00:01
Alright, for this question, we are told that the mean of some fish is 49 .9 centimeters with a seren deviation of 3 .74 centimeters and that it's normally distributed.
00:14
We're going to take a sample of 15 fish.
00:17
So the random variable in this case would be x bar.
00:23
Okay, x bar, the mean length of our 15 fish.
00:32
From our sample, that is the random variable.
00:45
Okay? so that's a.
00:50
For b, find the mean of the sample mean.
00:55
So what that means is the mean of the sampling distribution.
01:02
Okay.
01:07
And the mean of the sampling distribution is equal to the mean of the population mu, and that's always going to be true always.
01:13
Okay? so that's the 49 .9 centimeters.
01:18
For c, the standard deviation, of the sample mean is called the standard error and standard error is computed as sigma over the square root of n so that is 3 .74 over the square root of 15 which is all right for this question 0 .966 if i go to three standard deviation or three decimals.
01:52
For d, the distribution of the sample mean is normally distributed with a mean of 49 .9 and standard deviation of 0 .966.
02:04
And we know that it's going to be normally distributed because the population is normally distributed.
02:10
So we know that if the population is normal, then the sampling distribution is also normal, regardless of n.
02:24
And that is not because of the central limit theorem.
02:26
That's just because of that's how sampling distributions work...