Question

SOMEONE PLEASE HURRY AND HELP

To qualify for a certain marathon in Raleigh, a runner must be faster than 84% of their peers. If the average marathon pace was 3.5 hours with a standard deviation of 10 minutes, what is the qualifying time for the marathon? Assume the data is normally distributed.

Answer 1

Using the normal distribution, it is found that the qualifying time for the marathon is of 200 minutes.

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:

• Average of 3.5 hours, hence, in minutes, $\mu = 3.5(60) = 210$

• Standard deviation of 10 minutes, hence $\sigma = 10$.

To qualify for a certain marathon in Raleigh, a runner must be faster than 84% of their peers, hence the qualifying time for the marathon is the 16th percentile, as lower times are better, which is X when Z has a p-value of 0.16, so X when Z = -0.994.

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

$-0.994 = \frac{X - 210}{10}$

$X - 210 = -0.994(10)$

$X = 200$

The qualifying time for the marathon is of 200 minutes.

To learn more about the normal distribution, you can check https://brainly.com/question/24663213