Answer to Question #163285 in Statistics and Probability for Dush

Question

The switchboard in a Police law office gets an average of 5.7 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?

Answer 1

Solution:

a. It is a Poisson distribution, so $E(\xi)=var(\xi)=\lambda=5.7$

$stddev(\xi) = \sqrt{var(\xi)}=\sqrt{5.7} \approx 2.3875$

b. Using formula for Poisson probability and calculator:

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

$P_n(\xi\le 6)= P_n(0)+P_n(1)+...+P_n(6)= 0.00335+$

$+0.01907+0.05436+0.10327+0.14717+0.16777+0.15938\approx 0.65437$

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

d. $P(\xi\ge8)= 1-P(\xi\le 7) = 1 -P(\xi=0)-...-P(\xi=7) = 1- 0.78415 \approx 0.21585$

Answer:

a. $E(\xi)=\lambda=5.7$

$stddev(\xi) \approx 2.3875$

b. $P_n(\xi\le 6)\approx 0.65437$

c. $P_n(\xi=6)=0.15840$

d. $P(\xi\ge8)\approx 0.21585$