00:01
This problem, we're told that at a coffee shop, the wait time for a cup of plain coffee is normally distributed with a mean of 125 seconds and with the standard deviation of 22 seconds.
00:14
There's four different parts to this problem, so i'll begin here with part a.
00:18
We were asked to find the percentage of wait times for a plain cup of coffee that are less than 90 seconds.
00:24
So here we're finding the probability that x is less than.
00:28
Let me fix up that less than sign.
00:31
Less than 90.
00:33
So our first step here is to find a z score for 90 using this red equation i have on the screen.
00:39
So using the equation we'll have z is equal to x, which is 90, minus the mean, which is 125, and divided by the standard deviation, which is 22.
00:53
That'll give us a z score of negative 1 .59.
00:57
Now that we have our z score, we can find it in our z table here on the left side of the screen.
01:03
We'll go down in the left -hand column to negative 1 .5, which is right here, and over on the right to 0 .09, which is up here.
01:11
And if we meet in the middle, we get 0 .059.
01:18
Now our z table always gives us what is less than what we're trying to find.
01:22
So this probability here is a probability of x being less than 90, and that's what we want.
01:28
We want the probability of x being less than 90.
01:30
So this is our final answer here.
01:33
The question asks for a percentage, so as a percentage, this would be 5 .59%.
01:41
Oops, let me fix that here.
01:46
5 .59%.
01:47
There we go.
01:49
Moving on out to part b, which asks us to find the percentage of weight times that are greater than 125 seconds.
01:57
So now we're finding the probability that x is greater than 125.
02:02
So we can go ahead and find our z score for 125 now.
02:07
So z is equal to x, which is 1 ,000...