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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`
Solution
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`.
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