00:01
For this problem, the height of 18 -year -old men are approximately normally distributed with a mean of 68 inches and a standard deviation of 3 inches.
00:14
For problem a, we want to find the probability that an 18 -year -old man selected at random has a height between 67 and 69.
00:25
So the probability that x is in between 67 and and 7.
00:32
And 69.
00:35
So i have my normal distribution.
00:40
The center is the average 68 and i want to find the probability between 67 and 69.
00:53
So i want the area in between these two values.
01:01
Now to find the area, i had to first find the c score of 69 and the c score of 67.
01:09
So for the 69, i'm going to do 69 minus the average 68 divided by standard deviation of 3, and that equals a z score of .33.
01:23
And then for 67, i'm going to do 67 minus 68 divided by standard deviation 3, and that equals negative .33.
01:37
To find the area in between, these two z scores, i need to to find the area to the left of 0 .33 and the area to the left of negative .33.
01:46
I'm going to use my table.
01:48
So for negative .33, it's going to be .37, the area to the left.
01:58
And for positive .33, positive .33, it's going to be .6293.
02:10
So the area was .6293.
02:17
The area to the left of 0 .33, and then the area to the left of the negative 0 .33 was 0 .37.
02:26
So i want to know the area in between of the 67 and 69, so i want to need to subtract these two areas, and the difference is 0 .2586.
02:41
So the probability that a random man, selected man, from this population is in between 67 and 69 is 0 .256.
02:57
Now for part b, we want to find the probability that a random sample of size 9, so n equals 9 of 18 year old men, is selected at random, and the mean height of this group of 9 is between 67 and 69.
03:21
So, we'll we're still finding the probability that it is in between 67 and 69, but now we have a group of 9.
03:32
And so we're selecting from the same population, right? so the population was normal with an average of 60a and a standard deviation of 3.
03:42
But because we're now dealing with a group of 9, we need to compute the sampling distribution data.
03:48
And so the sampling distribution is still going to be normal with the same mean of 60...