00:01
So we know that there's approximately 5 % of people take the public transit to work.
00:07
And they said they're going to give a discount to at least 30 people at these companies.
00:13
And we have three conditions.
00:16
We have that the company has 250 people, the company has 500 people, or a company has 1 ,000 people.
00:28
And so we want to see what's the likelihood of getting.
00:32
Greater than are equal to 30 in each of these settings.
00:45
And first of all, we can see that n times p, if we find the mean of each of these, the 250 times .05, the mean number of people is 12 .5 here, and the standard deviation is times .95, and then to the .5, comes out to be 3 .446.
01:13
And so converting this value to, and let's just draw a line down here, creating this value to a z value, and we can quick do the continuity correction.
01:24
We'd have 29 .5 minus that mean divided by that standard deviation.
01:30
Let's see what that ends up being.
01:33
And click store that value.
01:35
So we'd have that 29 .5 minus 30.
01:39
Whoops, minus the 12 .5, divided by that standard deviation.
01:48
And that z value is like 4 .9.
01:52
So z is greater than or equal to 4 .93.
01:55
And this probability is going to be very close to zero.
01:58
So it's very little chance that that's going to happen.
02:01
Now, there is a probability there, but it's just very close to zero.
02:04
So that's not very likely.
02:06
Now, what if we have 500 employees? well, we know this distribution will also be approximately normal.
02:11
And if we take 5 % of 500, the mean number wouldn't be expected to be 25, and the standard deviation would be times 0 .95, and then square root that.
02:26
The standard deviation there has reduced down, or has changed to, gotten bigger than 4 .873.
02:35
And now we'll find that z value.
02:40
Again, i'll use the continuity correction just to be a little or accurate...