Question

A local police station receives an average 6.3 emergency telephone calls per hour. These calls are Poisson distributed.

The probability that the station will get at least 3 calls per hour is:

a.         0.0397

2.       0.0941
3.       0.9059
4.       0.9502
5.       0.9630

Answer 1

Let say X~Poisson($\dpi{100} \lambda$)

So,

PMF

$\dpi{100} P(X=x) = e^{- \lambda } \frac{\lambda ^x}{x!}$ , x=0,1,2,.....

Thus Mean of X:

$\dpi{100} E(X)= \sum_{x=0}^{\infty}xP(X=x) \\\\ = \sum_{x=0}^{\infty}xe^{- \lambda } \frac{\lambda ^x}{x!} \\\\ = e^{- \lambda } \lambda \sum_{x=1}^{\infty}\frac{\lambda^{x-1}}{(x-1)!}$.........at x=0 the term is 0

$\dpi{100} = e^{- \lambda } \lambda \sum_{y=0}^{\infty}\frac{\lambda^{y}}{(y)!}$.....................y=x-1

$\dpi{100} = e^{- \lambda } \lambda e^{\lambda }$..................by the formula of taylor series

$\dpi{100} =\lambda$

Thus the mean or average $\dpi{100} =\lambda$

For this problem ,

$\dpi{100} \lambda =6.3$

X: Number of local calls per hour

X~Poisson(6.3)

So, P(at least 3 calls per hour) = P(X>=3) :

$\dpi{100} P(X\geq 3) =1- P(X<3) \\\\ =1- \sum_{x=0}^{2} e^{-6.3} \frac{6.3^x}{x!} \\\\ = 1- 0.04984 =0.95016 \approx 0.9502$

Option "d"