Let f(x) = k(3x - x^2) if 0 ≤x ≤3 and f(x) = 0 if x 3 . (a) For what value of k is f

Let f(x) = k(3x - x^2) if 0 ≤x ≤3 and f(x) = 0 if x

Let $ f(x) = k(3x - x^2) $ if $ 0 \le x \le 3 $ and $ f(x) = 0 $ if $ x < 0 $ or $ x > 3 $.
(a) For what value of $ k $ is $ f $ a probability density function?
(b) For that value of $ k $, find $ P (X > 1) $.
(c) Find the mean.

Transcript

00:01 So the question is there is a function given here like that if x is between 0 and 3, the function takes this and otherwise the function is 0.

00:14 And it says that the question is a probability dense function, pdf.

00:21 By the definition of pdf, when you take the integral of pdf between a and b, it gives me the probability of x being between.

00:30 And b and when you replace a with minus infinity and b with infinity then it gives me axis between minus and infinity and for any real number that's true and this is one a lot of being in that range is one and this is equals to one for any pdf this is a future for pdf so in the first part it asks for what is k so when you take this integral and take it equal to 1 and solve it i will replace fx with r function but our boundaries has changed with 0 and 3 because when our x is less than 0 our fx is equals to 0 when r 3 x is greater than 3 our function is 0.

01:29 That means there is no effect on integral when x is not between 0 and 3.

01:37 So this part must be equal to 1.

01:41 Then i will take the integral.

01:43 When i take the integral 3x becomes 3x2 and x squared over 2 and x squared becomes x cubed over 3.

01:52 And our boundaries are here times k equals to 1.

01:57 Then i just put 3 into instead of x, this part comes and when i put 0 instead of x, this is 0 and there is a minus 0.

02:08 I didn't write.

02:10 So i just calculate 9 over 2 k equals to 1, then k equals to no 2 over 9...