An NCAA study reported that the average salary of the 300 major college football coaches is $1.47 million. Using a random sample of 30 coaches and a population standard deviation of $300,000, what is the probability that the sample mean is between $1.4 million and $1.5 million per year?

60.85% probability that the sample mean is between $1.4 million and $1.5 million per year

Step-by-step explanation:

To solve this question, 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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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

In this question, we have that:

In millions of dollars.

What is the probability that the sample mean is between $1.4 million and $1.5 million per year?

This is the pvalue of Z when X = 1.5 subtracted by the pvalue of Z when X = 1.4. So

X = 1.5

has a pvalue of 0.7088

X = 1.4

has a pvalue of 0.1003

0.7088 - 0.1003 = 0.6085

60.85% probability that the sample mean is between $1.4 million and $1.5 million per year

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23