00:01
For this question, the mean and standard deviation were given here.
00:05
So the mean, which is denoted by mu, that is equal to 82 .9.
00:10
And the standard division, this is denoted by sigma.
00:16
And that was equal to 12 .8 here.
00:20
So first of all, we need to get the z score for 118 .5.
00:24
So i can define the random variable x.
00:26
And the z score, remember, this is equal to x minus mu over standard division.
00:31
So for x is equal to 118 .5, the z score would be, which is 118 .5 and minus the mean, and divided by the standard division.
00:45
This is 12 .8, which is 118 .5 minus 82 .9, which is divided by 12 .8, and that should be 2 .78.
01:01
This is the z score we have.
01:04
So significant high values are, so for the high values, if the z score is greater than 2, we just define this as the high value, because this is more than the critical value here.
01:18
So if the z score is 2, let's accept, so for let z is equal to 2, what is the x value here? so if i plug in the value here, which is x minus the mean and divided by the standard division, 12...