MULTIPLE CHOICE QUESTIONS

CHAPTER 9 SAMPLING DISTRIBUTIONS MULTIPLE CHOICE QUESTIONS In the following multiple choice que s tions, plea s e circle the correc t answ e r. 1. As a gen er al rule, the nor m al distribution is use d sa m pling distribution of the sa m pl e proportion only if a. the sa m pl e size n is grea t e r than 30 b. the population proportion p is close to 0.50 c. the und erlying population is nor m al d. np and n (1 p ) are both gre a t e r than 5 d 2. Rando m sam pl e s of size 49 are take n from an infinite popula tion whos e me a n is 300 and stand a r d deviation is 21. The me a n and sta n d a r d error of the sa m pl e me a n, resp e c tiv ely, are: a. 300 and 21 b. 300 and 3 c. 70 and 230 d. 49 and 21 b 3. A nor m ally distribut e d popula tion with 200 ele m e n t s has a me a n of 60 and a stand a r d deviation of 10. The prob a bility that the me a n of a sa m pl e of 25 elem e n t s take n from this popula tion will be sm aller tha n 56 is a. 0.0166 b. 0.0228 c. 0.3708 d. 0.0394 a 4. Given an infinite population with a me a n of 75 and a stan d a r d deviation of 12, the prob a bility that the me a n of a sa m pl e of 36 obs erv a tion s, take n at rando m from this population, exce e d s 78 is a. 0.4332 b. 0.0668 c. 0.0987 d. 0.9013 b 5. A population that consist s of 500 obs erv a tion s has a me a n of 40 and a stand a r d deviation of 15. A sa m pl e of size 100 is take n at rando m from this population. The stan d a r d error of the sa m pl e me a n equ als: a. 2.50 139 to approxi m a t e the

140 Chapter Nine b. 12.50 c. 1.343 d. 1.50 c 6. An infinite popul ation has a me a n of 60 and a sta n d a r d deviation of 8. A sa m pl e of 50 obs erv a tion s will be take n at rando m from this popul ation. The prob a bility that the sam pl e me a n will be betw e e n 57 and 62 is a. 0.9576 b. 0.9960 c. 0.2467 d. 0.3520 a 7. If all possible sa m pl e s of size n are drawn from an infinite popula tion with a me a n of 15 and a stan d a r d deviation of 5, then the sta nd a r d error of the sa m pl e me a n equ als 1.0 only for sa m pl e s of size a. 5 b. 15 c. 25 d. 75 c 8. If the stand a r d error of the sa m pling distribution of the sa m pl e proportion is 0.0229 for sa m pl e s of size 400, then the popula tion proportion must be either a. 0.4 or 0.6 b. 0.5 or 0.5 c. 0.2 or 0.8 d. 0.3 or 0.7 d 9. As a gen er al rule in com p u ti ng the sta nd a r d error of the sa m pl e me a n, the finite population correction factor is use d only if the : a. sa m pl e size is sm aller tha n 10% of the popula tion size b. population size is sm aller tha n 10% of the sa m pl e size c. sa m pl e size is great e r tha n 1% of the popula tion size d. population size is great e r than 1% of the sa m pl e size c 10. Given that X is a bino mi al rando m variable, the binomi al prob a bility P(X x ) is approxi m a t e d by the area und er a nor m al curve to the right of a. x 0.5 b. x + 0. 5 c. x 1 d. x + 1

Sampling Distributions 141 a 11. Consider an infinite popula tion with a me a n of 160 and a sta n d a r d deviation of 25. A rando m sam pl e of size 64 is take n from this popula tion. The stand a r d deviation of the sa m pl e me a n equ als: a. 12.64 9 b. 25.0 c. 2.56 d. 3.125 d 12. A sam pl e of size 200 will be take n at rando m from an infinite popula tion. Given that the population proportion is 0.60, the proba bility that the sa m pl e proportion will be grea t e r tha n 0.58 is a. 0.281 b. 0.719 c. 0.580 d. 0.762 b 13. A sam pl e of size 40 will be take n from an infinite popula tion whos e me a n and stand a r d deviation are 68 and 12, resp e c tively. The prob a bility that the sa m pl e me a n will be larger tha n 70 is a. 0.3970 b. 0.4332 c. 0.1469 d. 0.0668 c 14. A sa m pl e of size n is select e d at rando m from an infinite popul a tion. As n incre a s e s, which of the following stat e m e n t s is true? a. The population stan d a r d deviation decr e a s e s b. The stand a r d error of the sa m pl e me a n decr e a s e s c. The population stan d a r d deviation incre a s e s d. The stand a r d error of the sa m pl e me a n incre a s e s b 15. The finite population correc tion factor should not be use d when: a. we are sam pling from an infinite popula tion b. we are sam pling from a finite popul ation c. sa m pl e size is great e r tha n 1% of the popula tion size d. None of the above stat e m e n t s is correct a

142 Chapter Nine 16. If the stand a r d error of the sa m pling distribution of the sa m pl e proportion is 0.0337 for sa m pl e s of size 200, then the popula tion proportion must be either: a. 0.25 b. 0.75 c. 0.20 or 0.80 d. 0.35 or 0.65 e. 0.30 or 0.70 c 17. Rando m sam pl e s of size 81 are take n from an infinite popula tion whos e me a n and stan d a r d deviation are 45 and 9, resp e c tively. The me a n and stand a r d error of the sam pling distribution of the sa m pl e me a n are: a. 9 and 45 b. 45 and 9 c. 81 and 45 d. 45 and 1 d 18. A sam pl e of 250 obs erv a tion s will be select e d at rando m from an infinite population. Given that the popula tion proportion is.25, the stan d a r d error of the sa m pling distribution of the sa m pl e proportion is : a. 0.0274 b. 0.50 c. 0.0316 d. 0.0548 a 19. A sam pl e of size 25 is selec t e d at rando m from a finite popul ation. If the finite population correction factor is 0.632 5, then the popula tion size is: a. 10 b. 41 c. 15 d. 35 b 20. If two popul ations are norm ally distribut e d, the sa m pling distribution of the sa m pl e me a n differenc e X 1 X 2 will be: a. approxi m a t el y norm ally distribut e d b. nor m ally distribut e d only if both sa m pl e sizes are gre a t e r tha n 30 c. nor m ally distribut e d d. nor m ally distribut e d only if both popul ation sizes are gre a t e r tha n 30 c

Sampling Distributions 143 21. Two sa m pl e s are select e d at rando m distribut e d popul ations. Sam pl e 1 has 49 and a stan d a r d deviation of 5. Sa m pl e me a n of 12 and a stan d a r d deviation sa m pling distribution of the sa m pl e me a n a. 0.1853 b. 0.7602 c. 0.7331 d. 0.8719 d from two indep e n d e n t nor m ally obs e rv a tion s and has a me a n of 10 2 has 36 obs erv a tion s and has a of 3. The stan d a r d error of the differe nc e X 1 X 2 is 22. Given a bino mi al distribution with n trials and proba bility p of a succ e s s on any trial, a conven tion al rule of thu m b is that the nor m al distribution will provide an adeq u a t e approxi m a ti on of the binomi al distribution if a. np 5 and n (1 p ) 5 b. np 5 and n (1 p ) 5 c. np 5 and n (1 p ) 5 d. np 5 and n (1 p ) 5 a 23. If two rando m sa m pl e s of sizes n1 and n 2 are selec t e d inde p e n d e n tl y from two population s with me a n s µ1 and µ2, then the me a n of the sa m pling distribution of the sam pl e me a n differe nc e, X 1 X 2, equ als: a. µ1 + µ2 b. µ1 µ2 c. µ1 / µ2 d. µ1 µ2 b 24. If two rando m sa m pl e s of sizes n1 and n 2 are selec t e d inde p e n d e n tl y from two popul ations with varianc e s σ 12 and σ 22, the n the stan d a r d error of the sa m pling distribution of the sa m pl e me a n differe nc e, X 1 X 2, equ als: a. (σ 12 σ 22 ) / n1 n 2 b. (σ 12 + σ 22 ) / n1 n2 c. σ 12 σ 22 n1 n2 d. σ 12 σ 22 + n1 n2 25. d Suppo s e that the proba bility p of a succ e s s on any trail of a binomi al distribution equ als 0.90. Then for which of the following num b e r of trials, n,

