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Unformatted text preview: STAT 390 – Homeworks 46 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 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 (6043)/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 (4043)/4.5 ≤ z ≤ (5043)/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. z*) = ....
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This note was uploaded on 10/17/2008 for the course STAT 390 taught by Professor Marzban during the Spring '08 term at Washington State University .
 Spring '08
 marzban

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