00:01
We're looking at a golf tournament and we have some information on the average number of strokes for the 13th pole.
00:08
Typically, the mean, mu, is 4 .25 and the standard deviation, sigma, is 1 .6.
00:16
We're looking at a sample of 64 golfers.
00:21
So we're not actually looking at the original distribution.
00:23
We're taking a sample of 64 and we're looking at the average number of strokes that they need.
00:30
So we're looking at a sample mean.
00:33
So it doesn't matter what the shape of the original distribution is.
00:36
We have a nice big sample here, oversized 30.
00:40
So the sample means are going to be approximately normal.
00:44
So if you take every possible sample of 64 golfers for this particular course, then you plot out the sample means, you get something approximately normal.
00:57
Okay, but we need to know the mean of the means and the standard deviation of the means.
01:01
While the mean of the means, mu x bar, is the same as the original distribution.
01:08
Some groups of golfers will be better than average, some will be worse than average, but the average of the averages should be the same as the population mean.
01:18
The standard deviation, however, of the average number of strokes for a sample of size 64 is going to actually be less than the original.
01:27
And the larger the sample, the less variation between sample means.
01:32
Because small outliers have less than effect.
01:37
And the formula is the standard deviation of sample means is equal to original standard deviation over root n.
01:44
So that's going to be 1 .6 divided by 8, which is 0 .2.
01:50
Okay, now i have an approximately normal distribution.
01:53
I can get rid of this and just look at my new distribution.
01:59
So i want the probability of the sample average being greater than 4 .75...