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HW 2 with Solutions

# HW 2 with Solutions - MGMT 305 HW2 SOLUTIONS 27 a At =...

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MGMT 305 HW2 SOLUTIONS 27. a. At = 178,000, P ( z ≤ 1.58) = .9429 At = 158,000, z = -1.58 P ( z < -1.58) = .0571, thus P (158,000 ≤≤ 178,000) = .9429 - .0571 = .8858 b. At = 127,000, P ( z ≤ 2.53) = .9943 At = 107,000, z = -2.53, P ( z < -2.53) = .0057, thus P (107,000 ≤ ≤ 127,000) = .9943 - .0057 = .9886 c. In part (b) we have a higher probability of obtaining a sample mean within \$10,000 of the population mean because the standard error is smaller. d. With n = 100, At = 164,000, P ( < 164,000) = P ( z < -1) = .1587 28. a. This is a graph of a normal distribution with = 95 and b. Within 3 strokes means 92 98 P (92 98) = P (-1.17 z 1.17) = .8790 - .1210 = .7580 OR .3790*2=.7580 The probability the sample means will be within 3 strokes of the population mean of 95 is .7580. c.

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Within 3 strokes means 103 109 P (103 109) = P (-1.44 z 1.44) = .9251 - .0749 = .8502 Or .4251*2=8502 The probability the sample means will be within 3 strokes of the population mean of 106 is .8502. d. The probability of being within 3 strokes for female golfers is higher because the sample size is larger. 39. a. Normal distribution with and b. P ( z ≤ 1.96) = .9750 P ( z < -1.96) = .0250 P (.71 .79) = P (-1.96 z 1.96) = .9750 - .0275 = .9500 Or .4750*2=.9500 c. Normal distribution with and d. P ( z ≤ 1.31) = .9049 P ( z < -1.31) = .0951 P (.71 .79) = P (-1.31 z 1.31) = .9049 - .0951 = .8098
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