Solution HW2_2010

Solution HW2_2010 - ≤ ≤ 109 P (103 ≤ ≤ 109) = P...

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MGMT 305: Homework 2 Due Date: February 4, 2010 Quiz2- Sampling Distribution 27. ( Total -15 pts ) 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 ( 5 pts ) 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 ( 4 pts ) 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. ( 2 pts ) d. With n = 100,
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At = 164,000, P (< 164,000) = P ( z < -1) = .1587 ( 4 Pts ) 28. ( Total 10 pts ) a. This is a graph of a normal distribution with = 95 and ( 2pts ) b. Within 3 strokes means 92 98 P (92 98) = P (-1.17 ≤ z ≤ 1.17) = .8790 - .1210 = .7580 The probability the sample means will be within 3 strokes of the population mean of 95 is .7580.( 4pts ) c. Within 3 strokes means 103
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Unformatted text preview: ≤ ≤ 109 P (103 ≤ ≤ 109) = P (-1.44 ≤ z ≤ 1.44) = .9251 - .0749 = .8502( 4pts ) 46. ( Total 15pts ) μ = 41,979 σ = 5000 a. ( 2pts ) b. P (> 41,979) = P ( z > 0) = .50( 4pts ) c. P ( z ≤ 1.41) = .9207 P ( z < -1.41) = .0793 P (40,979 ≤ ≤ 42,979) = P (-1.41 ≤ z ≤ 1.41) = .9207 - .0793 = .8414( 4pts ) d. P ( z ≤ 2.00) = .9772 P ( z < -2.00) = .0228 P (40,979 ≤ ≤ 42,979) = P (-2 ≤ z ≤ 2) = .9772 - .0228 = .9544( 5pts ) 37. ( Total 15pts ) a. Normal distribution E () = .50 ( 2pts ) b. P ( z ≤ 1.94) = .9738 P ( z < -1.94) = .0262 P (.46 ≤≤ .54) = .9738 - .0262 = .9476 ( 4pts ) c. P ( z ≤ 1.46) = .9279 P ( z < -1.46) = .0721 P (.47 ≤≤ .53) = .9279 - .0721 = .8558 ( 4pts ) d. P ( z ≤ .97) = .8340 P ( z < -.97) = .1660 P (.48 ≤≤ .52) = .8340 - .1660 = .6680 ( 5pts )...
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This note was uploaded on 01/31/2011 for the course MGMT 305 taught by Professor Priya during the Spring '08 term at Purdue University.

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Solution HW2_2010 - ≤ ≤ 109 P (103 ≤ ≤ 109) = P...

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