margin of error = high minus low in (a) so it's 4.55 Mbps. c. Because the sample size of 50 is greater than 30, the distribution of sample means can be treated as a normal distribution.
The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be 90% confident that his estimate is within seven percentage points of the true population percentage?
a. 139 b. 95 c. No, a sample of students at the nearest college is a convenience sample, not a simple random sample, so it is very possible that the results would not be representative of the population of adults.
In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use 0.03 E and 0.99%
a. 1842 b. 1769 c. No, using the additional survey information from part (b) only slightly reduces the sample size.
In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.02 margin of error and use a confidence level of 95%.
a. 2401 b. 2305 c. No, slightly.
The pulse rates of 154 randomly selected adult males vary from a low of 44bpm to a high of 116bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want 95% confidence that the sample mean is within 2bpm of the population mean.
a. 312 (sigma = range/4) b. 102 c. Sample size in part (a) is larger so the result from part (b) is likely to be better because it uses a better estimate of sigma.
The pulse rates of 151 randomly selected adult males vary from a low of 44 bpm to a high of 112 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want 98% confidence that the sample mean is within 2bpm of the population mean.
a. 391 b. 202 c. The result from part (a) is larger than the result from part (b). The result from part (b) is likely to be better because it uses a better estimate of sigma .
A survey found that women's heights are normally distributed with mean 62.6 in. and standard deviation 2.8 in. The survey also found that men's heights are normally distributed with mean 67.7 in. and standard deviation 3.2 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 63 in.
a. 7.05% men meet the height requirement (don't change the signs) Since most men do not meet the height requirement, it is likely that most of the characters are women. b. The new height requirements are a minimum of 62.4 in and a maximum of 67.7 in.
A survey found that women's heights are normally distributed with mean 63.5 in and standard deviation 2.5 in. A branch of the military requires women's heights to be between 58 in and 80 in.
a. 98.61% No, because only a small percentage of women are not allowed to join this branch of the military because of their height. b. All women are eligible except the shortest 1% and the tallest 2% which is at least 57.7 in and at most 68.6 in. (change the highest z score sign).
The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8 cm.
a. Find the probability that an individual distance is greater than 214.80 cm. 1-0.8775=0.1225 b. Find the probability that the mean for 25 randomly selected distances is greater than 203.70 cm. (σ/sqrtN) = 8/sqrt25 = 1.6 >> (203.7-205.5)/1.6 = -1.125 >> 1-0.1302 = 0.8697 c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? The normal distribution can be used because the original population has a normal distribution.
A ski gondola carries skiers to the top of a mountain. Assume that weights of skiers are normally distributed with a mean of 199lb and a standard deviation of 40lb. The gondola has a stated capacity of 25 passengers, and the gondola is rated for a load limit of 3750lb.
a. Given that the gondola is rated for a load limit of 3750lb, what is the maximum mean weight of the passengers if the gondola is filled to the stated capacity of 25 passengers? 3750/25=150 b. If the gondola is filled with 25 randomly selected skiers, what is the probability that their mean weight exceeds the value from part (a)? 40/sqrt25 = 8 >>> (150-199)/8 = -6.13. Thus, if the gondola is filled with 25 randomly selected skiers, the probability that their mean weight exceeds the value from part (a) is 1.0. c. If the weight assumptions were revised so that the new capacity became 20 passengers and the gondola is filled with 20 randomly selected skiers, what is the probability that their mean weight exceeds 187.5lb, which is the maximum mean weight that does not cause the total load to exceed 3750lb? 40/sqrt20=8.9443 >> 187.5-199/8.9443=-1.29. Thus, if the gondola is filled with 20 randomly selected skiers, the probability that their mean weight exceeds 187.5lb is 0.9015. d. The gondola will be overloaded if the mean weight of the passengers is above the maximum allowed mean weight. Not safe enough.
In a survey of 3013 adults aged 57 through 85 years, it was found that 83.1% of them used at least one prescription medication.
a. How many of the 3013 subjects used at least one prescription medication? 2504 b. Construct a 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication. 82%<p<84.2% c. What do the results tell us about the proportion of college students who use at least one prescription medication? The results tell us nothing about the proportion of college students who use at least one prescription medication.
