Question

An insurance company has determined that each week an average of 9 claims are filed. Assume that the claims are Poisson distributed. Find the probability that exactly 3 claims will be filed

.0000

.0150

.1273

.0050

none of the above

Answer 1

Let X be the number of claims filed in a week.

Given that X~P(mean=9)

The pmf of X is given by

f(x)=P[X=x]= $\frac{e^{-9}9^x}{x!}$    if x=0,1,2,......

                 = 0      otherwise

Required to find

P[X=3]= f(3) = $\frac{e^{-9}9^3}{3!} = 0.0150$

Hence option(b) is correct.