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
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This note was uploaded on 12/06/2011 for the course MGMT 305 taught by Professor Priya during the Fall '08 term at Purdue University.

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

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