Question

Suppose the monthly charges for cell phone plans in the U. S. Are normally distributed with a mean of $62 and a standard deviation of $10. What is the proportion of cell phone plans charging less than $42 a month for service?

Answer 1

Using the normal distribution, it is found that 0.0228 = 2.28% of cell phone plans charge less than $42 a month for service.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

In this problem, the mean and the standard deviation are given by, respectively: $\mu = 62, \sigma = 10$.

The proportion of cell phone plans charging less than $42 a month for service is the p-value of Z when X = 42, hence:

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

$Z = \frac{42 - 62}{10}$

$Z = -2$

$Z = -2$ has a p-value of 0.0228.

0.0228 = 2.28% of cell phone plans charge less than $42 a month for service.

More can be learned about the normal distribution at https://brainly.com/question/24663213