Question

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

`f(x) = x^2/18, -3 < x < 3`

            = 0,             otherwise

Answer 1

We have

P(X < 1) = `int_-3^1 f(x) dx`

= `int_-3^1 x^2/18dx`

= `[x^3/54]_-3^1`

= `1/54 - ((-27)/54)`

= `28/54`

= `14/27`

= 0.5185

Now |X| < 1

`\implies` ± X < 1

∴ X < 1, – X < 1,

i.e. X > – 1

i.e. – 1 < X < 1

∴ Required Probability = `int_-1^1 f(x)dx`

= `int_-1^1 x^2/18dx`

= `[x^3/54]_-1^1`

= `1/54 + 1/54`

= `2/54`

= `1/27`

= 0.03704