Question

An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean output of 451451 watts with a variance of 144144. What is the probability that the mean amplifier output would be greater than 449.8449.8 watts in a sample of 7676 amplifiers if the claim is true?Round your answer to four decimal places.

Answer 1

Answer:

The probability that the mean amplifier output would be greater than 449.8 watts in a sample of 76 amplifiers is 0.8078.

Step-by-step explanation:

According to the Central Limit Theorem if an unknown population is selected with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from this population with replacement, then the distribution of the sample means will be approximately normally.  

Then, the mean of the sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The information provided is as follows:

$\mu=451\\\sigma^{2}=144\\n=76\\\bar x=449.8$

Compute the probability that the mean amplifier output would be greater than 449.8 watts in a sample of 76 amplifiers as follows:

$P(\bar X>449.8)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{449.8-451}{\sqrt{144/76}})\\\\=P(Z>-0.87)\\\\=P(Z$

*Use a z-table.

Thus, the probability that the mean amplifier output would be greater than 449.8 watts in a sample of 76 amplifiers is 0.8078.