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# Solution - Given the P. D. F. (Probability Density Function) of a Continuous Random Variable X as - Probability Distribution - Probability Density Function (P.D.F.)

ConceptProbability Distribution Probability Density Function (P.D.F.)

#### Question

Given the p. d. f. (probability density function) of a continuous random variable x as :

f(x)=x^2/3, -1

= 0 , otherwise

Determine the c. d. f. (cumulative distribution function) of x and hence find P(x < 1), P(x ≤ -2), P(x > 0), P(1 < x < 2)

#### Solution

c.d.f. of the continous random variable is given by

F(x)=int_-1^xy^2/3dx

=[y^3/9]_-1^x

=(x^3+1)/9, x in R

Consider P(X<1)=F(1)=(13+1)/9=2/9

P(x<=-2)=0

P(X>0)=1-P(X<=0)

=1-F(0)

=1-(0/9+1/9)

=8/9

P(1<x<2)=F(2) - F(1)

=1-(1/9+1/9)

=7/9

Is there an error in this question or solution?

#### APPEARS IN

2016-2017 (March)
Question 6.2.3 | 4 marks
2014-2015 (March)
Question 6.2.3 | 4 marks

#### Video TutorialsVIEW ALL [1]

Solution for question: Given the P. D. F. (Probability Density Function) of a Continuous Random Variable X as concept: Probability Distribution - Probability Density Function (P.D.F.). For the courses HSC Arts, HSC Science (Computer Science), HSC Science (Electronics), HSC Science (General)
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