A new phone system was installed last year to help reduce the expense of personal calls that were being madeby employees. Before the new system was installed, the amount being spent on personal calls followed anormal distribution with an average of $500 per month and a standard deviation of $50 per month. Refer tosuch expenses as PCEʹs (personal call expenses). Using the distribution above, what is the probability that arandomly selected month had a PCE of between $375.00 and $590.00?
A) 0.9579
B) 0.0421
C) 0.9999
D) 0.0001

Question

Quizplus · Accepted Answer

To find the probability that a randomly selected month had a PCE between $375.00 and $590.00, we use the normal distribution properties. First, we calculate the Z-scores for $375 and $590 using the formula $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. For $375, Z = \frac{375 - 500}{50} = -2.5\), and for $590, Z = \frac{590 - 500}{50} = 1.8\). Looking these Z-scores up in the standard normal distribution table or using a calculator, we find the probabilities corresponding to these Z-scores and subtract the smaller from the larger to find the probability of being between these two values. This calculation gives us approximately 0.9579, indicating that there's about a 95.79% chance a randomly selected month had a PCE between $375.00 and $590.00.

A New Phone System Was Installed Last Year to Help