Section 7.2 Homework Answers

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1 Sample Mean P sum n b. The two z-scores are z 25 20(1.7) n sum n and z 30 20(1.7) n , 20 so the probability is approximately ( using Table A). P14. a. The sampling distribution of the sample total should be approximately normal because of the very large sample size. It has mean and standard error sum n 1000(0.9) 900 sum n The z-score for 1000 families is then sum n z n P Sample Total (n 1000) The probability of getting at least 1000 children is about Thus, there is almost no chance the network will get 1000 children. b. Yes, the probability goes from practically 0 to almost certain. The z-score for 1200 families is sum n z n P Sample Total (n 1200) The probability of getting at least 1000 children in a random sample of 1200 families is Note on P15: You may wish to have each student work only one part of this question. Then have the students compare results and notice that as n increases, the range of reasonably likely outcomes for the sample mean decreases. P15. a ( ) , or (0.469, 1.331) b ( ) , or (0.684, 1.116) c ( ) , or (0.832, 0.968) d ( ) , or (0.866, 0.934) Exercises E15. a. I. Histogram B; n 25 II. Histogram A; n 4 III. Histogram C; n 2 b. The theoretical standard error is n, which turns out to be 1.698, 1.201, and for the respective sample sizes of 2, 4, and 25. All of these are fairly close to the standard errors estimated from the simulations. c. For samples of sizes 2 and 4, the simulated sampling distributions of the mean reflect the skewness of the population distribution. For samples of size 25, the skewness is essentially eliminated and the simulated sampling distribution looks like a normal distribution. d. The rule that about 95% of the observations lie within two standard errors of the population mean works well for n 25, and slightly less well for the skewed distributions for n 4 and n 2. E16. a. I. Histogram B; n 4 II. Histogram C; n 25 III. Histogram A; n 2 b. The population standard deviation for this distribution is 3.5. This standard deviation divided by the square root of 2, 4, and 25, respectively, yields 2.47, 1.75, and These values are quite close to the observed standard deviations of the simulated sampling distributions. c. Despite the new peak centered at the mean, the simulated sampling distribution for n 2 still reflects much of the pattern of the population, showing the mounds at the extremes. For n 4, a little of the population pattern remains, but by n 25 it disappears and all that is seen is an essentially normal distribution. d. The rule works well for n 25, but not nearly so well for the smaller sample sizes. As usual, the rule works well for the sampling distribution of the sample mean as long as the sample size is reasonably large.

2 E17. The population is approximately like that in this table. Students estimates will vary. (Many other assignments for random numbers are possible.) Value Percentage Assignment of Random Numbers To get a sample of size 5, divide the random digits into groups of two and use the assignments given in the third column. E18. The population is approximately like that in this table. Students estimates will vary. (Many other assignments for random numbers are possible.) Value Percentage Assignment of Random Numbers To get a sample of size 5, divide the random digits into groups of two and use the assignments given in the third column. E19. a. No. Exactly two accidents happened in about 15% of the days. Two or more accidents happened in about 27% of the days. b. In Display 7.35, the first plot is the one for 8 days, and the second plot is the one for 4 days. c. No, if the days can be viewed as a random sample of all days. An average of 1.75 occurred about 16 times out of 200, so an average of 1.75 accidents is reasonably likely. d. Yes, if the days can be viewed as a random sample of all days. An average of 1.75 or more occurred about 4 times out of 200, so is a rare event. E20. a. e. The sampling distributions used in parts a to c are based on random samples of 4 and 8 days. The means for consecutive days may not look like a random sample at all because of the high dependency from day to day due to seasonal weather or holiday traffic, for example. Relative Frequency Exam Score b. A has n 25, B has n 1, and C has n 5. Remind students that repeated sampling with samples of size 1 should produce a distribution that looks very much like the population. c. If the class can be considered a random sample of the students who took this exam, then an average of 3.6 would be very unusual for a class size of 25. A reasonable conclusion is that the class size is 5. It is always possible, however, that 25 students in a well-taught class would do much better than a random sample of 25 students. Note on E21: Because the scores are normally distributed, students can use the normal approximation with all sample sizes. Point out that a larger sample size makes the denominator smaller, which makes the z-score larger and the probability smaller. E21. a. The z-score is z which gives a probability of Alternatively, on a TI-83 Plus or TI-84 Plus normalcdf (510,1E99,500,100) b. The z-score is x 510 P SAT Score (n 1) z which gives a probability of

