Chapter 4: Continuous Variables and Their Probability ...

For the standard normal distribution, P(Z < z0) = .01 for z0 = –2.33, thus = 500 + 50(–2.33) = 383.5. Referring to Ex. 4.66, let X = # of defective bearings. Thus, X is binomial with n = 5 and p = P(defective) = .073. Thus, P(X > 1) = 1 – P(X = 0) = 1 – (.927)5 = .3155. Let Y = lifetime of a drill bit. Then, Y has a normal distribution ... ................
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