Question

It is estimated that 0.3% of the callers to the billing department of a local telephone company will receive a busy signal.

What is the probability that of today's 1150 callers at least 5 received a busy signal? Use the poisson approximation to the binomial. (Round the final answer to 4 decimal places.)

Probability             

Answer 1

0.3% receive a busy call so, p = 0.003

n= 1150

The estimated mean of the distribution is or mean busy call receive are () = 1150*0.003 = 3.45

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 x = 0, 1, 2, 3 and 4 and subtract from 1.

x! is the factorial of x fac of = 1, fac of 2 = 2*1 =2, fac of 3 = 3*2*1 = 6, fac of 4 =4*3*2*1 =24
λ is a positive real number.

1-[( e-3.45*3.45^1)/1+( e-3.45 * 3.45^2) /2 + ( e-3.45*3.45^3)/6 + ( e-3.45*3.45^4)/24

= 1- 0.70

= 0.30