#16 page 199

“Assume that, during each second, a Dartmouth switchboard receives one call with probability .01 and no calls with probability .99. Use the Poisson approximation to estimate the probability that the operator will miss at most one call if she takes a 5-minute coffee break.”

The probability that the operator will miss at most one call is equal to the probability of her missing zero calls plus the probability of her missing exactly one call.

Since we know the expected rate is .01 calls per second, in a five minute interval, we set lambda at:

\[ \lambda = .01*60*5 = 3 \]

sum(dpois(0:1,3))
## [1] 0.1991483

There is approximately a 20% probability that she will miss at most one call during a five minute break.