# hw2_soln - STAT 390 Homeworks 4-6 Solutions 9/29 We have...

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STAT 390 – Homeworks 4-6 Solutions 9/29 We have that 2 2 1 1. 2 x e dx μ σ πσ - - - = So using the substitution z x μ σ = - gives dz dx σ = , so 2 2 1 1. 2 z e dz σ πσ - - = Hence 2 2 1 1, 2 z e dz π - - = giving the normal density with 0 μ = , 1 σ = . 1.34 (d) Let z* denote the 99.9 th percentile, Proportion(z z*) = .9990. From Table I, any z value between 3.08 and 3.10 will do; a z* of about 3.09 would be a good choice. 1.37 (a) Let x = yield strength. Then, x 40, when standardized, becomes z (40- 43)/4.5 = - .67. From Table I, Proportion(z -.67) = .2514. Similarly, standardizing 60 x yields z (60-43)/4.5 = 3.78. From Table I, Proportion(z > 3.78) = 1 - Proportion(z 3.78) = 1- .9999 = .0001. (b) Standardizing 40 x 50 gives (40-43)/4.5 z (50-43)/4.5 or, -.67 z 1.56. From Table I, this proportion equals Proportion(z 1.56) - Proportion(z -.67) = .9406 - .2514 = .6892. [The answer at the back of the text is still incorrect.] (c) We need to find the value of x* for which Proportion(x > x*) = .75, or equivalently, Proportion(x x*) = .25. The corresponding statement about a z curve is Proportion(z z*) = .25. From Table I, z* -.675; i.e., z * is .675 σ

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