c. This is a graph of a normal distribution with a mean of = .45 and d. Within ± .05 means .40 .50 ≤ ≤ .50) = 1.42) = .9222 - .0778 = .8444 P(.40 P(-1.42 z ≤ ≤ e. No, the probabilities are exactly the same. This is because , the standard error, and the width of the interval are the same in both cases. Notice the formula for computing the standard error. It involves . So whenever = 1 - the standard error will be the same. In part (b), = .45 and 1 = .55. In p p p p – part (d), = .55 and 1 = .45. p p – f. For = 400, n Within ± .05 means .50 .60 ≤ ≤ .60) = 2.01) = .9778 - .0222 = .9556 P(.50 P(-2.01 z ≤ ≤ The probability is larger than in part (b). This is because the larger sample size has reduced the standard error from .0352 to .0249. 37. a. Normal distribution b. 1.94) = .9838 P(z ≤ < -2.14) = .0162 P(z .15) = .9838 - .0162 = .9676 P(.09 ≤≤ c. 1.07) = .8577 P(z ≤ < -1.07) = .1423 P(z .135) = .8577 - .1423 = .7154 P(.105 ≤≤ 38. a. It is a normal distribution with = .42 E() b. 1.05) = .8531 P(z ≤ < -1.05) = .1469 P(z
c. This is a graph of a normal distribution with a mean of = .45 and d. Within ± .05 means .40 ≤ ≤ .50 P(.40 .50) = P(-1.42 ≤ z ≤ 1.42) = .9222 - .0778 = .8444 e. No, the probabilities are exactly the same. This is because , the standard error, and the width of the interval are the same in both cases. Notice the formula for computing the standard error. It involves . So whenever p = 1 - p the standard error will be the same. In part (b), p = .45 and 1 – p = .55. In part (d), p = .55 and 1 – p = .45. f. For n = 400, Within ± .05 means .50 ≤ ≤ .60 P(.50 .60) = P(-2.01 ≤ z ≤ 2.01) = .9778 - .0222 = .9556 The probability is larger than in part (b). This is because the larger sample size has reduced the standard error from .0352 to .0249. 37. a. Normal distribution b. P(z ≤ 1.94) = .9838 P(z < -2.14) = .0162 P(.09 ≤≤ .15) = .9838 - .0162 = .9676 c. P(z ≤ 1.07) = .8577 P(z < -1.07) = .1423 P(.105 ≤≤ .135) = .8577 - .1423 = .7154 38. a. It is a normal distribution with E() = .42 b. P(z ≤ 1.05) = .8531 P(z < -1.05) = .1469