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:

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

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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 [tex]\mu = 429, \sigma = 10[/tex]

Sample of 76:

This means that [tex]n = 76, s = \frac{10}{\sqrt{76}} = 1.147[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{432.1 - 429}{1.147}[/tex]

[tex]Z = 2.7[/tex]

[tex]Z = 2.7[/tex] 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.

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