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Ugo [173]
1 year ago
15

A new phone system was installed last year to help reduce the expense of personal calls that were being made by employees. Befor

e 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
Mathematics
1 answer:
Leya [2.2K]1 year ago
3 0

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.

<h3>Normal Probability Distribution</h3>

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 <u>p-value of Z when X = 790 subtracted by the p-value of Z when X = 575</u>, 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

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