00:01
For this question, we know that a manufacturer knows their items have lengths that are approximately normally distributed.
00:10
The mean, which i'll call mu, is given by 17 .3 inches, and the standard deviation, which i'll call sigma, is given by 4 .2 inches.
00:27
If we take 30 items chosen at random, so say n is equal to 30, we want to find the probability that their mean length is greater than 15 .9 inches.
00:42
So i'll call the variable for the sampling distribution x -bar.
00:49
So x -bar will be the average length of a sample of size 30 randomly selected from this population, in which case we're interested in calculating the probability that this mean length x -bar is greater than 15 .9.
01:09
So to calculate this probability, we need to work out a couple things first.
01:14
First, we need the mean of x -bar.
01:18
Very generally, this is just given by the population mean, so for us that's 17 .3 inches once again, and the standard deviation of x -bar, or standard error, that's given by the population standard deviation sigma divided by the square root of your sample size n, which for us that's going to be 4 .2 divided by 30.
01:44
And if we want to know what this is approximately as a decimal, let me just plug this into a calculator real quick.
01:53
So 4 .2 over the square root of 30.
02:00
Oops, i didn't include the square root here, let me put that in.
02:04
So 4 .2 over the square root of 30 is 0 .7668, but i will be using the exact answer when i plug things into my calculator.
02:17
Okay, so now we have the statistics of x -bar here.
02:23
Let's draw a little normal curve, because since these lengths are coming from a population which is itself normal, we can assume that the means of the lengths are also normal.
02:44
So our sampling distribution is normal, and we can draw it like this.
02:51
So the mean in the middle here is 17 .3, and this curve is supposed to be symmetric about the mean with total area under the curve equal to 1...
