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.