chap5S2

# Assumeindependenceofoutcomesfortheseinstallations

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Unformatted text preview: most one success in n trials) = P(X = 0) + P(X = 1) 20 Chapter 5 Example: Exercise 5.39, page 156. A company installs new heating furnaces. For any installation, the probability of a return visit for a repair is 0.15. Six installations are made in a given week. Assume independence of outcomes for these installations. Let the random variable X be the number of return visits. X follows a binomial distribution with: p = 0.15 and 1 − p = 0.85 Find the probability that a return visit will be needed in more than one of the installations. P(X &gt; 1) = 1 − P( X ≤ 1) = 1 − [P(X = 0) + P(X = 1)] = 1 − [(0.85)6 + (6)(0.15)(0.85)5 ] At this point, it is clear that the numerical calculations can be tedious. 21 Chapter 5 Numerical answers for binomial distribution probabilities can be obtained with Microsoft Excel. Select Insert Function BINOMDIST. The general usage is: BINOMDIST(x, n, p, cumulative) where cumulative = 0 for the probability distribution function, cumulative = 1 for the cumulative probability function For this exer...
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