An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean output of 364 watts with a standard deviation of 12 watts. What is the probability that the mean amplifier output would be greater than 364.8 watts in a sample of 52 amplifiers if the claim is true? Round your answer to four decimal places.

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

P(X>364.8) = 0.6844

Step-by-step explanation:

Step(i):-

Given mean of the Population = 364 watts

Given standard deviation of the Population = 12 watts

Let 'X ' be the Random variable in Normal distribution

x⁻ = 364.8

Given sample size 'n' = 52

Standard error (S.E) =  σ/√n = 1.664

Z = 0.481

Step(ii):-

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

P(X>364.8) = P(Z>0.481)

                   = 1- P( Z<0.481)

                  =  1- (0.5 -A(0.481)

                 = 0.5 + A(0.481)

                = 0.5 + 0.1844

              =  0.6844

Final -

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

P(X>364.8) = 0.6844

plz yuo

step-by-step explanation:

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!