Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 102 months, with a variance of 100. If he is correct, what is the probability that the mean of a sample of 82 computers would be less than 105.29 months? Round your answer to four decimal places.

Answer 1

To find the probability that the mean of a sample of 82 computers would be less than 105.29 months, we can use the z-score and standard normal distribution table. We get, approximately 0.9726.

Step-by-step explanation:

To find the probability that the mean of a sample of 82 computers would be less than 105.29 months, we need to calculate the z-score and use the standard normal distribution table.

The z-score is calculated as:

z = (x - μ) / (√(σ^2 / n))

Where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, the population mean (μ) is 102 months, the population variance (σ^2) is 100, and the sample size (n) is 82.

Plugging these values into the formula, we have:

z = (105.29 - 102) / (√(100 / 82))

Solving for z, we get z = 1.924. Referencing the standard normal distribution table, we find that the probability of getting a z-score less than 1.924 is approximately 0.9726.

Therefore, the probability that the mean of a sample of 82 computers would be less than 105.29 months is approximately 0.9726.