144 Chapter Nine would the norm al distribution provide a good approxi m a tio n to the bino mi al distribution? a. 25 b. 35 c. 45 d. 55 d 26. If two rando m sa m pl e s of sizes n1 and n 2 are selec t e d inde p e n d e n tl y from two non norm ally distribut e d popula tions, then the sa m pling distribution of the sa m pl e me a n differenc e, X 1 X 2, is a. always non nor m al b. always nor m al c. approxi m a t el y norm al only if n1 and n 2 are both larger tha n 30 d. approxi m a t el y norm al regar dl e s s of n1 and n 2 c 27. Given that X is a binomi al rando m variabl e, the bino mi al prob a bility P(X= x ) is approxi m a t e d by the are a und er a norm al curve betw e e n a. x 0.5 and 0.0 b. 0.0 and x +0. 5 c. 1 x and 1 + x d. x 0.5 and x +0. 5 d 28. The Centr al Limit Theor e m stat e s that, if a rando m sa m pl e of size n is drawn from a popul ation, then the sa m pling distribution of the sa m pl e me a n X : a. is approxi m a t el y nor m al if n > 30 b. is approxi m a t el y nor m al if n < 30 c. is approxi m a t el y nor m al if the und erlying popula tion is nor m al d. has the sam e varianc e as the popula tion a 29. The expect e d value of the sa m pling distribution of the sa m pl e me a n equ als the popul ation me a n µ : a. when the popul ation is nor m ally distribut e d b. when the popul ation is sym m e t ric c. when the popul ation size N > 30 d. for all population s d X

Sampling Distributions 145 30. If all possible sa m pl e s of size n are drawn from an infinite popula tion with a me a n of µ and a stan d a r d deviation of σ, then the sta n d a r d error of the sa m pl e me a n is invers ely propor tion al to: a. µ b. σ c. n d. n d 31. Given that X is a bino mi al rando m variable, the binomi al prob a bility P(X x ) is approxi m a t e d by the area und er a nor m al curve to the left of a. x b. x c. x + 0. 5 d. x 0.5 c 32. The stand a r d deviation of the sa m pling distribution of the sa m pl e me a n is also called the: a. centr al limit theor e m b. stand a r d error of the me a n c. finite population correction factor d. population stan d a r d deviation b 33. If a rando m sa m pl e of size n is drawn from a nor m al popula tion, then the sa m pling distribution of the sa m pl e me a n X will be: a. nor m al for all value s of n b. nor m al only for n > 30 c. approxi m a t el y norm al for all value s of n d. approxi m a t el y norm al only for n > 30 a 34. If all possible sam pl e s of size n are drawn from a popula tion, the proba bility distribution of the sam pl e me a n X is called the: a. stand a r d error of X b. expect e d value of X c. sa m pling distribution of X d. nor m al distribution c TRUE/FALSE QUESTIONS

146 Chapter Nine 35. In an effort 18 year s of the m wer e esti m a t e to to identify the true proportion of college fresh m a n who are unde r age, a rando m sa m pl e of 500 fresh m a n was take n. Only fifty of und er the age of 18. The value 0.10 would be use d as a point the true propor tion of unde r age 18 fresh m a n. T 36. The centr al limit theor e m is basic to the conc e p t of statistic al infere nc e, beca u s e it per mit s us to draw conclusion s about the popula tion bas e d strictly on sa m pl e dat a, and without having any knowle d g e about the distribution of the und erlying population. T 37. When a grea t many simple rando m sa m pl e s of size n are drawn from a population that is nor m ally distribut e d, the sa m pling distribution of the sa m pl e me a n s will be nor m al rega r dl e s s of sa m pl e size n. T 38. The me a n of the sam pling distribution of the sa m pl e proportion p, when the sa m pl e size n = 100 and the popula tion proportion p = 0.92, is 92.0. F 39. The stand a r d error of the sa m pling distribution of the sa m pl e proportion p, when the sam pl e size n = 100 and the popula tion proportion p = 0.30, is 0.0021. F 40. Recall the rule of thu m b use d to indicat e when the nor m al distribution is a good approxi m a tio n of the sa m pling distribution for the sa m pl e proportion p. For the com bin a tion n = 50; p = 0.05, the rule is satisfied. F 41. The stand a r d error of the me a n is the sta n d a r d deviation of the sa m pling distribution of X. T 42. The stand a r d deviation of the sa m pling distribution of the sa m pl e me a n is also called the centr al limit theor e m. F 43. Consider an infinite popula tion with a me a n of 100 and a sta n d a r d deviation of 20. A rando m sam pl e of size 64 is take n from this popula tion. The stand a r d deviation of the sa m pl e me a n equ als 2.5. T 44. If all possible sa m pl e s of size n are drawn from an infinite popula tion with a me a n of 60 and a stan d a r d deviation of 8, then the sta nd a r d error of the sa m pl e me a n equ als 1.0 only for sa m pl e s of size 64. T

