Question

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 75 months with a standard deviation of 5 months. If the claim is true, what is the probability that the mean monitor life would be greater than 74.4 months in a sample of 85 monitors

Answer 1

Answer:

86.65% probability that the mean monitor life would be greater than 74.4 months in a sample of 85 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

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 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 $s = \frac{\sigma}{\sqrt{n}}$.

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

In this question:

Population: $\mu = 75, \sigma = 5$

Sample of 85: $n = 85, s = \frac{5}{\sqrt{85}} = 0.5423$

If the claim is true, what is the probability that the mean monitor life would be greater than 74.4 months in a sample of 85 monitors

This is 1 subtracted by the pvalue of Z when X = 74.4. So

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

By the Central Limit Theorem

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

$Z = \frac{74.4 - 75}{0.5423}$

$Z = -1.11$

$Z = -1.11$ has a pvalue of 0.1335

1 - 0.1335 = 0.8665

86.65% probability that the mean monitor life would be greater than 74.4 months in a sample of 85 monitors