Question

The heights of men in the USA are normally distributed with a mean of 68 inches and a standard deviation of 4 inches. A random sample of five men is selected. What is the probability that the sample mean is greater than 69.5 inches?

Answer 1

`P(\bar X >69.5) =P(Z (69.5-68)/((4)/(√(5)))) = P(Z>0.839)`

And we can use the complement rule with the normal standard table or excel and we got:

`P(Z>0.839) =1-P(z<0.839) = 1-0.7993= 0.2007`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

`X \sim N(68,4)`

Where
`\mu=68` and
`\sigma=4`

We select a sample of size n =5. Since the distribution for X is normal then we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

We want this probability:

`P(\bar X >69.5)`

And we can use the z score formula given by:

`z =(\bar X -\mu)/((\sigma)/(√(n)))`

And using this formula we got:

`P(\bar X >69.5) =P(Z (69.5-68)/((4)/(√(5)))) = P(Z>0.839)`

And we can use the complement rule with the normal standard table or excel and we got:

`P(Z>0.839) =1-P(z<0.839) = 1-0.7993= 0.2007`

Answer 2

20.05% probability that the sample mean is greater than 69.5 inches

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 problem, we have that:

`\mu = 68, \sigma = 4, n = 5, s = (4)/(√(5)) = 1.79`

What is the probability that the sample mean is greater than 69.5 inches?

This is 1 subtracted by the pvalue of Z when X = 69.5. So

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

By the Central Limit Theorem

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

`Z = (69.5 - 68)/(1.79)`

`Z = 0.84`

`Z = 0.84` has a pvalue of 0.7995

1 - 0.7995 = 0.2005

20.05% probability that the sample mean is greater than 69.5 inches