Assume that the finishing times in a New York City 10-kilometer road race are normally distributed with a mean of 55 minutes and a standard deviation of 10 minutes. Let X be a randomly selected finishing time.

a. What percent of finish times was higher than 72 minutes? (Give your answers rounded to two decimal places)

b. What percent of finish times was between 52 and 70 minutes? (Give your answers rounded to two decimal places)

c. What is the 40th percentile finish time? (Round your answer to nearest tenth)

d. What is the 95th percentile finish time? (Round your answer to nearest tenth)

Using the normal distribution, it is found that:

a) 4.46% of finish times was higher than 72 minutes.

b) 55.11% of finish times was between 52 and 70 minutes.

c) The 40th percentile of finish times is 52.5 minutes.

d) The 95th percentile of finish times is 71.45 minutes.

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem:

Item a:

The proportion is 1 subtracted by the p-value of Z when X = 72, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{72 - 55}{10}[/tex]

[tex]Z = 1.7[/tex]

[tex]Z = 1.7[/tex] has a p-value of 0.9554

1 - 0.9554 = 0.0446

0.0446 x 100% = 4.46%

4.46% of finish times was higher than 72 minutes.

Item b:

Th proportion is the p-value of Z when X = 70 subtracted by the p-value of Z when X = 52, hence:

X = 70:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70 - 55}{10}[/tex]

[tex]Z = 1.5[/tex]

[tex]Z = 1.5[/tex] has a p-value of 0.9332.

X = 52:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{52 - 55}{10}[/tex]

[tex]Z = -0.3[/tex]

[tex]Z = -0.3[/tex] has a p-value of 0.3821.

0.9332 - 0.3821 = 0.5511

0.5511 x 100% = 55.11%

55.11% of finish times was between 52 and 70 minutes.

Item c:

The 40th percentile is X when Z has a p-value of 0.4, so X when Z = -0.253.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.253 = \frac{X - 55}{10}[/tex]

[tex]X - 55 = -0.253(10)[/tex]

[tex]X = 52.5[/tex]

The 40th percentile of finish times is 52.5 minutes.

Item d:

The 95th percentile is X when Z has a p-value of 0.95, so X when Z = 1.645.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.645 = \frac{X - 55}{10}[/tex]

[tex]X - 55 = 1.645(10)[/tex]

[tex]X = 71.45[/tex]

The 95th percentile of finish times is 71.45 minutes.

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