Q.5.8 Let be a random variable with pr... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 5: Q.5.8 (page 212)

Let be a random variable with probability density function
$f (x) = \{\begin{cases} c (1 - x^{2}) & - 1 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$
(a) What is the value of?
(b) What is the cumulative distribution function of?

Short Answer

Expert verified

(a) The value of c is $\frac{3}{4}$

(b) The cumulative distribution function of X is

$F_{x} (x) = \{\begin{cases} 0 & x < - 1 \\ \frac{3}{4} x - \frac{1}{4} x^{3} + \frac{1}{2} & - 1 鈮� x < 1 \\ 1 & x 鈮� 1 \end{cases}$

Step by step solution

Step 1

Given: Density function of a random variable X as f(x)= $\{\begin{cases} c (1 - x^{2}) - 1 < x < 1 \\ 0 o t h e r w i s e \end{cases}$

Step 2

We know that any random variable integrates to 1. Therefore, we have

${鈭�}_{- 鈭�}^{鈭�} f (x) d x = 1 {鈭�}_{{- 鈭�}}^{- 1} 0 d x + {鈭�}_{{- 1}}^{1} c (1 - x^{2}) d x + {鈭�}_{1}^{鈭�} 0 d x = 1 {鈭�}_{{- 1}}^{1} c (1 - x^{2}) d x = 1 c {[x - \frac{x^{3}}{3}]}^{1}_{\{- 1\}} = 1 c [1 - \frac{1}{3} - (- 1) + \frac{- 1}{3}] = 1 \frac{4}{3} c = 1 c = \frac{3}{4}$

Step 3

Thus,the density function is $f (x) = \{\begin{cases} \frac{3}{4} (1 - x^{2}) & - 1 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$

Step 4

Cummulative distribution function of X is given as $F_{X} (x)$ as

$F_{x} (x) = P [X 鈮� x] {鈭�}_{\{- 鈭�\}}^{蟺} 0 d t + {鈭�}_{\{- 1\}}^{蟺} \frac{3}{4} (1 - t^{2}) d t = \frac{3}{4} {[t - \frac{t^{3}}{3}]}^{x}_{\{- 1\}} = \frac{3}{4} [x - \frac{x^{3}}{3} - (- 1) + \frac{- 1}{3}] = \frac{3}{4} [x - \frac{x^{3}}{3} + \frac{2}{3}] = \frac{3}{4} x - \frac{1}{4} x^{3} + \frac{1}{2}$

Step 5

Thus, CDF of X is

$F_{X} (x) = \{\begin{matrix} 0 & x 鈮� - 1 \\ \frac{3}{4} x - \frac{1}{4} x^{3} + \frac{1}{2} & - 1 鈮� x < 1 \\ 1 & x 鈮� 1 \end{matrix} \begin{array}{l} \end{array}$

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91影视

Let be a random variable with probability density function
$f (x) = \{\begin{cases} c (1 - x^{2}) & - 1 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$
(a) What is the value of?
(b) What is the cumulative distribution function of?

Short Answer

Step by step solution

Step 1

Step 2

Step 3

Step 4

Step 5

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91影视

Let be a random variable with probability density functionfx=c1-x2-1<x<10otherwise(a) What is the value of?(b) What is the cumulative distribution function of?

Short Answer

Step by step solution

Step 1

Step 2

Step 3

Step 4

Step 5

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Most popular questions from this chapter

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Statistics

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Probability and Statistics

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Study anywhere. Anytime. Across all devices.

Let be a random variable with probability density function
$f (x) = \{\begin{cases} c (1 - x^{2}) & - 1 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$
(a) What is the value of?
(b) What is the cumulative distribution function of?