00:01
All right, so here we're dealing with a normal distribution of cell phone apps used by people with a mean of 18 and a standard deviation of 3, as you can see that are drawn here.
00:11
The first question that we have is if we have 25 people that we sample, what is the probability of the sample mean or the x bar being between 17 .5 and 18 .5.
00:25
So 17 .5 is going to be on this side of the curve.
00:30
18 .5 is going to be on this side of the curve.
00:34
And so note that they're the same distance away, so they'll have the same z score, except one will be positive and one will be negative.
00:41
So we're going to go ahead and do the z score by doing our x bar minus our mu over sigma over the square root of n.
00:53
So sigma is three squared.
00:56
Of n is 5 because n is 25 so that gives us the following numbers x bar we can just do either 17 .5 or 18 .5 we can just pick one um i'll do 17 .5 so 17 .5 minus the mean of 18 over 3 divided by 5 right because um that's our sigma over our square root of n 3 divided by 5 is 0 .6.
01:26
So we have 0 .6 here on the bottom.
01:33
So we have 17 .5 minus 18 is negative 0 .5, divided by 0 .6 is 0 .5, let's see, divided by 0 .6 is negative .83.
01:52
So that tells us that these values are negative 0 .83 standard deviations above and below.
01:57
And we're looking for the probability that the sample mean is between those two.
02:03
So what i'm going to end up doing is to find that area in between.
02:08
I'll find the area, the z score for negative .83.
02:15
I'll find its p value, and i'll find the p value of .83 for the z, and i'll subtract them.
02:21
So the p value for negative .83, the area below that, minus the area below 0 .83 when i look that up on the table.
02:38
So i look each of those up on the table, my z values of 0 .83 and negative 0 .83.
02:44
And for my p value, then, i end up with about 0 .6.
02:51
So i think we need to go to three decimal places.
02:57
So 1 minus 0 .4047...