0:00
We have a normal population.
00:02
So this is for, i'll call it individuals.
00:06
So for just one unit at a time, and the mean is 60, and the standard deviation is 12.
00:13
And then we're going to take samples of size 9.
00:17
And so that distribution will also be normal, and it will center at a different spot.
00:24
And it will have less dispersion.
00:27
It will still center and have the mean of the x bars would still be, 60, but the standard deviation of x bars will be that 12 divided by the square root of 9, which will end up being 12 divided by 3, which is 4.
00:41
And so we want to find in part a, what is the likelihood that an x bar? so again, this is called the sampling distribution of x bars.
00:53
Each of the x bars is coming from a sample size of nine, and that will have less dispersion than the original.
00:59
And what is the likelihood that an x bar? will end up being greater than 63 and so 63 approximately right there we convert that to a z value and we'll have the 63 minus 60 divided by that 12 divided by the square root of 9 which is 4 so we're going to get in 3 4s or z is greater than 0 .75 standard deviations and i'm actually going to look up the area below negative 0 .75 to get that answer and that comes that to be 0 .2266...