Question

An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean output of 118 watts with a variance of 121 . What is the probability that the mean amplifier output would differ from the population mean by greater than 0.6 watts in a sample of 77 amplifiers if the claim is true? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 118

standard deviation = = 11

n = 77

= = 118

= / n = 11 / 77 = 1.2536

P(117.4 < < 118.6) = P((117 - 118) / 1.2536<( - ) / < (118.6 - 118) / 1.2536))

=1 - (P(-0.48 < Z <0.48))

= 1 - ( P(Z < 0.48) - P(Z < -0.48))

= 1 - (0.6844 - 0.3156)

= 1 - 0.3688

= 0.6312

The probability is 0.6312