An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean output of 364364 watts with a standard deviation of 1212 watts. What is the probability that the mean amplifier output would be greater than 364.8364.8 watts in a sample of 5252 amplifiers if the claim is true? Round your answer to four decimal places.

We have to find the probability that the mean output for 52 randomly selected amplifiers will be greater than 364.8. Since the population is Normally Distributed and we know the value of population standard deviation, we will use the z-distribution to solve this problem.

We will convert 364.8 to its equivalent z-score and then finding the desired probability from the z-table. The formula to calculate the z-score is:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

x=364.8 converted to z score for a sample size of n= 52 will be:

[tex]z=\frac{364.8-364}{\frac{12}{\sqrt{52} } }=0.48[/tex]

This means, the probability that the output is greater than 364.8 is equivalent to probability of z score being greater than 0.48.

i.e.

P( X > 364.8 ) = P( z > 0.48 )

From the z-table:

P( z > 0.48) = 1 - P(z < 0.48)

= 1 - 0.6844

= 0.3156

Since, P( X > 364.8 ) = P( z > 0.48 ), we can conclude that:

The probability that the mean amplifier output would be greater than 364.8 watts in a sample of 52 amplifiers is 0.3156