Math, asked by jashwanthichowdary29, 1 month ago

For a continuous probability function f(x)=Kx^2 e^(-x) where x>0. Find (i)K (ii)mean (iii)variance

Answers

Answered by MaheswariS
18

\textbf{Given:}

\textsf{Probability function is}

\mathsf{f(x)=K\;x^2\;e^{-x}\;where\;x\;>\;0}

\textbf{To find:}

\textsf{(i)K}

\textsf{(ii)Mean}

\textsf{(iii)Variance}

\textbf{Solution:}

\underline{\mathsf{Finding\;K:}}

\textsf{since f(x) is probability function,}

\mathsf{\displaystyle\int\limits^{\infty}_{-\infty}\;f(x)\;dx=1}

\mathsf{\displaystyle\int\limits^{\infty}_0\;K\,x^2\;e^{-x}dx=1}

\mathsf{\displaystyle\;K\int\limits^{\infty}_0\,x^2\;e^{-x}dx=1}

\mathsf{Using,}

\boxed{\mathsf{\int\limits^{\infty}_0\,x^n\;e^{-ax}dx=\dfrac{n!}{a^{n+1}}}}

\mathsf{K\left(\dfrac{2!}{1^{2+1}}\right)=1}

\mathsf{K(2)=1}

\implies\boxed{\mathsf{K=\dfrac{1}{2}}}

\underline{\mathsf{Mean:}}

\mathsf{E(X)=\displaystyle\int\limits^{\infty}_{-\infty}x\;f(x)\;dx}

\mathsf{E(X)=\displaystyle\dfrac{1}{2}\int\limits^{\infty}_{0}x\;(x^2\,e^{-x})\;dx}

\mathsf{E(X)=\displaystyle\dfrac{1}{2}\int\limits^{\infty}_{0}x^3\,e^{-x}\;dx}

\mathsf{E(X)=\dfrac{1}{2}\left(\dfrac{3!}{1^{3+1}}\right)=\dfrac{1}{2}{\times}1{\times}2{\times}3}

\implies\boxed{\mathsf{Mean=3}}

\mathsf{Also,}

\mathsf{E(X^2)=\displaystyle\int\limits^{\infty}_{-\infty}x^2\;f(x)\;dx}

\mathsf{E(X^2)=\displaystyle\dfrac{1}{2}\int\limits^{\infty}_{0}x^2\;(x^2\,e^{-x})\;dx}

\mathsf{E(X^2)=\displaystyle\dfrac{1}{2}\int\limits^{\infty}_{0}x^4\,e^{-x}\;dx}

\mathsf{E(X^2)=\dfrac{1}{2}\dfrac{4!}{1^{4+1}}=\dfrac{1}{2}{\times}1{\times}2{\times}3{\times}4}

\implies\mathsf{E(X^2)=12}

\underline{\mathsf{Variance:}}

\mathsf{Variance=E(X^2)-[E(X)]^2}

\mathsf{Variance=12-3^2}

\mathsf{Variance=12-9}

\implies\boxed{\mathsf{Variance=3}}

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