Question

CLT

A telephone switchboard handles on the average 4 calls per minute. If the calls follow a Poisson distribution.

How many calls can we expect in a day?

What is the standard deviation of this number?

Using the normal approximation what is the bar for the number of calls for a day to be considered 10% of the busiest?

Answer 1

Here X follows Poisson distribution with mean = = 4 per minutes.

Therefore X_t follows Poisson distribution with mean =

There are 24 * 60 minutes in a day.

Therefore ,  X_t follows Poisson distribution with mean = 4 *(60*24) = 5760 calls.

So we expect 5760 calls in a day.

For poisson process the expected value and variance are same.

So standard deviation of this number is

Suppose that Y follows approximately normally distributed with mean = 5760 and standard deviation = 75.895

So we want to find y such that P( Y > y) = 0.10

That is P( Y < y) = 1 - 0.10 = 0.9

Let's use excel.

y = "=NORMINV(0.9,5760,75.895)" = 5857.26 = 5857

Therefore, the bar for the number of calls for a day to be considered 10% of the busiest are 5857 calls.