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

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

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)=(\sigma)/(√(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((\bar X-\mu_(\bar x))/(\sigma_(\bar x))>(449.8-451)/(√(144/76)))\\\\=P(Z>-0.87)\\\\=P(Z<0.87)\\\\=0.8078`

*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.