Chapter 18 Review Problems

Chapter 18 Review Problems - Chapter 18 Sampling...

Info iconThis preview shows pages 1–15. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 18 Sampling Distribution Models (Review) Name MliLTII-‘LE CHOiCE. Choose the one alternative that best completes the statement or answers the question Determine whether the Normal model may be used to describe the distribution of the sample proportions. If the Normal model may be used, list the conditions and explain why each is satisfied. If Normal model may not be used, explain which condition is not satisfied. 1) In a large statistics class, the professor has each student toss a coin12 times and calculate the 1) proportion of tosses that come up tails. The students then report their results, and the professor plots a histogram of these several proportions. May a Normal model be used here? A) A Normal model should not be used because the 10% condition is not satisfied: the sample size, 12, is larger than 10% of the population of all coins flips. B) A Normal model should not be used because the sample size is not large enough to satisfy the success/failure condition. For this sample size, np n 6 : nq a 6 which are both less than 10. C) A Normal model may be used: Coin flips are independent of each other~ no need to check the 10% condition Success/Failure condition is satisfied: np = nq = 6 which are both less than 10 D) A Normal model may not be used because the population distribution is not Normal. ‘8) A Normal model may be used: Coin flips are independent of each other — no need to check the 10% condition Success/Failure condition is satisfied: np = nq = 12 which are both greater than 10 2) A candy company claims that 6% of the jelly beans in its spring mix are pink. Suppose that the 2) candies are packaged at random in small bags containing about '70 jelly beans. A class of students opens several bags, counts the various colors of jelly beans, and calculates the proportion that are pink in each bag. Is it appropriate to use a Normal model to describe the distribution of the proportion of pink jelly beans? A) A Normal model is not appropriate because the randomization condition is not satisfied: the 70 jelly beans in the bag are not a simple random sample and cannot be considered representative of all jelly beans. B) A Normal model is not appropriate because the population distribution is not Normal. C) A Normal model is not appropriate because the 10% condition is not satisfied: the sample size, 70, is larger than 10% of the population of all jelly beans. D) A Normal model is not appropriate because the sample size is not large enough to satisfy the success/failure condition. For this sample size, np m 4.2 which is less than 10. E) A Normal model is appropriate: Randomization condition is satisfied: the '70 jelly beans in the bag are selected at random and can be considered representative of all jelly beans 10% condition is satisfied: the sample size, 70, is less than 10% of the population of all jelly beans. The successgfailure condition is satisfied: n x 70 which is greater than 10 3) A grower claims that only 10% of the strawberries he grows are unsatisfactory. When a truckload of his strawberries arrives at a farmers‘ market, one basket containing approximately 110 strawberries is examined for bruised or rotten fruit. May the Normal model be used to describe the distribution of the proportion of rotten strawberries in the basket? A) Normal model may not be used to describe distribution of sample proportions. Randomization condition is not satisfied: Since the strawberries are in the same basket, these strawberries are not a simple random sample and are not representative of all the grower's strawberries. Growing or storage conditions could affect all the strawberries in a basket. B) Normal model may not be used to describe distribution of sample proportions. 