00:01
For this problem, to begin, i'll note that since we have a lead time of two weeks, to determine if we will exceed the reorder point when the reorder quantity is set at 143 units, that would be equivalent to saying that the sample mean for those two weeks is greater than or equal to 143 over 2.
00:29
Oops, 143 over 2.
00:32
Oops, i'm fighting with my computer here.
00:35
Okay, so we're effectively looking for the probability that x bar is greater than 71 .5, where we know that since we have that the population is normally distributed with an average of 51 units per week and a standard deviation of 13 units, the sample means, and now this specifically comes from the fact that we know the population is normally distributed, this isn't an example of the central limit theorem per se, but we know that the standard deviation of the sample means then will be equal to the population standard deviation divided by the square root of the sample size, where the sample size is just 2.
01:14
So we have a standard deviation for the sampling distribution of 9 .19.
01:19
So the probability of x bar greater than 71 .5 is going to be equal to the probability of, or the proportion of, the standard normal distribution to the right of the z -score corresponding to 71 .5.
01:31
So that's 71 .5 minus mu x bar divided by sigma x bar, or when we plug in our values, that's 71 .5 minus 51 divided by 9 .19.
01:43
So that is a z -score of 2 .23.
01:48
The proportion, oops, 2 .23...