SPA 3e_ Teachers Edition _ Ch 6

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442 C H A P T E R 6 • Sampling Distributions Teaching Tip: Differentiate By this point in the chapter, advanced math students may notice similarities between the sampling distribution of p^ and the sampling distribution of x. This is not a coincidence. Thinking back to the “A penny for your thoughts?” activity in Lesson 6.1, we can let the random variable X be 0 if the penny is not from the 2000s, and let it be 1 if it is from the 2000s. That is, X is 0 for a “failure” and 1 for a “success.” Then in a sample of n 5 5 pennies, the sample proportion is just the sum of the values of X, divided by 5. In other words, the sample proportion can be thought of as a special case of the sample mean. 34.5 38 41.5 45 48.5 52 55.5 50 Sample mean number of texts Using Table A: 50 − 45 z = = 1.43 3.5 P (Z ≥ 1.43) 5 1 2 0.9236 5 0.0764 Using technology: Applet/normalcdf (lower: 50, upper: 100000, mean: 45, SD: 3.5) 5 0.0766 L e SSon APP 6. 6 Keeping things cool with statistics? 1. Draw a normal curve. 2. Perform calculations. (i) Standardize the boundary value and use Table A to find the desired probability; or (ii) Use technology. FOR PRACTICE TRY EXERCISE 5. Lesson App Answers 1. Because n = 70 ≥ 30, the sampling distribution of x is approximately normal by the central limit theorem. 2. Mean: m x = m = 1 hour; SD: s x = s !n = 1.5 = 0.179 hour !70 1.1 − 1.0 z = = 0.56; P( x > 1.1) 0.179 = P(Z > 0.56) = 1 − 0.7123 = 0.2877 Using technology: Applet/normalcdf (lower:1.1, upper:1000, mean:1, SD:0.179) 5 0.2882 3. No; there is a 29% chance that the time allotted will not be enough. Your company has a contract to perform preventive maintenance on thousands of air-conditioning units in a large city. Based on service records from the past year, the time (in hours) that a technician requires to complete the work follows a strongly right-skewed distribution with m 5 1 hour and s 5 1.5 hours. As a promotion, your company will provide service to a random sample of 70 airconditioning units free of charge. You plan to budget an average of 1.1 hours per unit for a technician to complete the work. Will this be enough time? 1. What is the shape of the sampling distribution of x for samples of size n 5 70 from this population? Justify. 2. Calculate the probability that the average maintenance time x for 70 units exceeds 1.1 hours. 3. Based on your answer to the previous problem, did the company budget enough time? Explain. simazoran/iStock/Getty Images Starnes_3e_CH06_398-449_Final.indd 442 TRM Quiz 6B: Lessons 6.4–6.6 You can find a prepared quiz for Lessons 6.4–6.6 by clicking on the link in the TE-book, logging into the Teacher’s Resource site, or accessing this resource on the TRFD. TRM chapter 6 Activity: Sampling Movies (The Sequel) This activity reviews the sampling distribution of x by sampling from a population of movies. Access this resource by clicking on the link in the TE-book, logging into the Teacher’s Resource site, or accessing this resource on the TRFD. 18/08/16 5:04 PMStarnes_3e_CH0 442 C H A P T E R 6 • Sampling Distributions Starnes_3e_ATE_CH06_398-449_v3.indd 442 11/01/17 3:58 PM
L E S S O N 6.6 • The Central Limit Theorem 443 Lesson 6.6 18/08/16 5:04 PMStarnes_3e_CH06_398-449_Final.indd 443 WhAT DiD y o U LeA rn? LEARNINg TARgET EXAMPLES EXERCISES Determine if the sampling distribution of x is approximately normal when sampling from a non-normal population. If appropriate, use a normal distribution to calculate probabilities involving x. Exercises Mastering Concepts and Skills 1. Songs on an iPod David’s iPod has about 10,000 songs. The distribution of the play times for these songs is heavily skewed to the right with a mean of 225 seconds and a standard deviation of 60 seconds. (a) Describe the shape of the sampling distribution of x for SRSs of size n 5 5 from the population of songs on David’s iPod. Justify your answer. (b) Describe the shape of the sampling distribution of x for SRSs of size n 5 100 from the population of songs on David’s iPod. Justify your answer. 2. Insurance claims An insurance company claims that in the entire population of homeowners, the mean annual loss from fire is m 5 $250 with a standard deviation of s 5 $5000. The distribution of losses is strongly right-skewed: Many policies have $0 loss, but a few have large losses. (a) Describe the shape of the sampling distribution of x for SRSs of size n 5 15 from the population of homeowners. Justify your answer. (b) Describe the shape of the sampling distribution of x for SRSs of size n 5 1000 from the population of homeowners. Justify your answer. 