Question

A telephone company has determined that during non-holidays the number of phone calls that pass through
the main branch office each hour has a relative frequency distribution with a mean of 80,000 calls and a
standard deviation of 35,000. Suppose that a random sample of 60 nonholiday hours is selected and the sample mean of the incoming phone calls is computed.

Eighty-five percent of the incoming calls are less than what sample mean?

Answer 1

The central limit theorem is

P( < x) = P( Z < x - / / sqrt(n) )

We have to calculate x such that,

P( < x) = 0.85

That is

P( Z < x - / / sqrt(n) ) = 0.85

From the Z table, z-score for probability of 0.85 is 1.0364

Therefore,

x - / / sqrt(n) = 1.0364

Put the values of , and n in above expression and solve for x

x - 80000 / 35000 / sqrt(60) = 1.0364

x - 80000 = 1.0364 * ( 35000 / sqrt(60) )

x = 84682.95

Eighty-five percent of the incoming calls are less than sample mean of 84682.95