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If the p.d.f. of c.r.v. X is f(x) = x218, for -3 < x < 3 and = 0, otherwise, then P(|X| < 1) = ______. - Mathematics and Statistics

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प्रश्न

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) = ______. 

विकल्प

  • `1/27`

  • `1/28`

  • `1/29`

  • `1/26`

MCQ
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उत्तर

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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अध्याय 7: Probability Distributions - Miscellaneous Exercise 1 [पृष्ठ २४२]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 7 Probability Distributions
Miscellaneous Exercise 1 | Q 3 | पृष्ठ २४२

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