IGNOU MST 3 2018 - The distribution function F of a continuous variable is given byF(

Question

The distribution function F of a continuous variable is given by

F(x) = 0, if x < 0

= x 2 , if 0 < x < 1

$= 1 - \frac{3(3-X)^{2}}{25}, if \frac{1}{2}\leq x< 3$

= 1, if x > 3

(i) Find the pdf of X,

(ii) Evaluate P[|X| <1/ 2], using distribution function F,

(iii) Evaluate P[|X| < 1/ 2], using pdf.

16 Feb 2018

Answer :

Word Count : 556

Let's break this down step by step based on the given distribution function $ F(x) $: ### (i) Find the pdf of $ X $: The probability density function (pdf) $ f(x) $ is the derivative of the cumulative distribution function (CDF) $ F(x) $. We will differentiate $ F(x) $ piece by piece. - For $ x < 0 $, $ F(x) = 0 $. The derivative of 0 is also 0, so: \[ f(x) = 0 \quad \text{for} \quad x < 0. \] - For $ 0 \leq x < 1 $, $ F(x) = x^2 $. The derivative of $ x^2 $ with respect to $ x $ is: \[ f(x) = 2x \quad \text{for} \quad 0 \leq x < 1. \] - For $ \frac{1}{2} \leq x < 3 $, $ F(x) = 1 - \frac{3(3-x)^2}{25} $. The derivative of this with respect to $ x $ is: \[ f(x) = \frac{6(3-x)}{25} \quad \text{for} \quad \frac{1}{2} \leq x < 3. \] - For

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