Question

For the following probability density function (p. d. f) of X, find : (i) P ( X<1), (ii) P |x| < 1

if $f(x) ="x"^2/18 , -3 < x < 3$

 = 0,      otherwise

Answer 1

Given, f(x) is pdf

(i) P(X< 1) = $int_-3^1 "f"("x") "dx"$

=$int_-3^1("x"^2/18)"dx"$

=$1/18["x"^3/3]_-3^1 = 14/27 = 0.5185$

(ii) P(|X|<1) = P(-1<X<1) = $int_-1^1 "f"("x")"dx"$

$=int_-1^1 "x"^2/18"dx"$

$= 1/18["x"^3/3]_-1^1 = 1/27 = 0.03704$