The quality control manager at a computer manufacturing company believes that the mean life of a computer is 83 months, with a standard deviation of 9 months. If he is correct, what is the probability that the mean of a sample of 78 computers would be less than 82.06 months? Round your answer to four decimal places.
Solution
Step 1 ：We are given that the population mean (\(\mu\)) is 83 months, the population standard deviation (\(\sigma\)) is 9 months, and the sample size (\(n\)) is 78. We are asked to find the probability that the sample mean is less than 82.06 months.
Step 2 ：We can use the formula for the z-score, which is \((X - \mu) / (\sigma / \sqrt{n})\), where X is the sample mean. The z-score tells us how many standard deviations an element is from the population mean.
Step 3 ：Substituting the given values into the z-score formula, we get a z-score of approximately -0.9224.
Step 4 ：Once we have the z-score, we can use a z-table or a statistical function to find the probability. The probability corresponding to this z-score is approximately 0.1782.
Step 5 ：Final Answer: The probability that the mean of a sample of 78 computers would be less than 82.06 months is \(\boxed{0.1782}\).
