It is estimated that 60 percent of the callers to the Custom

It is estimated that .60 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 1,200 callers at least 5 received a busy signal? Use the poisson approximation to the binomial

Solution

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=1200, p=0.0060, so L = np = 1200*0.0060 = 7.2

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 7.2, you should get a value of a bit over 0.3