10% condition is not satisfied: the 110 strawberries likely represent less than 10% of all the grower's strawberries. C) Normal model may not be used to describe distribution of sample proportions. Successflfailure condition is not satisfed: np w 11 is not greater than 20 D) Normal model may be used to describe distribution of sample proportions. Randomization condition is satisfied: the strawberries in the basket are a simple random sample and therefore representative of all the grower's strawberries 10% condition is satisfied: the 110 strawberries likely represent less than 10% of all the grower's strawberries. Successgfailure condition is satisfed: up n 11 and nq m 99 are both greater than 10. E) Normal model may not be used to describe distribution of sample proportions. Distribution of population is not normal 4) A national study found that 13% of college seniors regret their choice of major. Suppose a group oi 115 college sem‘ors is selected at random. May the Normal model be used to describe the distribution of the proportion of seniors in the sample who regret their choice of major? A) Normal model may not be used to describe distribution of sample proportions. Distribution of population is not normal 8) Normal model may be used to describe distribution of sample proportions. Randomization condition is satisfied: the seniors were selected at random and are therefore representative of all college seniors 10% condition is satisfied: the 115 seniors are less than 10% of all college seniors Successgfailure condition is satisfed: np a 14.95 and nq 2 100.05 are both greater than 10. C) Normal model may not be usedto describe distribution of sample proportions. 10% condition is not satisfied: the 115 seniors are less than 10% of all college seniors D) Normal model may not be used to describe distribution of sample proportions. Successffailure condition is not satisfed: np m 14.95 is not greater than 20 E) Normal model may not be used to describe distribution of sample proportions. Randomization condition is not satisfied: the seniors may not be representative of all college seniors, as some of those in the sample may have the same major. 3) 4) Find the mean of the sample proportion. 5) Based on past experience, a bank believes that 6% of the people who receive loans will not make payments on time. The bank has recently approved 300 loans. What is the mean of the proportion of clients in this group who may not make timely payments? A) nz3% B) pn18% C) p.==6% 13) a: 1.37% E) p=4.24% 6) A realtor has been told that 42% of home0wners in a city prefer to have a finished basement. She surveys a group of 400 homeowners randomly chosen from her client list. Find the mean of the proportion of homeowners in this sample who prefer a finished basement. A) s 0.42% a) = 4.2% e) = 1.2% o) M = 58% E) p m 2.5% s s $4 Find the standard deviation of the sample proportion. 7) Based on past experience, a bank believes that 5% of the people who receive loans wilt not make payments on time. The bank has recently approved 300 loans. What is the standard deviation of the proportion of clients in this group who may not make timely payments? - A) o m 1.29% B) o m 1.26% C) o m 1.58% D) o m 3.77% E) o n 3.87% 8) A realtor has been told that 44% of homeowners in a city prefer to have a finished basement. She surveys a group of 200 homeowners randomly chosen frorn her client list. Find the standard deviation of the proportion of homeowners inthis sample who prefer a finished basement. A) 024% B) 0'2 3.5% C) o: 1.7% 3:3) 0: 0.44% r.) or: 0.5 % 5) 6) 7) In a large class, the professor has each person toss a coin several times and calculate the proportion of his or her tosses that come up heads. The students then report their results, and the professor ptots a histogram of these proportions. Use the 68—95—9937 Rule to provide the appropriate response. 9) If each student tosses the coin 50 times, about 68% of the sample proportions should be between what two numbers? A) 0.49 and 0.51 B) 0.07 and 0.14 C) 0.34 and 0.67 D) 0.16 and 0.84 E) 0.43 and 0.57 9) 10) If each student tosses the coin 200 tirnes, about 99.7% of the sample proportions should be between 10) what two numbers? A) 0.106 and 0.1414 B) 0.394 and 0.606 C) 0.00075 and 0.99925 D) 00015 and 0.9985 E) 0.4925 and 0.5075 Find the specified probability, from a table of Normal probabilities. Assume that the necessary conditions and assumptions are met. ‘ 11) Based on past experience, a bank believes that 7% of the people who receive loans will not make 11) payments on time. The bank has recently approved 300 loans. What is the probability that over 8% of these clients will not make timeiy payments? - A) 0.504. B) 0.752 C) 0.248 D) 0.4% s) 0.68 12) A candy company claims that its jelly bean mix contains 15% blue jelly beans. Suppose that the 12) candies are packaged at random in small bags containing about 200 jeiiy beans. What is the probability that a bag will contain more than 18% blue jelly beans? A) 0.883 3) 0.755 C) 0.234. D) 0.245 a) 0.117 13) When a truckload of oranges arrives at a packing plant, a random _sample of 125 is selected