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=σ_{x̅}=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 |

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