6.53 Suppose that the number of accidents occurring in an industrial plant is de
ID: 3066335 • Letter: 6
Question
6.53 Suppose that the number of accidents occurring in an industrial plant is described by a Poisson distribution with an average of one accident per month. Let x denote the time (in months) between successive accidents. a. Find the probability that the time between successive accidents is: 1. More than two months. 2. Between one and two months. 3. Less than one week (1/4 of a month). b. Suppose that an accident occurs less than one week after the plant's most recent accident. Would you consider this event unusual enough to warrant special investigation? Explain.
Explanation / Answer
The time between successive accidents is exponentially distributed with parameter 1/month
The CDF of the expinential distribution, F(x) = 1 - e-x
a. Probability that gap between successive accidents > 2 months,P(X>2) = 1-P(X2) = 1-[1-e-2x1] = 0.1353
P(1 < X < 2) = F(2) - F(1) = [1-e-2x1] - [1-e-1x1] = e-1 - e-2 = 0.3679 - 0.1353 = 0.2325
P(X<0.25) = F(0.25) = 1 - e-0.25x1 = 1-0.7788 = 0.2212
b. Because of the memoryless property of the geometric distribution, probability that an accident occurs less than one week after the plant's most recent accident = P(X<0.25) =0.2212
So with a significance level of 0.05, the probability of P(X<0.25) =0.2212 lies in the acceptance region. So it's significantly probable to expect accidents within a gap of 1/4 weeks and this doesn't warrant special investigation.
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