Sampling Distributions 147 45. If all possible sam pl e s of size n are drawn from a popula tion, the proba bility distribution of the sam pl e me a n X is referr e d to as the nor m al distribution. F 46. As a gen er al rule, the nor m al distribution is use d to approxi m a t e the sa m pling distribution of the sa m pl e proportion only if the sa m pl e size n is great e r than or equ al to 30. F 47. A sam pl e of size n is select e d at rando m from an infinite popula tion. incre a s e s, the stan d a r d error of the sa m pl e me a n decr e a s e s. T 48. If the stand a r d error of the sa m pling distribution of the sa m pl e proportion is 0.0245 for sa m pl e s of size 400, then the popula tion propor tion must be 0.40. F 49. A sam pl e of size 25 is selec t e d at rando m from a finite popul ation. If the finite population correction factor is 0.822, the popula tion size mus t be 75. T 50. If a simple rando m sa m pl e of 300 obs erv a tion s is take n from a popul ation whos e propor tion p = 0.6, then the expe c t e d value of the sa m pl e proportion p is 0.40. F 51. The nor m al approxi m a ti on to the binomi al distribution works best whe n the num b e r of trials is large, and whe n the binomi al distribution is sym m e t ric al (like the nor m al). T 52. If two rando m sa m pl e s of size 36 eac h are selec t e d inde p e n d e n tl y from two population s with varianc e s 42 and 50, then the sta n d a r d error of the sa m pling distribution of the sa m pl e me a n differe nc e, X 1 X 2, equ als 2.555 6. F 53. If two rando m sam pl e s of sizes 30 and 32 are select e d indep e n d e n t ly from two population s with me a n s 109 and 121, the n the me a n of the sa m pling distribution of the sam pl e me a n differe nc e, X 1 X 2, equ als 12. T 54. As a gen er al rule, the nor m al distribution is use d to approxi m a t e the sa m pling distribution of the sa m pl e proportion only if the sa m pl e size n is great e r than 30. F TEST QUESTIONS As n

148 55. Chapter Nine A res e a r c h e r conduct e d a survey on a univer sity ca m p u s for a sa m pl e of 64 seniors and report e d that seniors read an aver a g e of 3.12 books in the prior acad e m i c sem e s t e r, with a stan d a r d deviation of 2.15 books. Deter mi n e the prob a bility that the sam pl e me a n is: a. above 3.45 b. betw e e n 3.38 and 3.58 c. below 2.94 a. 0.1093 b. 0.1224 c. 0.2514 56. An infinite popul ation has a me a n of 150 and a sta nd a r d deviation of 40. A sa m pl e of 100 obs erv a tion s will be select e d at rando m from the popula tion. a. What is the expect e d value of the sa m pl e me a n? b. What is the stand a r d deviation of the sa m pl e me a n? c. What is the shap e of the sa m pling distribution of the sa m pl e me a n d. What does the sam pling distribution of the sa m pl e me a n show? a. µx =1 5 0 b. σx = 4 c. Approxim a t ely nor m al with a me a n of 150 and a sta n d a r d deviation of 4. d. It shows the prob a bility distribution of all possible sa m pl e me a n s that can be obs erv e d with rando m sa m pl e s of size 100. This distribution can be used to calculat e the me a n and the sta n d a r d deviation of the sa m pl e me a n. It can also be use d to com p u t e the proba bility that the sa m pl e me a n is within a specified rang e from the popul ation me a n. 57. If the weekly dem a n d for cas e s of soda at a store is norm ally distribut e d with a me a n of 47.6 cas e s and a sta nd a r d deviation of 5.8 cas e s, what is the prob a bility that the aver a g e de m a n d for a sa m pl e of 10 stor e s will exc e e d 50 cas e s in a given week? 0.0952 58. The prob a bility of a succe s s on any trial of a bino mi al expe ri m e n t is 20%. Find the proba bility that the proportion of succ e s s in a sa m pl e of 400 is less than 18%. 0.1587