and 13) examined. The whole truckload wili be rejected if more than 8% of the sample is unsatisfactory. Suppose that in fact9°/o of the oranges on the truck do not meet the desired standard. What's the probability that the shipment will be accepted anyway? A) 0.3483 B) 0.6517 C) 0.7803 D) 0.6966 E) 0.2197 14) Researchers beiieve that 7% of children have a gene that may be linked to a certain childhood 14) disease. In an effort to track 50 of these children, researchers test 950 newborns for the presence of this gene. What is the probability that they find enough subjects for their study? A) 0.0358 B) 0.9821 C) 0.9216 D) 0.9581 E) 0.0179 15) A summer resort rents rowboats to customers but does not allow more than four people to a boat 15) Each boat is designed to hold no more than 800 pounds. Suppose the distribution of adult males who rent boats, including their clothes and gear, is normal with a mean of 195 pounds and standard deviation of 10 pounds. if the weights of individual passengers are independent, What is the probability that a group of four adult male passengers will exceed the acceptable weight limit of 800 pounds? A) 0.023 B) 0.977 C) 0.159 D) 0.317 E) 0.046 Answer the question. 16) in a large class, the professor has each person toss a coin 200 times and calculate the proportion 01 his or her tosses that were tails. The students then report their results, and the professor records the proportions. One student claims to have tossed her coin 200 times and {Ound58% tails. What do you think of this claim? Explain your response. A) This is a typicai result. Her proportion is only 2.26 standard deviations above the rnean. B) This is an unusuai result. Her proportion is about 2.26 standard deviations above the mean. C) This is a typical result. Her proportion is only 1.60 standard deviations above the mean. D) This is a fairly unusual result. Her proportion is about 1.60 standard deviations above the mean. E) This is an extremely unlikely result. Her proportion is about 5.2 standard deviations above the mean. 17) A nationai study reported that 74% of high school graduates pursue a college education immediately after graduation. A private high school advertises that 155 of their 196 graduates last year went on to college. Does this school have an unusually high proportion of students going to college? A) This school can boast an unusually high proportion of students going to college. Their proportion is 1.30 standard deviations above the mean. B) This school cannot boast an unusually high proportion of students going to coliege. Their proportiori is oniy 0.97 standard deviations above the mean. C) This school cannot boast an unusually high proportion of students going to college. Their proportion is oniy 1.62 standard deviatioris above the mean. D) This school cannot boast an unusually high proportion of students going to college. Their proportion is only 1.30 standard deviations above the mean. E) This school can boast an unusually high proportion of students going to college. Their proportion is 2.61 standard deviations above the mean. Solve the problem. 18) One hospital has found that 13.95% of its patients require speciaiiy equipped beds. If the hospital has 2.58 beds, What percentage of the beds should be specially equipped if the hospital wishes to be "pretty sure" of having enough of these beds? Assume that the hospital wants only a 5% chance that they could run short of these beds, even when the hospital is tuiiy occupied. A) 19.0% s) 19.5% C) 21.9% D) 17.5% s) 18.2% 16) i7) 18) Determine whether the Normal model may be used to describe the distribution of the sample means. If the Normal model may be used, list the conditions and explain why each is satisfied. If Normal model may not be used, explain which condition is not satisfied. 19) The weights of men in a certain city are normally distributed with a mean 013153 11) and a standard 19) deviation of 22 lb. Suppose a sample of 3 men is selected at random from the city and the mean weight, n): is determined for the men in the sampie. May the Normai model be used to describe the sampling distribution of the mean, 3:? A) Yes, Normai model may be used. Randomization condition- The men were selected at random indegendence assumgtion: it is reasonable to think that weights of randomly selected men are mutually independent. I Large enough sample condition: Since the original population is normally distributed, a sample of 3 is iarge enough, in fact any sample would be iarge enough. 