00:01
Ok, so for the first part, we just need to show that k equals a ninth.
00:04
Now, the way we do that is that we demand that since it's a probability function, the sum of the probability or the integral in this case, because it's continuous and the integral of the probability to be anywhere must be one.
00:18
Now, this is split up for this pdf between zero and three, where we get k x squared minus two x plus two.
00:26
And between 3 and 4 where we get 3k.
00:33
Integrating this we find that we get k times x cubed over 3 minus x squared plus 2x evaluated between 0 and 3 and 3k x evaluated between 3 and 4.
00:49
When x equals 0 this is just going to vanish and so we're just going to get the contribution when when x equals 3, which gives us 9 minus 9 plus 6, which is 6k.
00:58
And then here, this is just going to give us 3k times 4 minus 3.
01:03
4 minus 3 is 1, so we just get 3k.
01:05
So we get 9k.
01:07
Remember, that had to equal 1.
01:10
And so k is equal to 1 over 9.
01:15
Part b says find the cumulative distribution function...