Question

Among U.S. cities with a population of more than 250,000 the mean one-way commute time to work is 24.3 minutes. The longest one-way travel time is in New York City, where the mean time is 38.7 minutes. Assume the distribution of travel times in New York City follows the normal probability distribution and the standard deviation is 7.3 minutes.
a.
What percent of the New York City commutes are for less than 29 minutes? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)

Percent %

b.
What percent are between 29 and 36 minutes? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)

Percent %

c.
What percent are between 29 and 44 minutes? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)

Percent %

Answer 1

Answer:

a) 9.18% f the New York City commutes are for less than 29 minutes.

b) 26.39% are between 29 and 36 minutes.

c) 67.55% are between 29 and 44 minutes

Step-by-step explanation:

Problems of normally distributed samples can be 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 problem, we have that:

$\mu = 38.7, \sigma = 7.3$

a.

What percent of the New York City commutes are for less than 29 minutes?

This is the pvalue of Z when X = 29.

So

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

$Z = \frac{29 - 38.7}{7.3}$

$Z = -1.33$

$Z = -1.33$ has a pvalue of 0.0918.

So 9.18% f the New York City commutes are for less than 29 minutes.

b.

What percent are between 29 and 36 minutes?

This is the pvalue of Z when X = 36 subtracted by the pvalue of Z when X = 29.

X = 36

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

$Z = \frac{36 - 38.7}{7.3}$

$Z = -0.37$

$Z = -0.37$ has a pvalue of 0.3557.

X = 29

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

$Z = \frac{29 - 38.7}{7.3}$

$Z = -1.33$

$Z = -1.33$ has a pvalue of 0.0918.

So 0.3557 - 0.0918 = 0.2639 = 26.39% are between 29 and 36 minutes.

c.

What percent are between 29 and 44 minutes?

This is the pvalue of Z when X = 44 subtracted by the pvalue of Z when X = 29.

X = 44

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

$Z = \frac{44 - 38.7}{7.3}$

$Z = 0.73$

$Z = -0.37$ has a pvalue of 0.7673.

X = 29

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

$Z = \frac{29 - 38.7}{7.3}$

$Z = -1.33$

$Z = -1.33$ has a pvalue of 0.0918.

So 0.7673 - 0.0918 = 0.6755 = 67.55% are between 29 and 44 minutes