Question

Kathy and her brother Clay recently ran in a local marathon. The distribution of finishing time for women was approximately normal with mean 259 minutes and standard deviation 32 minutes. The distribution of finishing time for men was approximately normal with mean 242 minutes and standard deviation 29 minutes. (a) The finishing time for Clay was 289 minutes. Calculate and interpret the standardized score for Clay's marathon time. Show your work. (b) The finishing time for Kathy was 272 minutes. What proportion of women who ran the marathon had a finishing time less than Kathy's? Show your work. (c) The standard deviation of finishing time is greater for women than for men. What does this indicate about the finishing times of the women who ran the marathon compared to the finishing times of the men who ran the marathon?

Answer 1

Part a)

We have that,the distribution of finishing time for men was approximately normal with mean 242 minutes and standard deviation 29 minutes.

We want to calculate and interpret the standardized score for Clay's marathon time, if the finishing time for Clay was 289 minutes.

We use the formula:

$z = \frac{x - \bar x}{s}$

we substitute the values to get:

$z = \frac{289 - 242}{29}$

$z = 1.62$

This means Clay's finishing time is 1.62 standard deviation above the mean finishing time.

Part b)

This time, we have that, the distribution of finishing time for women was approximately normal with mean 259 minutes and standard deviation 32 minutes.

We want to find the proportion of women who ran the marathon that had a finishing time less than Kathy if the finishing time for Kathy was 272 minutes.

We first calculate the z-score to get:

$z = \frac{272 - 259}{32} = 0.41$

From the normal standard distribution table P(z<0.41)=0.6591.

This means 65.91% of women had a finishing time less than Kathy's finishing time.

Part c

The standard deviation of a data set tells us how far away the individual data are from the mean.

If the standard deviation of finishing time is greater for women than men, it means the finishing time for women are farther from the mean finishing time than that of men.

Answer 2

Using the normal distribution, it is found that:

a) His standardized score was of z = 1.6, which means that his finishing time is of 1.6 standard deviations above the mean finishing time for all men.

b) 0.6591 = 65.91% of women who ran the marathon had a finishing time less than Kathy's.

c) The higher standard deviation shows that the finishing times of the women who ran the marathon are more spread out compared to the finishing times of the men who ran the marathon.

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.

Item a:

• For men, mean of 242 minutes, thus $\mu = 242$.
• Standard deviation of 29 minutes, thus $\sigma = 29$.

• The z-score is found when X = 289, thus:

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

$Z = \frac{289 - 242}{29}$

$Z = 1.6$

His standardized score was of z = 1.6, which means that his finishing time is of 1.6 standard deviations above the mean finishing time for all men.

Item b:

• For women, mean of 259 minutes, thus $\mu = 259$
• Standard deviation of 32 minutes, thus $\sigma = 32$.

The proportion is the p-value of Z when X = 272, thus:

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

$Z = \frac{272 - 259}{32}$

$Z = 0.41$

$Z = 0.41$ has a p-value of 0.6591.

0.6591 = 65.91% of women who ran the marathon had a finishing time less than Kathy's.

Item c:

The higher standard deviation shows that the finishing times of the women who ran the marathon are more spread out compared to the finishing times of the men who ran the marathon.

A similar problem is given at https://brainly.com/question/24855678