10% condition: the 3 men in the sample certainly represent less than 10% of men in the city. 13) Randomization condition is not satisfied: The men in the sample do not represent a simple random sample and may not be representative of all men in the city C) Indegendence assumgtion is not satisfied: Since the men may be related, the chance of selecting a heavy man depends on who has aiready been selected. D) No, Norma}. modei may not be used: Large enough samgle condition is not satisfied: since the sample size is only 3 E) No, Normal model may not be used: 10% condition: is not satisfied since the 3 men in the samgale represent less than 10% of men in the city. 20) The mean annuai income for women in one city is $28,520 and the standard deviation of the 20) incomes is $5600. The distribution of incomes is skewed to the right. Suppose a sample of 12 women is seiected at random from the city and the mean income, it is determined for the women in the sample. May the Normal model be used to describe the sampling distribution of the mean, x ? A) No, Normai model may not be used since incomes of women in the city are not normaiiy distributed but are skewed to the right B) Yes, Normal model may be used. Randomization condition; The women were selected at random Independence assumption: It is reasonable to think that incomes of randomly seiected women are mutuaiiy independent. Large enough sample condition: a sample of 12. is large enough for the Central Limit Theorem to apply 10% condition is satisfied since the 12 women in the sample certainly represent iess than 10% of women in the city C) No, Normal modei may not be used: 10% condition is not satisfied since the 12 women in the sample represent less than 10% of women in the city D) No, Normal model may not be used: Independence assumption is not satisfied: since the women in the sample may live in the same neighborhood, the chance of picking a woman with a high income depends on who has already been selected. E) No, Normal model may not be used: Large enough sampie condition is not satisfied: since the distribution of incomes in the original population is skewed, a sample of 12 is not large enough Describe the indicated sampling distribution. 21) The weights of people in a certain popuiation are normally distributed with a mean of 158 lb and a 21) standard deviation 01325 ib. Describe the sampling distribution of the mean for samples of size9. In particular, state whether the distribution of the sample mean is normai or approximately normal and give its mean and standard deviation. A) Normal, mean = 158 lb, standard deviation : 25 Fe B) Approximately normal, mean r» 158 lb, standard deviation = 2.78 lb C) Normal, mean a 158 lb, standard deviation: 8.33 lb D) Approximately normal, mean a 158 lb, standard deviation = 8.33 lb ‘13) Normal, mean = 158 1b, standard deviation : 2.78 lb 22) The mean. annual income for adult women in one city is $28,520 and the standard deviation of the 22) incomes is $5700. The distribution of incomes is skewed to the right. Determine the sampling distribution of the mean for samples of size 110. In particular, state whether the distribution of the sample mean is normal or approximately normal and give its mean and standard deviation. A) Normal, mean = $28,520, standard deviation m $543 13) Normal, mean = $28,520, standard deviation 2 $52 C) Approximately normal, mean = $28,520, standard deviation 2 $52 D) Approximately normal, mean = $28,520, standard deviation a: 335700 E) Approximately normal, mean = $28,520, standard deviation :2 $543 23) For the population of one town, the distribution of the number of siblings, x, is skewed to the right 23) The mean number of siblings is 1.1 and the standard deviation is 1.5. Let ; denote the mean number of siblings for a random sample of size 38. Determine the sampling distribution of the mean, In particular, state whether the distribution of the sample mean is normal or approximately normal and give its mean and standard deviation. A) Approximately normal, mean a 352, standard deviation = 1.5 B) Approximately normal, mean : 1.1, standard deviation = 0.24 C) Normal, mean m 1.1, standard deviation : 0.24 D) Normal, mean a 1.1, standard deviation 2 1.5 E) Approximately normal, mean = 1.1, standard deviation = 1.5 24) A museum offers several levels of membership, as shown in the table. 