Question

A manufacturer knows that their items have a normally distributed length, with a mean of 18.2 inches, and standard deviation of 3.9 inches. If 2 items are chosen at random, what is the probability that their mean length is less than 21.9 inches

Answer 1

0.9099 = 90.99% probability that their mean length is less than 21.9 inches.

Explanation:

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

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 = (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 = (\sigma)/(√(n))`.

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

Mean of 18.2 inches, and standard deviation of 3.9 inches.

This means that
`\mu = 18.2, \sigma = 3.9`

2 itens:

This means that
`n = 2, s = (3.9)/(√(2))`

What is the probability that their mean length is less than 21.9 inches?

This is the p-value of Z when X = 21.9. So

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

`Z = (21.9 - 18.2)/((3.9)/(√(2)))`

`Z = 1.34`

`Z = 1.34` has a p-value of 0.9099.

0.9099 = 90.99% probability that their mean length is less than 21.9 inches.