SOLUTION: Let X be a random variable with pdf f(x) = kx^2 where 0 is less than or equal to x which is less than or equal to 1. (a) Find k. (b) Find E(X) and Var(X).

Algebra ->  Probability-and-statistics -> SOLUTION: Let X be a random variable with pdf f(x) = kx^2 where 0 is less than or equal to x which is less than or equal to 1. (a) Find k. (b) Find E(X) and Var(X).      Log On


   

Question 349782: Let X be a random variable with pdf f(x) = kx^2 where 0 is less than or equal to x which is less than or equal to 1.
(a) Find k.
(b) Find E(X) and Var(X).
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
a)int%28f%28x%29%2Cdx%2C0%2C1%29=1
int%28kx%5E2%2Cdx%2C0%2C1%29=1
%28kx%5E3%29%2F3=1 from x=0 to x=1
k%2F3=1
k=3
.
.
.
b) E%28x%29=int%28x%2Af%28x%29%2Cdx%2C0%2C1%29
E%28x%29=int%283x%5E3%2Cdx%2C0%2C1%29
E%28x%29=%283%2F4%29x%5E4 from x=0 to x=1
E%28x%29=mu=3%2F4
.
.
.
c) V%28x%29=int%28%28x-mu%29%5E2%2Af%28x%29%2Cdx%2C0%2C1%29
V%28x%29=int%28%28x-3%2F4%29%5E2%2A3x%5E2%2Cdx%2C0%2C1%29
V%28x%29=int%28%283x%5E4-%289%2F2%29x%5E3%2B%2827%2F4%29x%5E2%29%2Cdx%2C0%2C1%29
V%28x%29=%283%2F5%29x%5E5-%289%2F4%29x%5E4%2B%2827%2F8%29x%5E3%29 from x=0 to x=1
V%28x%29=3%2F5-9%2F4%2B27%2F8
V%28x%29=3%2F80