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 less than 0.6 W in a sample of 77 amplifiers if the claim is true?

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	118
std deviation   =σ=	11.000
sample size       =n=	77
std error=σx̅=σ/√n=	1.25

P( probability that the mean amplifier output would differ from the population mean by less than 0.6 W i

probability =P(117.4<X<118.6)=P((117.4-118)/1.254)<Z<(118.6-118)/1.254)=P(-0.48<Z<0.48)=0.6844-0.3156=0.3688