Assume that females have pulse rates that are normally distributed with a mean of μ=76.0 beats per minute and a standard deviation of σ=12.5 beats per minute.
a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 70 beats per minute and 82 beats per minute. p=0.3688 b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean between 70 beats per minute and 82 beats per minute. p=0.9451 c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.
Assume that females have pulse rates that are normally distributed with a mean of μ=75.0 beats per minute and a standard deviation of σ=12.5 beats per minute.
a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 79 beats per minute. (79-72)/12.5 then calculator >> p=0.6255 b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 79 beats per minute. (σ/sqrtN) 12.5/sqrt25 = 2.5... so then you do z=(79-75)/2.5 = 1.6 THEN you find the probability with the calculator (552) which is 0.9452 c. Does the population have pulse rates that are normally distributed? YES. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.
The weights of a certain brand of candies are normally distributed with a mean weight of 0.8612g and a standard deviation of 0.0515g. A sample of these candies came from a package containing 440 candies, and the package label stated that the net weight is 375.7g. (If every package has 440 candies, the mean weight of the candies must exceed 375.7440=0.8539g for the net contents to weigh at least 375.7g.)
a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8539g. p=0.5564 b. If 440 candies are randomly selected, find the probability that their mean weight is at least 0.8539g. p=0.9985 c. Yes, because the probability of getting a sample mean of 0.8539g or greater when 440 candies are selected is not exceptionally small.
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150lb and 201lb. The new population of pilots has normally distributed weights with a mean of 160lb and a standard deviation of 30.4lb.
a. If a pilot is randomly selected, find the probability that his weight is between 150lb and 201lb. p=0.5402 b. If 37 different pilots are randomly selected, find the probability that their mean weight is between 150lb and 201lb. p=0.9773 c. When redesigning the ejection seat, which probability is more relevant? Part (a) because single
Assume that females have pulse rates that are normally distributed with a mean of μ=72.0 beats per minute and a standard deviation of σ=12.5 beats per minute.
a. Rate less then 79 bpm p=0.7123 b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 79 beats per minute. p=0.9875 c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.
TInterval (13.046,22.15) -x-=17.598 Sx=16.01712719 n=50
a. df= 49 b. t(a/2)=2.01 c. The number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values.
The lengths of pregnancies are normally distributed with a mean of 269 days and a standard deviation of 15 days.
a. p=0.0038 (1-0.9962) b. 243 days
A gender-selection technique is designed to increase the likelihood that a baby will be a girl. In the results of the gender-selection technique, 969 births consisted of 505 baby girls and 464 baby boys. In analyzing these results, assume that boys and girls are equally likely.
a. p=0.0108 b. p=0.0994 No, not far. c. Part b d. No, not unlikely.
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 120lb and 161lb. The new population of pilots has normally distributed weights with a mean of 128lb and a standard deviation of 25.2 lb.
a. p=0.5294 b. p=0.9638 c. Part (a) because single
Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The aircraft can carry 39 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6,591lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than 6,591 lb39=169lb. What is the probability that the aircraft is overloaded?
a. p=0.758 b. The pilot needs to take action.
A boat capsized and sank in a lake. Based on an assumption of a mean weight of 135lb, the boat was rated to carry 60 passengers (so the load limit was 8,100lb). After the boat sank, the assumed mean weight for similar boats was changed from 135lb to 175lb.
a. p=1 b. 40.4/sqrt14 = 10.797354 >> z = 175-178.9/10.7974 = -0.36 = 1-0.3590 = p=0.6410 c. Because there is a high probability of overloading, the new ratings do not appear to be safe when the boat is loaded with 13 passengers.
Which of the following groups of terms can be used interchangeably when working with normal distributions?
areas, probability, and relative frequencies
Nicotine in menthol cigarettes 99% confidence; n=21, s=0.21 mg.
df = 20 X^2L = 7.434 (the value right to the left of it) X^2R = 39.997 (all the way to the right on the table) so the confidence interval estimate of sigma o is 0.15 & 0.34
Platelet Counts of Women 80% confidence; n=24, s=65.1.
df = 23 FIND 1-given percentage (which is alpha)(and TWO TAILED) and plug that into online calculator for both values... X^2L = 14.848 & X^R = 32.007. CI estimate = 55.2, 81.0
Platelet Counts of Women 95% confidence; n=24, s=65.8.
df = 23 X^2L = 11.689 & X^2R = 38.076 CI estimate = 51.1, 92.3
Nicotine in menthol cigarettes 98% confidence; n=26, s=0.21 mg.
