00:01
Here's a solution to 9 .7.
00:01
We're given two samples of size 100, and we're given that the mean is 16, and then the standard deviation is 5 of the population, and then the mean of the second population is 12 and the standard deviation is 4.
00:14
So we're supposed to find the mean and standard deviation of the two sample means.
00:19
And so the mean of the sample mean by the central limit theorem is just the population mean for mu of x1 bar.
00:26
So that's just going to be 16.
00:28
Now the standard deviation is, the population standard deviation divided by the square root of the sample size.
00:34
So in this case, it's going to be 5 over 10 or 1⁄2.
00:39
All right, same thing for, i forgot to change these to a 2, but same thing for the second distribution or second population.
00:47
So the mean would be 12, and then the standard deviation would be 4 divided by the square root of that sample size.
00:54
So that's 4 over 10, which is 2 5ths, or you can just put it as 0 .4.
00:59
So those are the mean and standard deviations of the sampling distributions of x1 bar and x2 bar...