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Question

Given the probability density function (p.d.f) of a continuous random variable $$X$$ as,
$$f(x) = \dfrac {x^{2}}{3}, -1 < x < 2 = 1$$, otherwise
Determine 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.


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