Sampling Distributions 59. Suppo s e that the time ne e d e d distribut e d with a me a n of 85 minut e s.. a. What is the proba bility that stud e n t s will not exce e d 8,200 b. What ass u m p tio n did you have 149 to com ple t e a final exa m is nor m ally minut e s and a stan d a r d deviation of 18 the total tim e take n by a group of 100 minut e s? to mak e in your com p u t a ti on s in part (a)? a. 0.0475 b. The stud e n t s tim es nee d e d to com pl e t e the exa m are indep e n d e n t of one anot h e r. 60. Height s of 10 year old childre n are nor m ally distribut e d with a me a n of 52 inche s and a stand a r d deviation of 4 inche s. a. Find the proba bility that one rando mly selec t e d 10 year old child is taller than 54 inche s. b. Find the proba bility that two rando mly select e d 10 year old childre n are both taller than 54 inche s. c. Find the proba bility that the me a n height of two rando mly select e d 10 year old children is grea t e r tha n 54 inche s. a. 0.3085 b. 0.0952 c. 0.2389 61. Find the sa m pling distribution of the sa m pl e me a n drawn from the following popula tion: x p (x ) 2 0.2 0 0.6 X if sa m pl e s of size 2 are 2 0.2 x 2 p ( x ) 0.04 62. 1 0.24 0 0.44 1 0.24 2 0.04 Let X be a binomi al rando m variabl e with n = 25 and p = 0.6. Approxim a t e the following proba bilities, using the nor m al distribution. a. P(X 20) b. P(X 15) c. P(X = 10) a. 0.0329 b. 0.5793 c. 0.0207

150 63. Chapter Nine An infinite popul ation has a me a n of 100 and a sta nd a r d deviation of 20. Suppo s e that the popul ation is not nor m ally distribut e d. What does the centr al limit theor e m say about the sa m pling distribution of the me a n if sa m pl e s of size 64 are drawn at rando m from this popul ation? The sam pling distribution of X is approxi m a t e ly nor m al with a me a n of 100 and a stan d a r d deviation of 2.5 64. Suppo s e that the aver a g e annu al incom e of a defen s e attorn e y is $150, 0 0 0 with a stan d a r d deviation of $40,00 0. Assum e that the incom e distribution is nor m al. a. What is the proba bility that the aver a g e annu al incom e of a sa m pl e of 5 defens e attorn e y s is mor e tha n $120, 0 0 0? b. What is the prob a bility that the aver a g e annu al incom e of a sa m pl e of 15 defens e attorn e y s is mor e tha n $120, 0 0 0? a. 0.9535 b. 0.9981 65. Assum e the tim e need e d by a worker to perfor m a maint e n a n c e oper a tion is nor m ally distribut e d with a me a n of 70 minut e s and a stan d a r d deviation of 6 minut e s. What is the prob a bility that the aver a g e tim e ne e d e d by a sa m pl e of 5 worker s to perfor m the maint e n a n c e in betw e e n 63 minut e s and 68 minut e s? 0.221 66. In order to estim a t e the me a n salary for a popula tion of 500 em ploy e e s, the preside n t of a cert ain com p a n y select e d at rando m a sa m pl e of 40 em ploy e e s. a. Would you use the finite popul ation correction factor in calcula ting the stand a r d error of the sa m pl e me a n? Explain. b. If the population stan d a r d deviation is $800, com p u t e the stan d a r d error both with and without using the finite popula tion correc tion factor. c. What is the proba bility that the sa m pl e me a n salary of the em ploye e s will be within ± $200 of the popula tion me a n salary: a. n/n = 0.08 > 0.01; ther efor e, the finite popul a tion correc tion factor is nece s s a r y. b. σx =1 2 1. 4 4 8 and 126.4 9 1 with and without the finite popula tion correction factor, resp e c tively. c. 0.901

