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

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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 $\mu = 83, \sigma = \sqrt{81} = 9$

Sample of 146:

This means that $n = 146, s = \frac{9}{\sqrt{146}}$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{81.2 - 83}{\frac{9}{\sqrt{146}}}$

$Z = -2.42$

$Z = -2.42$ 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.