41. a. = .17 E() Distribution is approximately normal because = 800(.17) = 136 > 5 np and = 800(.83) = 664 > 5 n(1 p) – b. 1.51) = .9345 P(z ≤ < -1.51) = .0655 P(z .19) = 1.51) = .9345 - .0655 = .8690 P(.15 P(-1.51 z c. 2.13) = .9834 P(z ≤ < -2.13) = .0166 P(z .19) = 2.13) = .9834 - .0166 = .9668 P(.15 P(-2.13 z 42. The random numbers corresponding to the first seven universities selected are 122, 99, 25, 55, 115, 102, 61 The third, fourth and fifth columns of Table 7.1 were needed to find 7 random numbers of 133 or less without duplicate numbers. Author’s note: The universities identified are: Clarkson U. (122), U. of Arizona (99), UCLA (25), U. of Maryland (55), U. of New Hampshire (115), Florida State U. (102), Clemson U. (61). 43. a. With = 100, we can approximate the sampling distribution with a normal distribution having n = 8086 E() b. .80) = .7881 P(z ≤ < -.80) = .2119 P(z 8286) = .80) = .7881 - .2119 = .5762 P(7886 P(-.80 z The probability that the sample mean will be within $200 of the population mean is .5762. c. At 9000, 9000) = 3.66) 0 P( P(z ≥ ≥ Yes, the research firm should be questioned. A sample mean this large is extremely unlikely (almost 0 probability) if a simple random sample is taken from a population with a mean of $8086. 44. a. Normal distribution with 406 E=
41. a. E() = .17 Distribution is approximately normal because np = 800(.17) = 136 > 5 and n(1 – p) = 800(.83) = 664 > 5 b. P(z ≤ 1.51) = .9345 P(z < -1.51) = .0655 P(.15 .19) = P(-1.51 z 1.51) = .9345 - .0655 = .8690 c. P(z ≤ 2.13) = .9834 P(z < -2.13) = .0166 P(.15 .19) = P(-2.13 z 2.13) = .9834 - .0166 = .9668 42. The random numbers corresponding to the first seven universities selected are 122, 99, 25, 55, 115, 102, 61 The third, fourth and fifth columns of Table 7.1 were needed to find 7 random numbers of 133 or less without duplicate numbers. Author’s note: The universities identified are: Clarkson U. (122), U. of Arizona (99), UCLA (25), U. of Maryland (55), U. of New Hampshire (115), Florida State U. (102), Clemson U. (61). 43. a. With n = 100, we can approximate the sampling distribution with a normal distribution having E() = 8086 b. P(z ≤ .80) = .7881 P(z < -.80) = .2119 P(7886 8286) = P(-.80 z .80) = .7881 - .2119 = .5762 The probability that the sample mean will be within $200 of the population mean is .5762. c. At 9000, P( ≥ 9000) = P(z ≥ 3.66) 0 Yes, the research firm should be questioned. A sample mean this large is extremely unlikely (almost 0 probability) if a simple random sample is taken from a population with a mean of $8086. 44. a. Normal distribution with E= 406
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