Question

A manufacturer knows that their items have a normally distributed length, with a mean of 7.4 inches, and standard deviation of 0.8 inches. If 18 items are chosen at random, what is the probability that their mean length is less than 7.9 inches? P ( ¯ x < 7.9 ) =

Answer 1

P ( ¯ x < 7.9 ) = 0.9960

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 = 7.4, \sigma = 0.8, n = 18, s = (0.8)/(√(18)) = 0.1886`

If 18 items are chosen at random, what is the probability that their mean length is less than 7.9 inches?

This is the pvalue of Z when X = 7.9.

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

By the Central Limit Theorem

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

`Z = (7.9 - 7.4)/(0.1886)`

`Z = 2.65`

`Z = 2.65` has a pvalue of 0.9960

P ( ¯ x < 7.9 ) = 0.9960