Sampling Distributions 67. 151 A sa m pl e of 50 obs erv a tion s is drawn at rando m from a nor m al popula tion whos e me a n and stan d a r d deviation are 75 and 6, resp e c tiv ely. a. What does the centr al limit theor e m say about the sa m pling distribution of the sam pl e me a n? Why? b. Find the me a n and stan d a r d error of the sa m pling distribution of the sa m pl e me a n. c. Find P( X > 7 3) d. Find P( X < 7 4) a. X is norm al beca u s e the par e n t popul ation is norm al. b. µx =7 5 and σx =.8485 c. 0.9909 d. 0.1190 68. Suppo s e it is known that 60% of stud e n t s at a particular college are smok e r s. A sa m pl e of 500 stud e n t s from the college is select e d at rando m. Approxim a t e the prob a bility that at least 280 of the s e stud e n t s are smok e r s. 0.9664 69. Al Gore, the form er Vice Preside n t of the USA, believe s that the proportion of voter s who will vote for a de m o c r a t candida t e in the year 2004 preside n ti al elections is 0.65. A sam pl e of 500 voters is select e d at rando m. a. Assum e that Gore is correct and p = 0.65. What is the sa m pling p distribution of the sam pl e proportion? Explain. b. Find the expect e d value and the sta nd a r d deviation of the sa m pl e proportion p. c. What is the proba bility that the num b e r of voters in the sa m pl e who will vote for a dem ocr a t pre side n ti al candida t e in the year 2004 will be betw e e n 340 and 350? a. Approxim a t ely nor m al, since np = 25 and n( 1 p )= 175 are both gre a t e r than 5. b. E( p ) = 0.65, and σ p = 0.0213 c. 0.0699 70. Find the sa m pling distribution of the sa m pl e me a n drawn from the following popula tion: x p (x ) 3 0.6 5 0.4 X if sa m pl e s of size 2 are

152 Chapter Nine x p( x ) 71 3 0.36 4 0.48 5 0.16 Suppo s e that the starting salarie s of male mat h profes s or s are nor m ally distribut e d with a me a n of $56,0 0 0 and a sta nd a r d deviation of $12,00 0. The starting salaries of fem al e ma t h profe s s or s are nor m ally distribut e d with a me a n of $50,00 0 and a stan d a r d deviation of $10,0 0 0. A rando m sa m pl e of 50 male mat h profes s or s and a rando m sa m pl e of 40 fem al e ma t h profes s or s are select e d. a. What is the sa m pling distribution of the sa m pl e me a n differe nc e X 1 X 2? Explain. b. Find the expect e d value and the sta nd a r d error of the sa m pl e me a n differenc e. c. What is the proba bility that the sa m pl e me a n salary of fem al e mat h profes s or s will not exce e d that of the male ma t h profes s or s? a. X 1 X 2 is nor m ally distribut e d, since the pare n t popul a tion s are nor m ally distribut e d. b. E( X 1 X 2 ) = 6,000, and σ xi x2 = 2319. 4 8 3 c. 0.9952 72. Let X be a binomi al rando m variabl e with n = 100 and p = 0.7. Approxim a t e the following proba bilities, using the nor m al distribution. a. P(X = 7 5) b. P(X 70) c. P(X > 60) a. 0.0484 b. 0.5438 c. 0.9808 73. A sam pl e of size 400 is drawn from a popul ation whos e me a n and varianc e are 5,000 and 10,00 0, resp e c tiv ely. Find the following prob a bilities: a. P( X < 4,990) b. P(4,995 < X < 5,010) c. P( X = 5,000) a. 0.0228 b. 0.8185 c. 0.0 74. A fair coin is toss e d 500 time s. Approxim a t e the prob a bility that the num b e r of hea d s obs erv e d is betw e e n 240 and 270 (inclusive).

Sampling Distributions 153 0.7928 75. Historical dat a collect e d at First of America bank in Michiga n reve al e d that 80% of all custo m e r s applying for a loan are acc e p t e d. Suppos e that 50 new loan application s are select e d at rando m. a. Find the expect e d value and the sta nd a r d deviation of the num b e r of loans that will be acce p t e d by the bank. b. What is the proba bility that at least 42 loans will be acc e p t e d? c. What is the prob a bility that the num b e r of loans reject e d is betw e e n 10 and 15, inclusive? a. Let X be the num b e r of loans out of 50 that are acc e p t e d. Then X is a binomi al rando m variabl e with n = 50, and p = 0.80. Therefor e, E(X)=40, and σ = 2.828. b. 0.2981 c. 0.5452 76. A sam pl e of 25 obs erv a tion s is drawn from a nor m al popula tion with me a n of 900 and a stan d a r d deviation of 300. Suppos e the popula tion size is 600. a. Find the expect e d value of the sa m pl e me a n X. b. Find the stand a r d error of the sa m pl e me a n X. c. Find P( X > 1000) d. Find P( X < 960) e. Find P(980 < X < 1050) a. µ x = µ = 900 σ n c. 0.0436 d. 0.8461 e. 0.0815 b. σ x = 77. N n = 58.785 7 N 1 An auditor knows from past history that the aver a g e accoun t s receiva bl e for a com p a n y is $521. 7 2 with a stan d a r d deviation of $584. 6 4. If the auditor take s a simple rando m sa m pl e of 100 accou nt s, what is the proba bility that the me a n of the sam pl e will be within $120 of the popula tion me a n? 0.9596 78. In a given year, the aver a g e annu al salary of a NFL footb all player was $205, 0 0 0 with a stan d a r d deviation of $24,5 0 0. If a simple rando m sa m pl e of 50 player s was taken, what is the proba bility that the sa m pl e me a n will exce e d $210, 0 0 0?

