Question

At a large state university, the heights of male students who are interscholastic athletes is approximately Normally distributed with a mean of 74.3 inches and a standard deviation of 3.5 inches. The heights of male students who don’t play interscholastic sports (we’ll call them "non-interscholastics") is approximately Normally distributed with a mean of 70.3 inches and a standard deviation of 3.2 inches. You select an SRS of 10 interscholastic athletes and 12 non-interscholastics. What is the probability that the sample mean of non-interscholastics is greater than the sample mean of interscholastic athletes

Answer 1

Answer:

0.0027 = 0.27% probability that the sample mean of non-interscholastics is greater than the sample mean of interscholastic athletes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution, the central limit theorem, and subtraction of normal variables.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

At a large state university, the heights of male students who are interscholastic athletes is approximately Normally distributed with a mean of 74.3 inches and a standard deviation of 3.5 inches. Sample of 10:

This means that $\mu_I = 74.3, s_I = \frac{3.5}{\sqrt{10}}$

Those who are "non-interscholastics" have mean of 70.3 inches and a standard deviation of 3.2 inches. Sample of 12.

This means that $\mu_N = 70.3, s_N = \frac{3.2}{\sqrt{12}}$

What is the probability that the sample mean of non-interscholastics is greater than the sample mean of interscholastic athletes?

This is the probability that:

N - I > 0

Distribution N - I:

$\mu = \mu_N - \mu_I = 70.3 - 74.3 = -4$

$s = \sqrt{s_I^2+s_N^2} = \sqrt{(\frac{3.5}{\sqrt{10}})^2+(\frac{3.2}{\sqrt{12}})^2} = 1.44$

Probability:

One subtracted by the p-value of Z when X = 0.

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

By the Central Limit Theorem

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

$Z = \frac{0 - (-4)}{1.44}$

$Z = 2.78$

$Z = 2.78$ has a p-value of 0.9973

1 - 0.9973 = 0.0027

0.0027 = 0.27% probability that the sample mean of non-interscholastics is greater than the sample mean of interscholastic athletes

Answer 2

The probability for the sample mean of non-interscholastics is greater than the sample mean of interscholastic athletes in state university is 0.0027.

What is normally distributed data?

Normally distributed data is the distribution of probability which is symmetric about the mean.

The mean of the data is the average value of the given data. The standard deviation of the data is the half of the difference of the highest value and mean of the data set.

• At a large state university, the heights of male students who are interscholastic athletes is approximately Normally distributed with a mean of 74.3 inches and a standard deviation of 3.5 inches.
• The heights of male students who don't play interscholastic sports is approximately Normally distributed with a mean of 70.3 inches and a standard deviation of 3.2 inches.

For the above data, the value of mean would be

$m=70.3-74.3\\m=-4$

The selected sample size are 10 interscholastic athletes and 12 non-interscholastics. The value of standard deviation is,

$\sigma=\sqrt{(\dfrac{3.5}{\sqrt{10}})^2+(\dfrac{3.2}{\sqrt{10}})^2}\\\sigma=1.44$

The Z score can be find out using the following formula.

$Z=\dfrac{X-\mu}{\sigma}\\Z=\dfrac{0-(-4)}{1.44}\\Z=2.78$

The z score of 2.78 has the p value equal to 0.9973. For sample mean of non-interscholastics to be greater, the probability,

$P=1-0.9973\\P=0.0027$

Thus, the probability for the sample mean of non-interscholastics is greater than the sample mean of interscholastic athletes in state university is 0.0027.



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