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sveticcg [70]
4 years ago
13

An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean outpu

t of 364 watts with a standard deviation of 12 watts. What is the probability that the mean amplifier output would be greater than 364.8 watts in a sample of 52 amplifiers if the claim is true? Round your answer to four decimal places.
Mathematics
1 answer:
Westkost [7]4 years ago
5 0

Answer:

<em>The probability that the mean amplifier output would be greater than 364.8 watts in a sample of 52 amplifiers if the claim is true</em>

<em>P(X>364.8) = 0.6844</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given mean of the Population = 364 watts</em>

<em>Given standard deviation of the Population = 12 watts</em>

<em>Let 'X ' be the Random variable in Normal distribution </em>

<em>x⁻ = 364.8</em>

<em>Given sample size 'n' = 52 </em>

<em />Z = \frac{x-mean}{S.E}<em />

<em>Standard error (S.E) =  σ/√n = 1.664</em>

<em />Z = \frac{364.8-364}{1.664}<em />

Z = 0.481

<u><em>Step(ii):-</em></u>

<em>The probability that the mean amplifier output would be greater than 364.8 watts in a sample of 52 amplifiers if the claim is true</em>

<em>P(X>364.8) = P(Z>0.481)</em>

<em>                    = 1- P( Z<0.481)</em>

                  =  1- (0.5 -A(0.481)

                 = 0.5 + A(0.481)

                = 0.5 + 0.1844

              =  0.6844

<u><em>Final answer:-</em></u>

<em>The probability that the mean amplifier output would be greater than 364.8 watts in a sample of 52 amplifiers if the claim is true</em>

<em>P(X>364.8) = 0.6844</em>

<em />

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