154 Chapter Nine 0.0749 QUESTIONS 79 THROUGH 82 ARE BASED ON THE FOLLOWING INFORMATION: The height s of men in the USA are norm ally distribut e d with a me a n of 68 inche s and a stan d a r d deviation of 4 inche s. 79. What is the proba bility that a rando mly select e d ma n is taller tha n 70 inche s? 0.3085 80. A rando m sam pl e of five me n is selec t e d. sa m pl e me a n is grea t e r tha n 70 inche s? What is the prob a bility that the 0.1314 81. What is the proba bility that the me a n height of a rando m sa m pl e of 36 me n is grea t e r than 70 inche s? 0.0013 82. If the population of men s height s is not nor m ally distribut e d, which, if any, of the ques tion s 59 throug h 61 can you answ e r? We can answ er Questions 80 and 81. We cannot answ e r Ques tion 79. QUESTIONS 83 THROUGH 85 ARE BASED ON THE FOLLOWING INFORMATION: The am ou n t of tim e spen t by America n adults playing sport s per day is nor m ally distribut e d with a me a n of 4 hours and sta n d a r d deviation of 1.25 hours. 83. Find the proba bility that a rando mly select e d America n adult plays sport s for mor e than 5 hours per day 0.2119 84. Find the proba bility that if four America n adults are rando mly selec t e d, their aver a g e num b e r of hours spe nt playing sports is mor e than 5 hours per day. 0.0548

Sampling Distributions 85. 155 Find the proba bility that if four America n adults are rando mly selec t e d, all four play sport s for mor e than 5 hours per day. 0.0020 QUESTIONS 86 THROUGH 90 ARE BASED ON THE FOLLOWING INFORMATION: The following dat a give the num b e r of pet s owne d for a popula tion of 4 families Family Numb er of Pets Owned 86. B 1 C 4 D 3 Find the me a n and the stan d a r d deviation for the popula tion. µ = 2. 5 and 87. A 2 σ = 1. 1 1 8 Sam pl e s of size 2 will be drawn at rando m from the popula tion. Use the answ er s to Question 86 to calculat e the me a n and the sta n d a r d deviation of the sa m pling distribution of the sa m pl e me a n s. µ x = µ = 2.5 and σ x = 88. σ n N n = 0.645 5 N 1 List all possible sam pl e s of 2 families that can be selec t e d without replac e m e n t from this popula tion, and com p u t e the sa m pl e me a n X for each sa m pl e. Sam pl e x 89. AB 1.5 AC 3.0 AD 2.5 BC 2.5 BD 2.0 Find the sa m pling distribution of X. x p( x ) 1.5 1/6 2.0 1/6 2.5 1/6 3.0 1/6 3.5 1/6 CD 3.5

156 90. Chapter Nine Use the sam pling distribution in Ques tion 89 and directly recalcula t e the me a n and stan d a r d deviation of X. Comp a r e the answ e r s to that of Ques tion 87. What do you conclud e? µx = x p ( x ) = 2.50, and σx = x 2 p ( x ) ( µx ) 2 = 0.4167 = 0.6455 The sa m e answ er s. QUESTIONS 91 AND 92 ARE BASED ON THE FOLLOWING INFORMATION: Two indep e n d e n t rando m sam pl e s of 25 obs erv a tion s eac h are drawn from two nor m al population s. The par a m e t e r s of the s e popula tion s are: Population 1: Population 2: 91. µ =1 5 0 µ =1 3 0 σ σ =20 =16 Find the proba bility that the me a n of sa m pl e 1 will exc e e d the me a n of sa m pl e 2. 0.7823 92. Find the proba bility that the me a n of sa m pl e 1 is gre a t e r than the me a n of sa m pl e 2 by mor e than 15. 0.0158 QUESTIONS 93 THROUGH 95 ARE BASED ON THE FOLLOWING INFORMATION: A videoc a s s e t t e rent al stor e want s to know what proportion of its custo m e r s are und er age 21. A simple rando m sa m pl e of 500 custo m e r s was take n, and 375 of the m were und er age 21. Pres u m e that the true popul a tion propor tion of custo m e r s und er age 21 is 0.68. 93. Describe the sam pling distribution of propor tion of custo m e r s who are unde r age 21. Since np 5, and n (1 p ) 5, the sa m pling distribution of p is approxi m a t e ly nor m al. 94. Find the me a n and stan d a r d deviation of p

