00:01
So for the part a, we have a figure.
00:07
Looks like this.
00:10
So basically we're getting some symmetric point.
00:17
This is negative 1 .28.
00:20
This is 1 .28.
00:22
Actually, we can use the previous techniques that we first get this z1 and z2.
00:33
And then we use the pz2 minus.
00:36
Pza 1 to get the area.
00:39
This is one solution.
00:40
That was not convenient enough because you need to check two values.
00:47
Note that in here, this one is a symmetric.
00:54
So we can use the property.
00:59
Suppose both of them are this is a.
01:02
So we know that these two parts of the area that are the same.
01:07
But for this part area, it should be 1 minus pz.
01:12
Okay.
01:13
So this one.
01:13
One should also be 1 minus p z.
01:17
Once we have this two area, we can just calculate the middle parts by using one and subtract them from 1.
01:31
And this would be 2 p z minus 1.
01:36
So now we just need to check one value.
01:43
It's more efficient actually.
01:45
So previously we checked two values, but now we just need to check one values.
01:49
Alright, so in this example, z is 1 .28, where we can get the p .z value is 0 .897.
02:06
So the area would be 2p .0 .0 .794...