An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean output of 495
watts with a standard deviation of 12
watts.
What is the probability that the mean amplifier output would be greater than 496.1
watts in a sample of 88
amplifiers if the claim is true? Round your answer to four decimal places.
To calculate this probability, we need to use the z-score formula and then look up the corresponding z-score in a standard normal distribution table.
First, calculate the z-score:
z = (X - μ) / (σ / sqrt(n))
z = (496.1 - 495) / (12 / sqrt(88))
z = 1.1 / (12 / 9.38083152)
z = 1.1 / 1.27613399
z = 0.862828053
Next, look up the z-score of 0.8628 in the standard normal distribution table. The corresponding probability is approximately 0.8061.
Therefore, the probability that the mean amplifier output would be greater than 496.1 watts in a sample of 88 amplifiers if the claim is true is approximately 0.8061.