It is estimated that 46 percent of the callers to the Custom

It is estimated that .46 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. (Round your answer to 4 decimal places.)

Solution

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 = 5.06

P( X < 5) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= e^-5.06 * 3 ^ 4 / 4! + e^-5.06 * ^ 3 / 3! + e^-5.06 * ^ 2 / 2! + e^-5.06 * ^ 1 / 1! + e^-5.06 * ^ 0 / 0!   
= 0.43

P( X > = 5 ) = 1 - P (X < 5) = 0.57