Maharashtra State BoardHSC Arts 12th Board Exam
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The following is the p.d.f. (ProbabiIity Density Function) of a continuous random variable X : f(x)=x/32,0 = 0 otherwise - Mathematics and Statistics

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The following is the p.d.f. (ProbabiIity Density Function) of a continuous random variable X :

`f(x)=x/32,0<x<8`

= 0 otherwise

(a) Find the expression for c.d.f. (Cumulative Distribution Function) of X.

(b) Also find its value at x = 0.5 and 9.

 

 

 

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Solution 1

(a) c.d.f. of a continuous random variable X is given by

`F(x)=int_(-oo)^xf(y)dy`

In the given density function f(x), range of X starts at '0'.

`therefore F(x)=int_0^xf(y)dy=int_0^xy/32dy=[y^2/64]^2=x^2/64`

`"Thus ",F(x)=x^2/64, AA x in R`

(b)  `F(0.5) =(0.5)^2/64=1/256`

For any value of x ≥ 8, F(x)=1

∴ F(9)=1

Solution 2

Given p.d.f

`f(x)=x/32,0<x<8`

`c.d.f. = F(X)=int_0^xf(x)dx`

`=int_0^x x/32dx`

`=[x^2/64]_0^x`

`=X^2/64`

`"at "x = 0.5`

`F(X)=F(0.5)=(0.5)^2/64=0.25/64=0.0039`

For any value of x greater than 8

`F(X)=1`

`therefore F(9)=1`

Concept: Probability Distribution - Probability Density Function (P.D.F.)
  Is there an error in this question or solution?

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