3. How many in a car? A study of rush-hour traffic in San Francisco counts the number of people in each car entering a freeway at a suburban interchange. Suppose that the number of people per car in the population of all cars that enter at this interchange during rush hours has a mean of m 5 1.5 and a standard deviation of s 5 0.75. (a) Could the distribution of the number of people per car be normal for the population of all cars entering the interchange during rush hours? Explain. (b) Describe the shape of the sampling distribution of x for SRSs of size n 5 100 from the population of all cars that enter this interchange during rush hours. Justify your answer. pg 440 Lesson 6.6 TRM full Solutions to Lesson 6.6 Exercises You can find the full solutions for this lesson by clicking on the link in the TE-book, logging into the Teacher’s Resource site,or accessing this resource on the TRFD. Answers to Lesson 6.6 Exercises 1. (a) Because n = 5 < 30, the sampling distribution of x will also be skewed to the right but not quite as strongly as the population. (b) Because n = 100 ≥ 30, the sampling distribution of x is approximately normal by the central limit theorem. p. 440 1–4 p. 441 5–8 4. Flawed carpets A supervisor at a carpet factory randomly selects 1-square-yard pieces of carpet and counts the number of flaws in each piece. The number of flaws per square yard in the population of 1-square-yard pieces varies with mean m 5 1.6 and standard deviation s 5 1.2. (a) Could the distribution of the number of flaws be normal for the population of all 1-square-yard pieces of carpet? Explain. (b) Describe the shape of the sampling distribution of x for SRSs of size n 5 60 from the population of all 1-square-yard pieces of carpet. Justify your answer. 5. More songs on an iPod Refer to Exercise 1. What pg 441 is the probability that the mean length in a random sample of 100 songs is less than 4 minutes (240 seconds)? 6. More insurance claims Refer to Exercise 2. Suppose that the insurance company charges $300 for each policy. What is the probability that the insurance company will make money on a random sample of 1000 homeowners? That is, what is the probability that the mean loss for a random sample of homeowners is less than $300? 7. More people in a car Refer to Exercise 3. What is the probability that the mean number of people in a random sample of 100 cars that enter at this interchange during rush hours is at least 1.7? 8. More flawed carpets Refer to Exercise 4. What is the probability that the mean number of flaws in a random sample of sixty 1-square-yard pieces of carpet is at least 1.7? Applying the Concepts 9. Where does lightning strike? The number of lightning strikes on a square kilometer of open ground in a year has a mean of 6 and standard deviation of 2.4. The National Lightning Detection Network (NLDN) uses automatic sensors to watch for lightning in a random sample of fifty 1-square-kilometer 18/08/16 5:04 PM 2. (a) Because n = 15 < 30, the sampling distribution of x will also be skewed to the right but not quite as strongly as the population. (b) Because n = 1000 ≥ 30, the sampling distribution of x is approximately normal by the central limit theorem. 3. (a) No; a count only takes on wholenumber values, so it cannot be normally distributed. (b) Because n = 100 ≥ 30, the sampling distribution of x is approximately normal by the central limit theorem. 4. (a) No; a count only takes on wholenumber values, so it cannot be normally distributed. (b) Because n = 60 ≥ 30, the sampling distribution of x is approximately normal by the central limit theorem. 5. Mean: m x = m = 225 seconds; SD: s x = s !n = 60 !100 = 6 seconds Shape: Because n = 100 ≥ 30, the sampling distribution of x is approximately normal by the central limit theorem. 240 − 225 z = = 2.5; 6 P( x < 240) = P(Z < 2.5) = 0.9938 Using technology: Applet/normalcdf (lower:21000, upper:240, mean:225, SD:6) 5 0.9938 6. Mean: m x = m = $250; SD: s x = s !n = 5000 !1000 = $158.11 Shape: Because n = 1000 ≥ 30, the sampling distribution of x is approximately normal by the central limit theorem. 300 − 250 z = 158.11 = 0.32; P( x < 300) = P( Z < 0.32) = 0.6255 Using technology: Applet/normalcdf (lower:21000, upper:300, mean:250, SD:158.11) 5 0.6241 7. Mean: m x = m = 1.5 people; SD: s x = s !n = 0.75 = 0.075 people !100 Shape: Because n = 100 ≥ 30, the sampling distribution of x is approximately normal by the central limit theorem. 1.7 − 1.5 z = = 2.67; P( x > 1.7) = 0.075 P( Z > 2.67) = 1− 0.9962 = 0.0038 Using technology: Applet/normalcdf (lower:1.7, upper:1000, mean:1.5, SD:0.075) 5 0.0038 8. Mean: m x = m = 1.6 flaws; SD: s x = s !n = 1.2 = 0.155 flaws !60 Shape: Because n = 60 ≥ 30, the sampling distribution of x is approximately normal by the central limit theorem. 1.7 − 1.6 z = = 0.65; P( x > 1.7) = 0.155 P( Z > 0.65) = 1− 0.7422 = 0.2578 Using technology: Applet/normalcdf (lower:1.7, upper:1000, mean:1.6, SD:0.155) 5 0.2594 Lesson 6.6 L E S S O N 6.6 • The Central Limit Theorem 443 Starnes_3e_ATE_CH06_398-449_v3.indd 443 11/01/17 3:58 PM
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Page 34 and 35: Answers continued 16. No; assuming
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SPA 3e_ Teachers Edition _ Ch 6

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