3 b. The z-score is c. The z-score is x 510 P Mean Score (n 4) z which gives a probability of x 510 P Mean Score (n 25) d. The probability that one randomly selected score is 510 or greater is about Group the random digits in pairs and assign the digits 01 through 46 to be a score of 510 or greater. The other pairs of digits represent a score less than that. Take four pairs of random digits and see whether all four represent scores of 510 or greater. If so, this run is a success. If any of the pairs represents a score less than 510, this would be a failure. Repeat this process many times. The estimate of the probability is the proportion of runs that are successes. Note on E21d: The probability can be computed exactly using the Multiplication Rule for Independent Events: (0.4602) E22. Because the weights are normally distributed you can use the normal approximation with all sample sizes. a. The z-score is z which gives a probability of Alternatively, on a TI-83 Plus or TI-84 Plus normalcdf( 1E99,.148,.15,.003) P x Weight (n 1) z which gives a probability of P c. The z-score is x Mean Weight (n 4) z which gives a probability of P x Mean Weight (n 10) E23. a. You want the probability that the mean score will be between 490 and 510. For a mean score of ( ) 490, z ( and the probability 40 ) that the mean is less than or equal to 490 is about ( ) For a mean score of 510, z ( ) and the probability that the mean is less than or equal to 510 is about So the probability that the mean score is between 490 and 510 is Sample Mean Score P Or, normalcdf(490,510,500,100 (40) ) Note on E23b: This exercise contains material that will be covered in Chapter 9 of the student book and is optional for now. You can handle this exercise two ways: Students can work on finding the method themselves, or they can be given the formula below. Whichever way you choose, be sure students understand that the smaller the interval, the larger the sample size must be. They should notice that to cut the interval in half, they must quadruple the sample size.

4 b. You know that if the sampling distribution is approximately normal, about 95% of all sample means are in the interval _ x 1.96 n. Thus, to be 95% sure that the sample mean is within a value E of the population mean, you must have E 1.96 n. Solving for the square root gives n 1.96 E. Squaring both sides and simplifying gives n E You need a sample size of about 384. Note on E24: See the note about E23b E24. a. The z-score is z and the 10 probability that the mean increase is greater than 7 is about There is about a 45.1% chance that the mean increase in her stock prices will exceed 7%. Or, normalcdf(7,1e99,6.5,12 10 ) Sample Mean Increase P b. Jenny wants P( X 5) P X n 12.8 n Because the sample mean is approximately normally distributed and the z-score cutting off an area of 0.95 to the right is 1.645, it must be that n Solving this equation gives n Jenny should choose about 197 randomly selected stocks. E25. a. You would expect to see children. b. Since the numbers of children seen on a summer weekend were approximately normally distributed, the sampling distribution of the mean, even with n 2, will be approximately normally distributed. The z-score is z and the probability is almost 0. c. Conclude that the low number of children seen in the emergency room is not due to chance. Note on E25c: The article goes on to say: We observed a significant fall in the numbers of attendees to the emergency department on the weekends that the two most recent Harry Potter books were released. Both these weekends were in mid-summer with good weather. E26. a. 16,597,552 $ $1,712,867,366 b. With a sample of 100 people, the sampling distribution of the mean will be approximately normal with sum 100 $ $10,320 and sum $ $ ,000 10,320 The z-score is z 1, and the probability is about ,320 11,000 Sample Mean Increase P ,320 c. z 1, P(sum 0) is very close to zero. There is virtually no chance that the casino would lose money on a randomly selected group of 100 customers. Note on E27: This exercise is difficult and contains optional material. E27. a. Solve this problem using the sampling distribution of the sample sum, which has a mean of 4.3n and a standard error of 1.4 n. The point that cuts off the lower 0.02 (so 98% is above it) of the standard normal distribution is z If the total number of people is above 100, then choose the sample size, n, so that the point 100 lies standard errors below 4.3n, or n n 4.3n n This equation is quadratic in form and can be solved for n by using the quadratic formula and then squaring the positive solution to get n. The negative solution for n can be ignored. Alternatively, the solution can be estimated by plotting the equation on a graphing calculator. Because you need only the nearest integer solution, either method works well. The solution is about n 27, so the director should select about 27 names. b. By drawing 27 names, the director expects to get people. The 16.1 extra people will cost him 16.1 $250 $4025. This was a pretty costly oversight! Note on E28: This exercise is difficult and contains optional material. E28. From Display 7.38, 2.236, Solve this problem using the sampling distribution of the sample sum, which has a mean of 2.236n and a standard error of n. or

