Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 81 months, with a variance of 64. If he is correct, what is the probability that the mean of a sample of 60 computers would be less than 82.47 months? Round your answer to four decimal places.

Answer 1

To find the probability that the mean of a sample of 60 computers would be less than 82.47 months, we need to standardize the sample mean using the z-score formula and use a standard normal distribution table or calculator.

Step-by-step explanation:

To find the probability that the mean of a sample of 60 computers would be less than 82.47 months, we need to standardize the sample mean using the z-score formula. The formula for the z-score is:

z = (x - μ) / (σ / √n)

Where:

• x is the sample mean (82.47)
• μ is the population mean (81)
• σ is the population standard deviation (√64 = 8)
• n is the sample size (60)

Plugging in the values, we get:

z = (82.47 - 81) / (8 / √60) = 0.216

Using a standard normal distribution table or calculator, we can find the probability associated with this z-value. The probability that the mean of a sample of 60 computers would be less than 82.47 months is approximately 0.5864 (rounded to four decimal places).