Question

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 82 months with a standard deviation of 7 months. If the claim is true, what is the probability that the mean monitor life would be greater than 83.8 months in a sample of 71 monitors

Answer 1

Answer:

0.015 = 1.5% probability that the mean monitor life would be greater than 83.8 months in a sample of 71 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.

Mean life of 82 months with a standard deviation of 7 months.

This means that $\mu = 82, \sigma = 7$

Sample of 71

This means that $n = 71, s = \frac{7}{\sqrt{71}}$

What is the probability that the mean monitor life would be greater than 83.8 months?

1 subtracted by the p-value of Z when X = 83.8. So

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

By the Central Limit Theorem

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

$Z = \frac{83.8 - 82}{\frac{7}{\sqrt{71}}}$

$Z = 2.17$

$Z = 2.17$ has a p-value of 0.985.

1 - 0.985 = 0.015

0.015 = 1.5% probability that the mean monitor life would be greater than 83.8 months in a sample of 71 monitors