00:01
There are three parts to this problem.
00:02
So with part a, we have the mean daily takings for a store are $9 ,21515 with a standard deviation of 1 ,568, and the takings are normally distributed.
00:19
So as soon as i hear those words, i'm going to draw that bell -shaped curve, and i'm going to place the mean of 9 ,215 right in the center.
00:30
And i'm just going to record my standard deviation off to the side.
00:36
And for part a, we want to calculate the probability that the next day's taking is at least $9 ,000.
00:45
So $9 ,000 would be like right here.
00:51
And we want that to be the lowest, so we're looking for being greater than that value.
00:56
So the probability that the takings are greater than or equal to 9 ,000 is going to be comparable to what we get when we look at the z score associated with 9 ,000.
01:09
So we're going to find the z score for $9 ,000 by applying the z score formula x minus mu over sigma.
01:17
So we'll do 9 ,000 minus the average of 9 ,21515 over the standard deviation of 1 ,568, and we will get a z score of negative 0 .1 .1 .1 .518.
01:30
And we will get a z score of negative 0 .1 so being greater than 9 ,000 is the same as being greater than a z score of negative 0 .14.
01:44
So what we're going to need to do is use the standard normal table in the back of your textbook.
01:50
And what you'll do is you'll look for the units place and the tens or sorry, the units place and the tens place along the left side of your distribution chart.
02:02
And you'll look for the hundredth digit across the top.
02:06
And when you do that, you're going to find an area of 0 .4443.
02:12
And that area is going to be descriptive of the area that's in the left tail, to the left of that boundary, not to the right.
02:24
So for us to find our green shaded area, we have to use the concept that the entire bell is equivalent to one.
02:32
So when i add the left and the right together, they have to total up to one.
02:38
So for me to solve this problem, i'm going to have to do 1 minus the probability that z was less than negative 0 .14, or in essence, 1 minus the 0 .4443 that you got out of the table.
02:54
And the probability is going to be 0 .5557.
03:01
In part b, we're going to use that same exact picture.
03:09
We're still going to have our mean in the center, 9 ,215, and we're still going to recognize that our standard deviation is going to be 1 ,568.
03:22
But in part b, we want to calculate the probability that the next day's staking is less than 10 ,500.
03:34
So 10 ,500 is going to be to the right of the mean, and we're doing the probability that the takings are less than 10 ,500.
03:48
So again, we're going to calculate our z score that corresponds to 10 ,500 by doing 10 ,500 minus the mean, divide by the standard deviation, and you will get a z score of 0 .8 ,000.
04:07
So being less than 10 ,500 is no different than the z score being less than 0 .82...