It is estimated that .44 percent of the callers to the Customer Service departme

Question

It is estimated that .44 percent of the callers to the Customer Service department of Dell, Inc. will receive a busy signal. What is the probability that of today's 1,100 callers at least 5 received a busy signal? Use the poisson approximation to the binomial.

Explanation / Answer

You should have seen the poisson formula: p(k events) = L^k / (k! e^L) where L, usually written lambda, is the limit of np as p goes to 0. So in this case, n=1100, p=0.0046, so L = np = 1100*0.0046 = 5.52 Now you need the probability that *at least* 5 callers receive a busy signal. i.e. the probability that k>=5. To do this, work out the probability of k= 0, 1, 2, 3 and 4 (using the formula above), and subtract from 1. i.e. p(k>=5) = 1 - p(0 events) - p(1 events) - p(2 events) - p(3 events) - p(4 events) Since the mean is 5.52, you should get a value of a bit over 0.5

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