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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.

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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 =
(standard\ deviation)/(√(59))

=
(11)/(√(59))

= 1.43

z =
(x - \mu)/(1.43)

z =
(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

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