Questions: An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean output of 331 watts with a variance of 144. What is the probability that the mean amplifier output would be greater than 332.8 watts in a sample of 49 amplifiers if the claim is true? Round your answer to four decimal places.

Solution Steps

We are given the following parameters for the amplifier output:

• Population mean (\( \mu \)) = 331 watts
• Population variance (\( \sigma^2 \)) = 144
• Sample size (\( n \)) = 49
• Threshold for sample mean (\( \bar{x} \)) = 332.8 watts

The population standard deviation (\( \sigma \)) is calculated as follows: \[ \sigma = \sqrt{\sigma^2} = \sqrt{144} = 12 \]

To find the probability that the sample mean is greater than 332.8 watts, we first calculate the Z-score for the threshold value: \[ Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{332.8 - 331}{12 / \sqrt{49}} = \frac{1.8}{12 / 7} = \frac{1.8 \cdot 7}{12} = \frac{12.6}{12} = 1.05 \]

The probability that the sample mean is greater than 332.8 watts can be expressed as: \[ P(\bar{X} > 332.8) = P(Z > 1.05) = 1 - P(Z \leq 1.05) \] Using the cumulative distribution function \( \Phi \): \[ P(Z > 1.05) = 1 - \Phi(1.05) \] From the output, we have: \[ P(Z > 1.05) = 0.1469 \]

Final Answer