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.

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,

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

The information provided is as follows:

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

*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.

for b its (t) and c its (d)

step-by-step explanation:

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent