Question

An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean output of 482 watts with a variance of 121. What is the probability that the mean amplifier output would be less than 479.4 watts in a sample of 59 amplifiers if the claim is true? Round your answer to four decimal places.

Answer 1

Answer

given,

mean output (μ) = 482 watt

variance = 121

standard deviation = √121 = 11

the probability that the mean amplifier output would be less than 479.4 watts

sample = 59

standard error = $\dfrac{standard\ deviation}{\sqrt{59}}$

                        =  $\dfrac{11}{\sqrt{59}}$

                        = 1.43

z =  $\dfrac{x - \mu}{1.43}$

z =  $\dfrac{479.4 - 482}{1.43}$

z = -1.81

we have to find probability of  z > -1.81

probability for z > -1.81 = 1 - 0.0351

                                      = 0.9649