Question

X is a continuous random variable with probability density function

f(x) = `x^3/16,` 0 ≤ x ≤ 1

= 0, otherwise

Then, the value of P(0.2 ≤ X ≤ 0.3) is

Options

• `0.65/64`

• `0.0065/64`

• `0.065/64`

• `0.0064/65`

Answer 1

`0.065/64`

Explanation:

P(0.2 ≤ X ≤ 0.3) = `int_0.2^0.3 x^3/16 "dx" = [x^4/64]_0.2^0.3`

`= 1/64 [(0.3)^4 - (0.2)^4]`

`= (0.0081 - 0.0016)/64`

`= 0.00065/64`