Sampling Distributions 157 µ p = p = 0.68, and σ pˆ = p (1 p) / n = 0.0208 6 95. What is the proba bility that the sa m pl e proportion p will be within 0.03 of the true proportion of custo m e r s who are und er age 21? 0.8502 96. Suppo s e that the starting salarie s of financ e gradu a t e s from university A are nor m ally distribut e d with a me a n of $36, 75 0 and a stan d a r d deviation of $5,32 0. The starting salarie s of financ e gra du a t e s from university B are nor m ally distribut e d with a me a n of $34, 62 5 and a stan d a r d deviation of $6,54 0. If simple rando m sa m pl e s of 50 financ e grad u a t e s are select e d from each university, what is the proba bility that the sa m pl e me a n of university A gradu a t e s will exce e d that of univer sity B gradu a t e s? P ( X 1 X 2 > 0) = P ( Z > 1.78) = 0.962 5 97. The publisher of a daily news p a p e r claim s that 90% of its subscrib e r s are und er the age of 30. Suppos e that a sa m pl e of 300 subs cribe r s is selec t e d at rando m. Assuming the claim is correc t, approxi m a t e the proba bility of finding at least 240 subs criber s in the sa m pl e unde r the age of 30. 0.9783 98. A local news p a p e r sells an aver a g e of 2100 pap e r s per day, with a sta n d a r d deviation of 500 pap er s. Consider a sa m pl e of 60 days of oper a tion. a. What is the shap e of the sa m pling distribution of the sa m pl e me a n num b e r of pap er s sold per day? Why? b. Find the expect e d value and the sta nd a r d error of the sa m pl e me a n. c. What is the prob a bility that the sa m pl e me a n will be betw e e n 2000 and 2300 pap er s.? a. Approxim a t ely nor m al since n > 3 0. b. µx = 2100 and σx = 77.46 c. 0.8966 99. Given a bino mi al rando m variable with n = 1 5 and p =. 4 0, find the exact prob a bilities of the following eve nt s and their norm al approxi m a tio n s. a. X = 6 b. X 9 c. X 10 a. Exact and approxi m a t e d proba bilities are 0.207 and 0.2052, resp e c tiv ely. b. Exact and approxi m a t e d proba bilities are 0.095 and 0.0934, resp e c tiv ely.

158 100. Chapter Nine c. Exact and approxi m a t e d proba bilities are 0.991 and 0.9911, resp e c tiv ely. A simple rando m sam pl e of 300 obs erv a tion s is take n from a popula tion. Assum e that the popul ation proportion p = 0.6. a. What is the expect e d value of the sa m pl e proportion p? b. What is the stand a r d error of the sa m pl e proportion p? c. What is the proba bility that the sa m pl e proportion p will be within ± 0.02 of the popul ation pro portion p? a. 0.6000 b. 0.0283 c. 0.5222 101. The chair m a n of a statistics dep a r t m e n t in a cert ain college believe s that 70% of the dep ar t m e n t s grad u a t e assist a n t s hi p s are given to inter n a tion al stud e n t s. A rando m sa m pl e of 50 gra du a t e assist a n t s is take n. a. Assum e that the chair m a n is correc t and p = 0.70. What is the sa m pling distribution of the sam pl e proportion p? Explain. b. Find the expect e d value and the sta nd a r d error of the sa m pling distribution of p. c. What is the prob a bility that the sa m pl e proportion p will be betw e e n 0.65 and 0.73? d. What is the prob a bility that the sa m pl e proportion p will be within ±.05 of the popul ation proportion p? a. Approxim a t ely nor m al, since np =3 5 and n( 1 p )=1 5 are both gre a t e r tha n 5. b. E( p ) = 0.70, and σ p = 0.064 8 c. 0.4566 d. 0.5588