Question

The time required for the first stage of an assembly line is normally distributed with a mean of 28 minutes and variance of 5 minutes, while the time required for an independent second stage is normally distributed with a mean of 17 minutes with a variance of 4 minutes. what is the probability that the two stages take more than 50 minutes?

Answer 1

The mean for both processes is 28+17 = 45 minutes.
The variance is additive, so the total variance is 5+4 = 9. The standard deviation is the square root of variance, which is 3 minutes.
Then the z-score associated with a time of x = 50 minutes is:
z = (x - mean)/SD = (50 - 45)/3 = 1.67
From a z-table, the probability that (z > 1.67) is 0.0475.