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.

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Answer:

The point is [tex]\mu - 98[/tex], in which [tex]\mu[/tex] is the average per month.

Step-by-step explanation:

Normal Probability Distribution:

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu[/tex], standard deviation [tex]\sigma = 50[/tex]

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.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.96 = \frac{X - \mu}{50}[/tex]

[tex]X - \mu = -1.96*50[/tex]

[tex]X = \mu - 98[/tex]

The point is [tex]\mu - 98[/tex], in which [tex]\mu[/tex] is the average per month.