Question

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 83 months with a variance of 81. If the claim is true, what is the probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors? Round your answer to four decimal places.

Answer 1

0.9922 = 99.22% probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors.

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.

The production manager claims they have a mean life of 83 months with a variance of 81.

This means that
`\mu = 83, \sigma = √(81) = 9`

Sample of 146:

This means that
`n = 146, s = (9)/(√(146))`

What is the probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors?

This is 1 subtracted by the p-value of Z when X = 81.2. So

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

By the Central Limit Theorem

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

`Z = (81.2 - 83)/((9)/(√(146)))`

`Z = -2.42`

`Z = -2.42` has a p-value of 0.0078.

1 - 0.0078 = 0.9922.

0.9922 = 99.22% probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors.