Question

An operation manager at an electronics company wants to test their amplifiers. the design engineer claims they have a mean output of 256 watts with a standard deviation of 10 watts. what is the probability that the mean amplifier output would be greater than 254.7 watts in a sample of 53 amplifiers if the claim is true? round your answer to four decimal places.

Answer 1

The probability that the mean amplifier output would be greater than 254.7 watts in a sample of 53 amplifiers, if the claim is true, is approximately 0.2063.

Step-by-step explanation:

To find the probability that the mean amplifier output would be greater than 254.7 watts in a sample of 53 amplifiers if the claim is true, we can use the Central Limit Theorem. According to the theorem, the distribution of sample means will approach a normal distribution regardless of the shape of the population distribution.

First, we need to calculate the standard error of the mean, which is the standard deviation of the population divided by the square root of the sample size. In this case, the standard error of the mean (SE) would be 10/sqrt(53) = 1.375.

Next, we can calculate the z-score using the formula (X - µ) / SE, where X is the sample mean, µ is the population mean, and SE is the standard error of the mean. Plugging in the values, we get (254.7 - 256) / 1.375 = -0.8182.

Now, we can use a standard normal distribution table or a calculator to find the probability associated with the z-score. The probability can be found by looking up the z-score -0.8182, which is approximately 0.2063.

Therefore, the probability that the mean amplifier output would be greater than 254.7 watts in a sample of 53 amplifiers, if the claim is true, is approximately 0.2063.