5 The point that cuts off the lower 0.05 (so 95% is above it) of the standard normal distribution is z If the total number of color television sets is above 1000, then choose the sample size, n, so that the point 1000 lies standard errors below 2.235n, or n n 2.236n n This equation is quadratic in form and can be solved for n by using the quadratic formula and then squaring the positive solution to get n. The negative solution for n can be ignored. Alternatively, the solution can be estimated by plotting the equation on a graphing calculator. Because you need only the nearest integer solution, either method works well. The solution is about n 465 households. E29. a. The mean of the sampling distribution of the sample mean is equal to the population mean for all sample sizes. However, as the sample size increases, the standard error of the sampling distribution of the sample mean decreases by a factor of 1 divided by the square root of n. Specifically, x _ and x n. b. The mean of the sampling distribution of the sample total increases by a factor of n. The standard error of the sampling distribution of the sample mean increases by a factor of the square root of n. Specifically, sum n and sum n. E30. a. As long as n is greater than 1, the numerator will be smaller than the denominator, so multiplying by N n N 1 will decrease the standard error. This makes sense when sampling because extreme values of the mean are more difficult to get when you sample without replacement. For example, consider selecting two numbers from the set {1, 2, 3, 4, 5}. With replacement it is possible to draw 1 twice, for a mean of 1. Without replacement the smallest mean you can get is 1.5 from drawing 1 and 2. The distribution becomes less spread out. b. 76. _ 6, or c. Remaining Test Scores Mean 55, 65, 75, , 65, 75, , 65, 75, , 65, 80, , 65, 80, , 65, 90, , 75, 80, , 75, 80, , 75, 90, , 80, 90, , 75, 80, , 75, 80, , 75, 90, , 80, 90, , 80, 90, The sampling distribution is shown here Mean Score d. The mean of the sampling distribution of the sample mean, _ x, is 76. _ 6, the same as. e. Mean Score x x u _ x Sum f. The first formula gives _ x The second formula gives _ x n 4 n N n N 1

6 g. The second formula gives the correct standard error for sampling without replacement. (The small difference is purely due to rounding error.) E31. a. vm v m vm 2 v 2 m b. The sampling distribution of the sum will be approximately normal because both distributions are normal. The z-score is z e. The mean of the distribution of the sum will still be The standard error is unpredictable because the two scores are certainly not independent. Students who score high on the critical reading section also tend to score high on the math section. The shape is also unpredictable. E32. a. mf m f inches mf 2 m 2 f inches b. The distributions of heights of both males and females are approximately normal, so the distribution of the sum will be as well. P sum 800 ( ) z Sum of Scores (n 2) The probability is about ( using Table A). c (161.23), or between approximately 705 and 1337 d. You need c m 100. The sampling distribution of the difference will be approximately normal because both distributions are normal. The mean and standard error of the sampling distribution of the difference are cm c m cm 2 2 c m The z-score for a difference of 100 is 100 (15) z difference 100 P Difference of Scores (n 2) The probability is ( using Table A). Note on E31d: This does not imply that 23.78% of students have an SAT critical reading score of at least 100 points higher than their SAT math score. See part e. P sum Sum of Heights (n 2) P(sum 125) c. The middle 95% of sums will be between (4.01), or approximately between and d. You want P(male height female height 2). The distribution of the difference has mean mf m f inches. mf (2 5.2) z difference 2 P Difference of Heights (n 2) P(difference 2) e. The mean of the distribution of the sum will still be inches. The standard error is unpredictable because the two scores are certainly not independent. Relatives of taller than average people tend to also be taller than average. The shape is also unpredictable. 2 E33. a. sum 3(3.5) 10.5; sum 3(2.917) b. sum 7(3.5) 24.5; sum 7(2.917) c. The sampling distribution of the sum is approximately normal because E31 says that

7 the distribution of critical reading scores is approximately normal. It has mean and variance sum ,060 2 sum n ,380 The z-score for a total of 10,000 is 10,000 10,060 z ,380 The z-score for a total of 750 is z The probability that the total weight is more than 750 kg is sum 750 P P sum 10, Sum of Weights (n 10) 9,050 10,060 11,070 Sum of Scores (n 20) The probability is about ( using Table A). E34. a. Because the distribution of the weights is approximately normal, the sampling distribution of the sum of two (or any number of) weights will be approximately normal as well. The mean and standard error are sum sum b. Using the results from part a, the z-score is and the probability is d. The first ram is more than 5 kg heavier than the second ram if ram 1 ram 2 5. The sampling distribution of the difference ram 1 ram 2 has mean The standard error is the same as that in part a, The z-score for a difference of 5 is z difference 5 P Difference (n 2) P sum 145 The probability that the difference is greater than 5 is about Sum of Weights (n 2) c. Using the formulas in E33, the mean and standard error of the sampling distribution are sum n 10(75.4) 754 sum n 2 10(7.38)

Chapter 7 Discussion Problem Solutions D1 D2. D3.

Chapter 7 Discussion Problem Solutions D1. The agent can increase his sample size to a value greater than 10. The larger the sample size, the smaller the spread of the distribution of means and the more