Question

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?

Answer 1

`z= (1.4-1.47)/((0.3)/(√(30)))= -1.278`

`z= (1.5-1.47)/((0.3)/(√(30)))= 0.548`

And we can find the probability with this difference:

`P(-1.278<z<0.548) = P(z<0.548) -P(z<-1.278) =0.708-0.101= 0.607`

So then the probability that the sample mean is between $1.4 million and $1.5 million per year is 0.607

Explanation:

For this case we have the following info given:

`\mu = 1.47` the true mean for the problem

n =30 represent the sample size

`\sigma = 0.3 millions` represent the population deviation

And we want to find this probability

`P(1.4< \bar X <1.5)`

And we can use the z score given by:

`z= (\bar X -\mu)/((\sigma)/(√(n)))`

And if we find the z scores for the limits we got:

`z= (1.4-1.47)/((0.3)/(√(30)))= -1.278`

`z= (1.5-1.47)/((0.3)/(√(30)))= 0.548`

And we can find the probability with this difference:

`P(-1.278<z<0.548) = P(z<0.548) -P(z<-1.278) =0.708-0.101= 0.607`

So then the probability that the sample mean is between $1.4 million and $1.5 million per year is 0.607

Answer 2

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

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
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (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 = (\sigma)/(√(n))`.

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.

`\mu = 1.47, \sigma = 0.3, n = 30, s = (0.3)/(√(30)) = 0.0548`

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

`Z = (X - \mu)/(\sigma)`

`Z = (1.5 - 1.47)/(0.0548)`

`Z = 0.55`

`Z = 0.55` has a pvalue of 0.7088

X = 1.4

`Z = (X - \mu)/(\sigma)`

`Z = (1.4 - 1.47)/(0.0548)`

`Z = -1.28`

`Z = -1.28` 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