Let f(x) = k(9x − x^2) if 0 ≤ x ≤ 9

Question

Let f(x) = k(9x − x²) if 0 ≤ x ≤ 9 and f(x) = 0 if x < 0 or x > 9.

(a) For what value of k is f a probability density function? 

k= ?

(b) For that value of k, find P(X > 1).

 P(X > 1) = ??

(c) Find the mean. 

μ = ??

Answer 1

a) You want the integral from 0 to 9 of (k(9x − x²)dx to equal 1

int(k(9x − x²))dx= k*int(9x − x²)dx =k(9x²/2 - x³/3) from x=0 to x=9 = k[(729/2 - 729/3)-(0)] = 729k(1/2 - 1/3) = 729k(1/3) = 243k = 1. So k = 1/243

b) int (1/243)(int(9x − x²)dx from x=1 to x=9 = (1/243)(9x²/2 - x³/3) from x=1 to x=9 = ..... or compute 1 - int (1/243)(int(9x − x²)dx from x=0 to x=1

c)int x*((1/243)(9x − x²))dx from x=0 to x=9