Question

The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let X = the number of calls received at noon.

a. Find the mean and standard deviation of X.

b. What is the probability that the office receives at most six calls at noon on Monday?

c. Find the probability that the law office receives six calls at noon. What does this mean to the law office staff who get, on average, 5.5 incoming phone calls at noon?

d. What is the probability that the office receives more than eight calls at noon?

Short answer

a. mean is$E\left(\xi \right)=5.7$standard deviation is $2.3875$

b. The probability that the office receives at most six calls at noon on Monday role="math" localid="1649174863530" $P_n(\xi \le 6)=0.65437$

c. The probability that the law office receives six calls at noon.$P_n(\xi =6)=0.15840$

d. The probability that the office receives more than eight calls at noon $P_n(\xi \ge 8)=0.21585$

Step-by-step solution

Content Introduction

In a large population, the Poisson distribution is used to characterize the distribution of unusual events.

Explanation (part a)

It is a Poisson distribution,

mean is $E\left(\xi \right)=\lambda$

Therefore, mean is $E\left(\xi \right)=5.7$

standard deviation is $\delta =\sqrt{var\left(\xi \right)}$

role="math" localid="1649174541169" $$\begin{gathered} \delta =\sqrt{var\left(\xi \right)} \\ \delta =\sqrt{5.7} \\ \delta =2.3875 \end{gathered}$$

Explanation (part b)

Using formula for Poisson probability is

$P_n(\xi =m)=\frac{{\lambda }^m}{m!}{e}^{-\lambda }$

We are given the information,

$$\begin{gathered} P_n(\xi \le 6)=P_n\left(0\right)+P_n\left(1\right)+...........+P_n\left(6\right) \\ {P}_n(\xi \le 6)=0.00335+0.01907+0.05436+0.10327+0.14717+0.16777+0.15938 \\ {P}_n(\xi \le 6)=0.65437 \end{gathered}$$

Explanation (part c)

$P_n(\xi =6)=0.15840$It means nothing to that staff, since this is the probability that staff receives 6 calls with average $5.7$ but not $5.5$

Explanation (part d)

The probability that the office receives more than eight calls at noon is

$$\begin{gathered} P(\xi \ge 8)=1-P(\xi \le 7) \\ P(\xi \ge 8)=1-P(\xi =0)-...-P(\xi =7) \\ P(\xi \ge 8)=1-0.78415 \\ P(\xi \ge 8)=0.21585 \end{gathered}$$