df = 25 X^2L = 11.524 (left next to 13.120) X^2R = 44.314 (all the way to the right but not the last one) so the CI estimate of sigma is 0.16, 0.31
Find the critical value za/2 that corresponds to the confidence level 84%
find alpha (1-.84) and then divide by 2. 1-(a/2)=0.92 so find z(0.08)... menu553(0.92,0,1) = 1.41
Here are summary statistics for randomly selected weights of newborn girls: n=289, x=28.4hg, s=7.5hg. The confidence level is 90%.
menu 5, 5, 6: t(a/2) = 1.65
Pulse rates of women are normally distributed with a mean of 77.5 beats per minute and a standard deviation of 11.6 beats per minute. Answer the following questions.
meu = 0 & sigma = 1 The z scores are numbers without units of measurement.
If np≥5 and nq≥5, estimate P(at least 10) with n=13 and p=0.5 by using the normal distribution as an approximation to the binomial distribution; if np<5 or nq<5, then state that the normal approximation is not suitable.
np = (13*0.5) = 6.5 & nq = (13*0.5) = 6.5 μ=6.5 & sigma=sqrtnpq=1.803 x<10-0.5=9.5 z=(9.5-6.5)/1.803=1.66 P(z<1.66)=0.952 = 1-0.952 = 0.048
Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 167 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted. Probability that fewer than 39 voted.
np=167*0.22=36.74 nq=167*0.78=130.26 meu=np=36.74 sigma=sqrtnpq=5.353 x=39 so z=(38.5-36.74)/5.353=0.33 so the area left of z=0.33 is 0.6293 MINUS 0.0005 equals 0.6288
An elevator has a placard stating that the maximum capacity is 2490lb—15 passengers. So, 15 adult male passengers can have a mean weight of up to 2490/15=166 pounds. If the elevator is loaded with 15 adult male passengers, find the probability that it is overloaded because they have a mean weight greater than 166 lb. (Assume that weights of males are normally distributed with a mean of 176 lb and a standard deviation of 27 lb.)
p(overloaded)=0.9243. No, there is a good chance that 15 randomly selected adult male passengers will exceed the elevator capacity.
Assume that adults have IQ scores that are normally distributed with a mean of 99.6 and a standard deviation of 24.5. Find the probability that a randomly selected adult has an IQ greater than 146.1.
p=0.0288 (1-0.9712)
Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 198 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted. Probability that exactly 48 voted
p=0.0512
Assume that adults have IQ scores that are normally distributed with a mean of μ=105 and a standard deviation σ=15. Find the probability that a randomly selected adult has an IQ between 95 and 115.
p=0.4950
Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 170 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted. Probability that exactly 39 voted.
p=0.6517-0.5793=0.0724
Assume that adults have IQ scores that are normally distributed with a mean of μ=105 and a standard deviation σ=20. Find the probability that a randomly selected adult has an IQ less than 125.
p=0.8413 (convert to z score then calc552)
A _______ is a single value used to approximate a population parameter.
point estimate
The _______ is the best point estimate of the population mean.
sample mean
The best point estimate of the population variance σ2 is the _____________.
sample variance, s^2
The confidence level is 95%, σ is not known, and the histogram of 63 player salaries (in thousands of dollars) of football players on a team is as shown.
t(a/2) = 2
The confidence level is 99%, σ is not known, and the histogram of 58 player salaries (in thousands of dollars) of football players on a team is as shown.
t(a/2) menu, 5, 5, 6, but erase the negative. = 2.66
Which is NOT a criterion for distinguishing between results that could easily occur by chance and those results that are highly unusual?
the sample size is less than 5% of the size of the population
A critical value, zα, denotes the _______.
z dash score with an area of alpha to its right.
The confidence level is 90%, σ=3815 thousand dollars, and the histogram of 56 player salaries (in thousands of dollars) of football players on a team is as shown.
z(a/2) = 1.65
The confidence level is 95%, σ=3593 thousand dollars, and the histogram of 57 player salaries (in thousands of dollars) of football players on a team is as shown.
z(a/2) = 1.96 because of table
Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. A=0.9162
z=1.28 (calculator: menu, 6, 5, 3)
Find the critical value za/2 that corresponds to the confidence level 87%
za/2 = 1.51