Mathematics, 31.03.2020 02:48, coolvic345

For collegiate women soccer players the concussion rate is .63 per 1,000 player participation hours (practice and game participation hours), for collegiate men soccer players the concussion rate is .41 per 1,000 player participation hours (the concussion rate in college football is .61 per 1,000 player participation hours). Over an entire season a typical collegiate soccer team (men and women) has about 3,200 player participation hours. This results in the number of concussions on a women's collegiate soccer team having an expected value 2.02 and standard deviation 1.42; for a men's collegiate soccer team the number of concussions has expected value 1.31 and standard deviation 1.15.

Question 1. At the conclusion of a recent soccer season a random sample of 40 men's soccer teams is selected and the number of concussions experienced by each team is recorded. What is the probability that the mean number of concussions experienced by the 40 men's soccer teams is greater than 1.5? (Use 3 decimal places).

Question 2. At the conclusion of a recent soccer season a random sample of 40 women's soccer teams is selected and the number of concussions experienced by each team is recorded. What is the probability that the mean number of concussions experienced by the 40 women's soccer teams is greater than 1.5? (Use 3 decimal places).

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Answers

a) 0.149 = 14.9% probability that the mean number of concussions experienced by the 40 men's soccer teams is greater than 1.5

b) 0.99 = 99% probability that the mean number of concussions experienced by the 40 women's soccer teams is greater than 1.5

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Question 1. At the conclusion of a recent soccer season a random sample of 40 men's soccer teams is selected and the number of concussions experienced by each team is recorded. What is the probability that the mean number of concussions experienced by the 40 men's soccer teams is greater than 1.5? (Use 3 decimal places).

For a men's collegiate soccer team the number of concussions has expected value 1.31 and standard deviation 1.15.

So we have:

$\mu = 1.31, \sigma = 1.15, n = 40, s = \frac{1.15}{\sqrt{40}} = 0.1818$

The probability if 1 subtracted by the pvalue of Z when X = 1.5. So

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

By the Central Limit Theorem

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

$Z = \frac{1.50 - 1.31}{0.1818}$

Z = 1.04

Z = 1.04 has a pvalue of 0.8508

1 - 0.8508 = 0.149

0.149 = 14.9% probability that the mean number of concussions experienced by the 40 men's soccer teams is greater than 1.5

Question 2. At the conclusion of a recent soccer season a random sample of 40 women's soccer teams is selected and the number of concussions experienced by each team is recorded. What is the probability that the mean number of concussions experienced by the 40 women's soccer teams is greater than 1.5? (Use 3 decimal places).

Expected value 2.02 and standard deviation 1.42;

So we have:

$\mu = 2.02, \sigma = 1.42, n = 40, s = \frac{1.15}{\sqrt{40}} = 0.2245$