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The pdf of a random variable X is f x = 3 1 - 2 x 2 , 0 < x < 1 = 0 otherwise. The P 1 4 < X < 1 3 = …

Options

  1. A179 864
  2. B159 864
  3. C169 864
  4. D189 864

Correct answer

A. 179 864

Step-by-step solution

We have, p.d.f of a random varlabe X is f x = 3 1 - 2 x 2 , 0 < x < 1 = 0 , otherwise ∴ p 1 4 < X < 1 3 = ∫ 1 / 4 1 / 3 f x d x = ∫ 1 / 4 1 / 3 3 1 - 2 x 2 d x = 3 x - 2 3 x 3 1 / 4 1 / 3 = 3 1 3 - 2 3 1 3 3 - 1 4 - 2 3 1 4 3 = 3 1 3 - 1 4 - 2 3 1 3 3 - 1 4 3 = 3 1 12 - 2 3 × 37 1728 = 3 × 179 2592 = 179 864

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