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 $400 per month and a standard deviation of $50 per month. Refer to such expenses as PCE's .
Find the point in the distribution below which 2.5% of the PCE's fell.



a. $302



b. $10



c. $390



d. $498

Answer 1

Answer:

a. $302

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 400, \sigma = 50$

Point in the distribution below which 2.5% of the PCE's fell.

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

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

$-1.96 = \frac{X - 400}{50}$

$X - 400 = -1.96*50$

$X = -1.96*50 + 400$

$X = 302$