CEE 202 Homework 9 solution

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Unformatted text preview: lity that the sample mean lies between 64 and 66 ksi? Ans: Mean tensile strength, µ = 65 ksi Standard deviation, σ = 4.2 ksi Number of samples, n = 51 a) Expected value of the sample mean, µX (bar) = µ = 65 ksi b) Variance, σ2X(bar) = σ2/n =4.22/51 = 0.3458 Standard deviation of the sample mean, σX(bar) = 0.588 c) Z = (X (bar)  ­ µ)/( σ/√n) = (64 ­65)/(4.2/√51) = ­1.7 Therefore, P ( X <64) = P (Z< ­1.7) = 0.0446 (Using Table A.3) d) 95% probability range χ2values correspond to 95% distribution 0.025 in lower tail and 0.025 in upper tail For 2.5% probability to lie in lower tail Using Table A.5 for ν=50 and α =0.975, the χ2value is 32.357 (n − 1) s 2 2 We know, χ = 2 σ So, 32.357 = 50*s2/4.22; s =3.379 For 2.5% probability to lie in upper tail Using Table A.5 for ν=50 and α =0.025, the χ2value is 71.42...
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This document was uploaded on 03/06/2014 for the course CEE 202 at University of Illinois, Urbana Champaign.

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