24) Member Percent of Category of Donation (33) Members Individual 43 Family 28 Family Plus 16 Sponsor 9 Patron 4 Suppose that 20 museum members are selected at random. Describe the sampling distribution of the mean donation for these 20 members. In particular, state whether the distribution of the sample mean is normal or approximately normal, and give its mean and standard deviation. A) Approximately normal, mean : $101.10, standard deviation : $92.47 B) Normal, mean = $113.15, standard deviation : $20.68 C) Approximately normal, mean = $101.10, standard deviation : $20.68 D) Normal, mean = $113.15, standard deviation 5: $92.47 B) Approximately normal, mean u $101.10, standard deviation : $4.62 25) You pay $10 and r011 a die. if you get a five or six, you win $30. if not, you get to roll again. If you 25) get a 5 or 6 on the second roll, you get your $10 back. Suppose you play this game 30 times. Describe the sampling distribution of your mean winnings. In particular, state whether the distribution of the sample mean is normal or approximately normal, and give its mean and standard deviation. A) Approximateiy normal, mean :z $2.22, standard deviation 2 $0.44 B) Approximateiy normal, mean a $2.22, standard deviation 2 $2.40 C) Norma}, mean a $2.22, standard deviation : $13.15 D) Normal, mean 3 $2.22, standard deviation 3 $2.40 E) Approximately normai, mean u $2.22, standard deviation 3 $13.15 At a large university, students have an average credit card debt of $2500, with a standard deviation of $1200. A random sample of students is seiected and interviewed about their credit card debt. Use the68—95—99.7 Rule to answer the question about the mean credit card debt for the students in this sample. 26) If we imagine all the possible random samples of 250 students at this university, 68% of the 26) samples should have means between What two numbers? A) $2424.11 and $2575.89 3) $2272.33 and $2727.57 C) $2424.11 and $2651.78 1)) $2348.22. and $2651.78 3) $1300 and $3700 27) If we imagine all the possible random sampies of 150 students at this university, 99.7% of the 27) samples should have means between what two numbers? A) $150.00 and $2597.98 B) $2304.04 and $2695.96 C) $100 and $6100 33) $150.00 and $2695.96 E) $2206.06 and $2793.94 Find the specified probability, from a table of Normal probabilities. Assume that the necessary canditions and assumptions are met. 28) The number of hours per week that high school seniors spend on computers is normaily 28) distributed, with a mean of 6 hours and a standard deviation of 2 hours. 80 students are chosen at random. Lety represent the mean number of hours spent on the computer for this group. Find the probability that 3? is between 6.2 and 6.9. A) 0.187 B) 0.134 C) 0.371 D) 0.817 E) 3.130 29) The weight of crackers in a box is stated to be 16 ounces. The amount that the packaging machine puts in the boxes is beiieved to have a Normal model with mean 16.15 ounces and standard deviation 0.3 ounces. What is the probability that the mean weight of a 10—box case of crackers is below 16 ounces? A) 09429 B) 0.0571 C) 0.2273 D) 0.1142 E) 0.8858 30) A restaurant's receipts show that the cost of Customers dinners has a skewed distribution with a mean of $54; and a standard deviation of $18. What is the probabiiity that the next 100 customers will spend an average of at least $50 on dinner? A) 0.0132 B) 0.4121 C) 0.9614 D) 0.5879 E) 0.9868 31) The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 91 inches, and a standard deviation of 12 inches. What is the probability that the mean annual snowfall during 36 randomly picked years will exceed 93.8 inches? A) 0.4192. 13) 0.5808 C) 0.9192 D) 0.0026 E) 0.0808 32) The weights of the fish in a certain lake are normally distributed with a mean of13 lb and a standard deviation of 12. If 16 fish are randome selected, what is the probability that the mean weight will be between 10.6 and 16.6 lb? A) 0.3270 B) 0.5808 C) 0.6730 13) 0.0968 E) 0.4032 Find the indicated probability. 33) A museum offers several levels of membership, as shown in the table. Member Percent of Category of Donation ($) Members individual 43 Famin 28 Family fins 16 Sponsor 9 Patron 4 During a membership drive, a volunteer enrolls 20 new members. if these 20 new members can be considered a random sample of all the museum's members, what is the probability that the mean donation from the new members is at least $150? A) 0.0090 13) 0.1056 C) 0.0054 D) 0.0126 E) 0.2981 34) A restaurant's receipts show that the cost of customers dinners has a skewed distribution with a mean of $54 and a standard deviation of $18. What is the probabiiity that the next 100 customers will spend a total of at least $5800 on dinner? A) 0.9868 B) 0.0562 C) 0.4121 D) 0.5879 E) 0.0132 10 29) 30) 31) 32) 33) Find the specified probability, from a table of Normal probabilities. Assume that the necessary conditions and assumptions are met. 34.) Solve the problem. 