Question

An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean output of 429 watts with a standard deviation of 10 watts. What is the probability that the mean amplifier output would be greater than 432.1 watts in a sample of 76 amplifiers if the claim is true

Answer 1

0.0035 = 0.35% probability that the mean amplifier output would be greater than 432.1 watts.

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 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.

Mean output of 429 watts with a standard deviation of 10 watts.

This means that
`\mu = 429, \sigma = 10`

Sample of 76:

This means that
`n = 76, s = (10)/(√(76)) = 1.147`

What is the probability that the mean amplifier output would be greater than 432.1 watts?

This is 1 subtracted by the pvalue of Z when X = 432.1. So

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

By the Central Limit Theorem

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

`Z = (432.1 - 429)/(1.147)`

`Z = 2.7`

`Z = 2.7` has a pvalue of 0.9965

1 - 0.9965 = 0.0035

0.0035 = 0.35% probability that the mean amplifier output would be greater than 432.1 watts.