Question

# Given the probability density function (p.d.f) of a continuous random variable $$X$$ as,$$f(x) = \dfrac {x^{2}}{3}, -1 < x < 2 = 1$$, otherwiseDetermine the cumulative distribution function (c.d.f.) of $$X$$ and hence find $$P(X < 1), P(X > 0), P(1 < X < 2)$$

Solution

## Given the PDF of a continuous random variable $$X.$$$$\Rightarrow f\left( x \right) =\dfrac { { x }^{ 3 } }{ 3 } ,-1<x<2=1$$     $$f\left( x \right) =\int _{ -\infty }^{ \alpha }{ f\left( x \right) dx } =\int _{ \alpha }^{ \beta }{ \dfrac { 1 }{ \beta -\alpha } dx } =\left[ \dfrac { 1 }{ \beta -\alpha } x \right] _{ \infty }^{ x }=\dfrac { x-\alpha }{ \beta -\alpha }$$     $$f\left( x \right) =\int { \dfrac { 1 }{ 2-(-1) } ,-1<x<2}$$       $$=\int { \dfrac{1}{3},-1<x<2}$$                     $${0,otherwise }$$                                   $${0,otherwise }$$$$\Rightarrow$$ When, $$P(1<x<2)=\int _{ 1 }^{ 2 }{ \dfrac{1}{3} }\;dx=\left[ \dfrac { 1 }{ 3 } x \right] _{ 0 }^{ 1 }=\dfrac { 1 }{ 3 }$$     When $$P(x>0)=0$$     When $$P(1<x<2)=\int _{ 1 }^{ 2 }{ \dfrac{1}{3} }\;dx=\left[ \dfrac { x }{ 3 } x \right] _{ 1 }^{ 2 }=\dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 } =\dfrac { 1 }{ 3 }$$Hence, solved.Maths

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