35) At a shoe factory, the time taken to poiish a finished shoe has a mean of3.7 minutes and a standard deviation of 0.48 minutes. if 4.14 shoes are polished, there is a 5% chance that the mean time to polish the shoes is below what value? A) 3.82 minutes B) 3.53 minutes C) 3.58 minutes D) 3.51 minutes 13) 3.89 minutes 36) Bill’s hourly earnings have a skewed distribution with a mean of $84.20 and a standard deviation oi $6.90. In a typical week he works 30 hours. In the best 2.5% of such weeks, his average hourly earnings are above what amount? {In other words, there is a 2.5% chance that his average hourly earning for the 30 hours will be greater than what value?) A) $81.73 B) $85.82 C) $82.58 D) $86.67 E) $87.53 Provide an appropriate response. 37) A researcher seiects sampies of size n from a population and determines the mean test score for each sample. The mean and standard deviation of the sampling distribution are 60 and 12 respectively. If the sample size is multiplied by a factor of 4, what will the mean and standard deviation of the new sampling distribution be? A) mean wiil be 30, standard deviation will be 6 B) mean will be 30, standard deviation will be 3 C) mean wiil be 60, standard deviation will be 24 D) mean will be 60, standard deviation will be 3 E) mean will be 60, standard deviation will be 6 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 38) Some experts estimate that the length of time people iive in a house has a mean of 7.2 years 38) and a standard deviation of2.2 years. The distribution of the times is believed to be skewed to the right. Explain why you could estimate the probability that 100 people selected at random had lived in their houses for an average of 8 years or more, but you could not estimate the probability that an individual had done so. 39) The distribution of incomes of employees at one company is strongly skewed to the right 39) and the mean and standard deviation of the incomes are known. Is it possible to determine the probability that a randomly selected employee earns more than $80,000? Is it possible to determine the probability that the mean income for 10 randomly selected employees is more than $80,000? Is it possible to determine the probability that the mean income for 50 randomly selected employees is more than $80,000? Explain your responses. 11 35) 36) 37)m MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 40) In which of the following situations does the Central Limit Theorem allow use of a Normal mode] 40) for the sampling distribution model: A: Weights of students are normally distributed. We wish to determine the probability that the mean weight for a random sample of 4 students is greater than 150 pounds. B: The distribution of test scores of students is slightly skewed to the right. We wish to determine the probability that the mean score for a random sample of 8 students is greater than 80. C: The distribution of incomes of students is strongly skewed to the right. We wish to determine the probability that the mean income for a random sample of 100 students is greater than $28,000. A) A and C B) A, B, and C C) C only D) A and B E) A oniy 41) A sample is chosen randomly from a population that was strongly skewed to the right. Describe 41) the sampling distribution modei for the sample mean if the sample size is small. A) Skewed right, center at u, standard deviationxlgln B) Normal, center at u, standard deviation o/«fn C) Normal, center at u, standard deviationmlo—ln D) There is not enough information to describe the sampling distribution model. E) Skewed right, center at u, standard deviation an]; 42) A certain population is bimodal. We want to estimate its mean, so we will collect a sampie. Which 42) should be true if we use a large sample rather than a small one? I. The distribution of our sample data will be more clearly bimodal H. The sampling distribution of the sample means will be approximately normal. III. The variability of the sample means will be smaller. A) I only B) I, II, and III C) 11 only D) II and III E) 111 only SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 43) A box of Raspberry Crunch cereai contains a mean of 13 ounces with. a standard deviation 43) of 0.5 ounce. The distribution of the contents of cereal boxes is approximately Normal. What is the probability that a case of 12 cereal boxes contains a total of more than 160 ounces? 