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 followed a normal distribution mean of $700 per month and a standard deviation of $50 per month

Answer 1

Using the normal distribution, it is found that there was a 0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 700, \sigma = 50$.

The probability of a month having a PCE between $575 and $790 is the p-value of Z when X = 790 subtracted by the p-value of Z when X = 575, hence:

X = 790:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{790 - 700}{50}$

Z = 1.8

Z = 1.8 has a p-value of 0.9641.

X = 575:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{575 - 700}{50}$

Z = -2.5

Z = -2.5 has a p-value of 0.0062.

0.9641 - 0.0062 = 0.9579.

0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1