Mathematics, 28.05.2021 01:20 1tzM3

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 donâ€™t play interscholastic sports (weâ€™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

Answers: 1

Show answers

Answers

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 from: Quest

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.

Answer from: Quest

true

step-by-step explanation: