Question

The effective life of a component used in an aircraft engine is normally distributed with mean 5000 hours and standard deviation 40 hours. The engine manufacturer introduces an improvement into the manufacturing process for this component that increases the mean life to 5050 hours and decreases the standard deviation to 30 hours. Suppose that a random sample of 16 components is selected from the "old" process and a random sample of 25 components is selected from the "improved" process. Assume that the "old" and "improved" processes can be regarded as inde- pendent populations. i. What is the probability that the mean life of the "improved" process is between 5045 and 5100 hours? ii. What is the probability that the difference between the mean life of the "im- proved" process and the mean life of the "old" process is more than 25 hours?

Answer 1

i. The probability that the mean life of the improved process is between 5045 and 5100 hours. We know that the mean and the standard deviation of the improved process are 5050 and 30 hours, respectively. Now, we need to find the z-values for the lower and upper bounds using the formula, z = (x - μ) / σ.

z for 5045 hours = (5045 - 5050) / 30 = -0.1667
z for 5100 hours = (5100 - 5050) / 30 = 1.6667

Using a standard normal distribution table, we find the probabilities associated with the z-values.

P(z < -0.1667) = 0.4332
P(z < 1.6667) = 0.9525

Now, to find the probability that the mean life of the improved process is between 5045 and 5100 hours, we subtract the probability associated with the lower bound from the probability associated with the upper bound.

P(5045 < x < 5100) = P(z < 1.6667) - P(z < -0.1667) = 0.5193

Therefore, the probability that the mean life of the improved process is between 5045 and 5100 hours is 0.5193.

ii. The probability that the difference between the mean life of the improved process and the mean life of the old process is more than 25 hours. We know that the mean and the standard deviation of the old process are 5000 and 40 hours, respectively. We need to find the mean and the standard deviation of the sampling distribution of the difference in means.

μ(difference) = μ(improved) - μ(old) = 5050 - 5000 = 50
σ(difference) = sqrt(σ(improved)^2/n(improved) + σ(old)^2/n(old))
= sqrt(30^2/25 + 40^2/16) = 12.40

Now, we need to find the z-value for the difference between the means using the formula, z = (x - μ) / σ.

z = (25 - 50) / 12.40 = -2.02

Using a standard normal distribution table, we find the probability associated with the z-value.

P(z < -2.02) = 0.0228

Now, to find the probability that the difference between the mean life of the improved process and the mean life of the old process is more than 25 hours, we subtract the probability associated with the z-value from 1.

P(difference > 25) = 1 - P(z < -2.02) = 0.9772

Therefore, the probability that the difference between the mean life of the improved process and the mean life of the old process is more than 25 hours is 0.9772.

Conclusion: In conclusion, the probability that the mean life of the improved process is between 5045 and 5100 hours is 0.5193. The probability that the difference between the mean life of the improved process and the mean life of the old process is more than 25 hours is 0.9772.