12 44) A can of pumpkin pie mix contains a mean of 30 ounces and a standard deviation of 2 44) ounces. The contents of the cans are normally distributed. What is the probability that four randomly selected cans of pumpkin pie mix contain a total of more than 126 ounces? 4:5) The weights of hens’ eggs are normally dietributed with a mean of 56 grams and a 45) standard deviation of 4.8 grams. What is the probability that a dozen randomly selected eggs Weighs over 690 grams? 13 Answer Key Testname: CHAPTER 18 SAMPLING DISTRIBUTION MODELS (REVIEW) 1) B 2) D 3) A 4) B 5) C 6) B 7) B 8) B 9) E 10) B 11) C 12) E 13) A 14) B 15) A 16) B 17) C 18) D 19) A 20) s 21) C 22) E 23) B 24) C 25) B 26) A 27) E 28) A 29) B 30) E 31) E 32) C 33) A 34) E 35) C 36) D 37) E 38) The Central Limit Theorem guarantees that the distribution of the mean time is Normally distributed for large sample sizes, even if the original population is not normally distributed. The Central Limit Theorem doesn't help us with the distribution of the individual times. 39) Since the distribution of incomes is skewed, we can't uSe the Normal distribution to determine the probability that ar individual income is more than $80,000. Since the distribution is strongly skewed, a sample of 10 employees is probably not large enough for the Central Limit Theorem to allow use of a Normal model for the sampling distribution model. So we cannot determine the probability that the mean income for 10 randomly selected employees is more than $80,000. A sample of 50 employees is large enough for the Central Limit Theorem to allow use of a Normal model for the sampling distribution model, even though the original distribution is strongly skewed. So it is possible to determine the probability that the mean income for 50 randomly selected employees is more than $80,000. 40) A 41) E 14 Answer Key Testname: CHAPTER 18 SAMPLING DISTRIBUTION MODELS (REVIEW) 42) B 43) Two methods can be used to solve this problem: Method 1: Let B = weight of one box of cereal and T a weight of 12 boxes of cereal. We are told that the contents of the boxes are approximately Normal, and we can assume that the content amounts are independent from box to box. E(T) = E2031 4- B3 4» + 1312) = 13031) + E(B2) + + E(B12) = 156 ounces Since the content amounts are independednt from box to box. Var(T) = Var(81 + Bz + + 1312) = Var(l31)+ Var(Bg) + 4- Varwrg) = 3 513(1) = ox/Var(T) : J2? = 1.73 ounces We model T with N(156, 1.73). z = 16%??- : 2.31 and PCT > 160 ) e P(z > 2.31) = 00104 There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160 ounces. Method 2: Using the Central Limit Theorem approach, let“); : average content of boxes in the case. Since the contents are Normally distributed, "3; is modeled by N[13, 13.33 — 13 -- 160 — >-—- =1? >13.33 = Z>W =P .>2.3i seem. Ply 12] {y ) P[ SEA/12] (? ) There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160 ounces. 44) Two methods can be used to solve this problem: Method 1: Let P 2: one can of pumpkin pie mix and T: 4 cans of pumpkin pie mix. We are told that the contents of the cans are normally. distributed, and we can assume that the content amounts are independent from can to can. E(T) = £031 + P2 + P3 4- P4) = £(P1) + E(Pz) + E(P3) + E(P4) = 120 ounces Since the content amounts are independednt from box to box. Var(T) = Var(P1 + P2 4- P3 + P4) = Varfl’i) + Varfl’g) + Var(P3) + Var(i’4) = 16 513(1) = m/Vara‘) 2 41—6 = 4 ounces We model T with N820, 4). a #126 l 120 = 1.5 and P(T > 126) e Pia > 1-5) = 0067 There is a 6.7% chance four randomly selected cans of pumpkin pie mix will contain more than 126 ounces. Z Method 2: Using the Central Limit Theorem approach, let 3;: average content of cans in the sample. Since the contents are Normally distributed, 3: is modeled by 30, «IE p[§>iE—6—] = P§> 31.5) a [{z >§2éffl = P(z > 1.5) = 0.067. There is a 6.7% chance four randomly selected cans of pumpkin pie mix will contain more than 126 ounces. - 690 57.5~56 45v! —— = >—-——— :1” 1.08 214% ) Y) 12] {Z 4.8/«I12] {z} ) 15 ...
View Full Document

This note was uploaded on 11/10/2009 for the course CAL 3452 taught by Professor Mr.sun during the Spring '09 term at Sungkyunkwan.

Page1 / 15

Chapter 18 Review Problems - Chapter 18 Sampling...

This preview shows document pages 1 - 15. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online