A study of 12,000 able-bodied make students at - id120005705972, MajentaSnow11601

Ответ:

kelly1027

22.01.2022

Mathematics

a) About 802 students ran the mile in less than 6 minutes

b) It took at least 8.06 minutes for the slowest 10% of the students to run the mile

c) 67.62% of students ran the mile in between 400 and 500 seconds.

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 7.11, \sigma = 0.74$

a. About how many students ran the mile in less than 6 minutes?

Proportion who ran in less than 6 minutes is the pvalue of Z when X = 6. So

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

$Z = \frac{6 - 7.11}{0.74}$

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

Out of 12000

0.0668*12000 = 801.6

About 802 students ran the mile in less than 6 minutes.

b. Approximately how long did it take the slowest 10% of the students to run the mile?

The slowest, the higher the time.

So this is the 100-10 = 90th percentile of times.

Which are times of X when Z = 1.28 and greater

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

$1.28 = \frac{X - 7.11}{0.74}$

X - 7.11 = 1.28*0.74

X = 8.06

It took at least 8.06 minutes for the slowest 10% of the students to run the mile

c. Suppose that these mile run times are converted from minutes to seconds. Estimate the percent of students who ran the mile in between 400 and 500 seconds.

Each minute has 60 seconds. So

$\mu = 60*7.11 = 426.6, \sigma = 44.4$