If the p.d.f. of c.r.v. X is f(x) = x^$\frac{2}{18}$, for -3

Question

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

KollegeApply · Accepted Answer

If the p.d.f. of c.r.v. X is f(x) = x^$\frac{2}{18}$, for -3 < x < 3 and = 0. otherwise, then P(|X| < 1) = bbunderline($\frac{1}{27}$). Explanation: To solve this problem, we need to calculate P(|X| < 1) for the given probability density function (p.d.f.): f(x) = x^$\frac{2}{18}$, -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^$\frac{2}{18}$, the integral becomes: P(|X| < 1) = int_-1^1 x^$\frac{2}{18}$ dx We can now compute this integral. The value of P(|X| < 1) is $\frac{1}{27}$.

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