Question

A new phone system was installed last year to help reduce the expense of personal calls that were being made by employees. Before the new system was​ installed, the amount being spent on personal calls follows a normal distribution with an average of per month and a standard deviation of​ $50 per month. Refer to such expenses as​ PCE's (personal call​ expenses). Find the point in the distribution below which​ 2.5% of the​ PCE's fell.

Answer 1

The point is
`\mu - 98`, in which
`\mu` is the average per month.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normal distribution with an average of per month and a standard deviation of​ $50 per month.

Average of
`\mu`, standard deviation
`\sigma = 50`

Find the point in the distribution below which​ 2.5% of the​ PCE's fell.

This is below the 2.5th percentile, which is the X when Z has a pvalue of 0.025, so X when Z = -1.96.

`Z = (X - \mu)/(\sigma)`

`-1.96 = (X - \mu)/(50)`

`X - \mu = -1.96*50`

`X = \mu - 98`

The point is
`\mu - 98`, in which
`\mu` is the average per month.