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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 113 watts with a variance of 100 . what is the probability that the mean amplifier output would differ from the population mean by less than 2.7 watts in a sample of 43 amplifiers if the claim is true? round your answer to four decimal places.

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Final answer:

To find the probability, calculate the standard error of the mean, determine the z-score, and find the probability using the z-score.

Step-by-step explanation:

To find the probability that the mean amplifier output in a sample of 43 amplifiers differs from the population mean by less than 2.7 watts, we can use the normal distribution and the given variance. Since the design engineer claims the mean output is 113 watts with a variance of 100, we can assume a normal distribution with a mean of 113 and a standard deviation of √100 = 10. Using the formulas for the standard error of the mean and the z-score, we can calculate the probability.

Step 1: Calculate the standard error of the mean.
SE = standard deviation / √sample size.
SE = 10 / √43 = 1.52

Step 2: Calculate the z-score.
z = (sample mean - population mean) / standard error.
z = (113 - 113) / 1.52 = 0

Step 3: Find the probability using the z-score.
Using a z-table or a calculator, find the probability of a z-score of 0 or less.
For a z-score of 0, the probability is 0.5000.

Therefore, the probability that the mean amplifier output would differ from the population mean by less than 2.7 watts in a sample of 43 amplifiers, if the claim is true, is 0.5000.

User Axel Podehl
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