b. 1.50) = .9332 P(z ≤ < -1.50) = .0668 P(z 421) = 1.50) = .9332 - .0668 = .8664 P(391 P(-1.50 z c. At = 380, 380) = -2.60) = .0047 P( P(z ≤ ≤ Yes, this is an unusually low performing group of 64 stores. The probability of a sample mean annual sales per square foot of $380 or less is only .0047. 45. With = 60 the central limit theorem allows us to conclude the sampling distribution is n approximately normal. a. This means 14 16 At = 16, 1.94) = .9738 P(z ≤ < -1.94) = .0262 P(z 16) = 1.94) = .9738 - .0262 = .9476 P(14 P(-1.94 z b. This means 14.25 15.75 At = 15.75, 1.45) = .9265 P(z ≤ < -1.45) = .0735 P(z 15.75) = 1.45) = .9265 - .0735 = .8530 P(14.25 P(-1.45 z 46. = 27,175 = 7400 a. b. 27,175) = > 0) = .50 P(> P(z Note: This could have been answered easily without any calculations ; 27,175 is the expected value of the sampling distribution of . c. 1.05) = .8531 P(z ≤ < -1.05) = .1469 P(z 28,175) = 1.05) = .8531 - .1469 = .7062 P(26,175 P(-1.05 z d. 1.35) = .9115 P(z ≤ < -1.35) = .0885 P(z
b. P(z ≤ 1.50) = .9332 P(z < -1.50) = .0668 P(391 421) = P(-1.50 z 1.50) = .9332 - .0668 = .8664 c. At = 380, P( ≤ 380) = P(z ≤ -2.60) = .0047 Yes, this is an unusually low performing group of 64 stores. The probability of a sample mean annual sales per square foot of $380 or less is only .0047. 45. With n = 60 the central limit theorem allows us to conclude the sampling distribution is approximately normal. a. This means 14 16 At = 16, P(z ≤ 1.94) = .9738 P(z < -1.94) = .0262 P(14 16) = P(-1.94 z 1.94) = .9738 - .0262 = .9476 b. This means 14.25 15.75 At = 15.75, P(z ≤ 1.45) = .9265 P(z < -1.45) = .0735 P(14.25 15.75) = P(-1.45 z 1.45) = .9265 - .0735 = .8530 46. = 27,175 = 7400 a. b. P(> 27,175) = P(z > 0) = .50 Note: This could have been answered easily without any calculations ; 27,175 is the expected value of the sampling distribution of . c. P(z ≤ 1.05) = .8531 P(z < -1.05) = .1469 P(26,175 28,175) = P(-1.05 z 1.05) = .8531 - .1469 = .7062 d. P(z ≤ 1.35) = .9115 P(z < -1.35) = .0885
CH7.docx