Question

he mean of the commute time to work for a resident of a certain city is 27.8 minutes. Assume that the standard deviation of the commute time is 6.6 minutes to complete parts​ (a) through​ (c).

Answer 1

Using the normal distribution, it is found that:

a) The minimum percentage of commute times within 2 standard deviations of the mean is of 95.44%.

b) The minimum percentage of commute times within 2.5 standard deviations of the mean is of 98.76%.

c) The minimum percentage of commute times within 3 standard deviations of the mean is of 99.74%.

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 is of $\mu = 27.8$.

• The standard deviation is of $\sigma = 6.6$.

Item a:

It asks the minimum percentage of commute times within 2 standard deviations of the mean. We consider that:

• Z = 2 has a p-value of 0.9772.

• Z = -2 has a p-value of 0.0228.

0.9772 - 0.0228 = 0.9544 = 95.44%.

The minimum percentage of commute times within 2 standard deviations of the mean is of 95.44%.

Item b:

Within 2.5 standard deviations, hence:

• Z = 2.5 has a p-value of 0.9938.

• Z = -2.5 has a p-value of 0.0062.

0.9938 - 0.0062 = 0.9876 = 98.76%.

The minimum percentage of commute times within 2.5 standard deviations of the mean is of 98.76%.

Item c:

It asks the minimum percentage between 8 and 47.6 minutes, which is within 3 standard deviations of the mean, hence:

• Z = 3 has a p-value of 0.9987.

• Z = -3 has a p-value of 0.0013.

0.9987 - 0.0013 = 0.9974.

The minimum percentage of commute times within 3 standard deviations of the mean is of 99.74%.

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