Question

If the p.d.f. of c.r.v. X is f(x) = `x^2/18`, for -3 < x < 3 and = 0, otherwise, then P(|X| < 1) = ______. 

Options

• `1/27`

• `1/28`

• `1/29`

• `1/26`

Answer 1

If the p.d.f. of c.r.v. X is f(x) = `x^2/18`, for -3 < x < 3 and = 0. otherwise, then P(|X| < 1) = `bbunderline(1/27)`.

Explanation:

To solve this problem, we need to calculate P(|X| < 1) for the given probability density function (p.d.f.):

f(x) = `x^2/18,   -3<x<3`

First, let's find the probability P(|X| < 1). This is the probability that the random variable X lies between -1 and 1, i.e., 

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

Substituting the given p.d.f. f(x) = `x^2/18,` the integral becomes:

P(|X| < 1) = `int_-1^1 x^2/18 dx`

We can now compute this integral.

The value of P(|X| < 1) is `1/27`.