Question

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

${\displaystyle f\left(x\right)=\frac{x^2}{18},-3<x<3}$

            = 0,             otherwise

Answer 1

We have

P(X < 1) = ${\displaystyle {\int }_{-3}^{1}f\left(x\right)dx}$

= ${\displaystyle {\int }_{-3}^1\frac{x^2}{18}dx}$

= ${\displaystyle {\left[\frac{x^3}{54}\right]}_{-3}^1}$

= ${\displaystyle \frac{1}{54}-\left(\frac{-27}{54}\right)}$

= ${\displaystyle \frac{28}{54}}$

= ${\displaystyle \frac{14}{27}}$

= 0.5185

Now |X| < 1

${\displaystyle \Rightarrow }$ ± X < 1

∴ X < 1, – X < 1,

i.e. X > – 1

i.e. – 1 < X < 1

∴ Required Probability = ${\displaystyle {\int }_{-1}^{1}f\left(x\right)dx}$

= ${\displaystyle {\int }_{-1}^1\frac{x^2}{18}dx}$

= ${\displaystyle {\left[\frac{x^3}{54}\right]}_{-1}^1}$

= ${\displaystyle \frac{1}{54}+\frac{1}{54}}$

= ${\displaystyle \frac{2}{54}}$

= ${\displaystyle \frac{1}{27}}$

= 0.03704