Question

8. A probability density function (PDF) is given by: f(x)-k(8x-x2) for 0cx<8 What value of 'k' will make this a PDF? 9. A probability density function (PDF) is given by: f(x)-e.( 4) for x>a What value of a will make this a PDF? 10. A probability density function (PDF) is given by: f(x)-1.5x2 for -acx<a What value of a will make this a PDF?

Answer 1

8) Given the PDF $f\left ( x \right )=k\left ( 8x-x^2 \right );0<x<8$ . The conditions for a function to be PDF are

$f\left ( x \right )\geqslant 0 \textup{ and } \int_{0}^{8}f\left ( x \right )dx=1$. Here

$\int_{0}^{8}f\left ( x \right )dx=\int_{0}^{8}k\left ( 8x-x^2 \right )dx\\ \int_{0}^{8}f\left ( x \right )dx=k\left ( 8x-x^2 \right ]_{0}^{8}\\ \int_{0}^{8}f\left ( x \right )dx=k\left ( 256-512/3\right ]$

Now setting conditions,

$k\left ( 256-512/3\right ]=1\Rightarrow{\color{Blue} k=3/256}$

9) Given the PDF $f\left ( x \right )=e^{x-4};x>a$ . The conditions for a function to be PDF are

$f\left ( x \right )\geqslant 0 \textup{ and } \int_{a}^{\infty }f\left ( x \right )dx=1$. Here

$\int_{a}^{\infty }f\left ( x \right )dx=\int_{a}^{\infty }e^{-\left (x-4\right )}dx\\ \int_{a}^{\infty }e^{-\left (x-4 \right )}dx=\left [ -e^{-\left (x-4 \right )} \right ]_{a}^{\infty }=e^{4-a}={\color{Blue} a=4}$

10) Given the PDF $f\left ( x \right )=1.5x^2;-a<x<a$ . The conditions for a function to be PDF are

$f\left ( x \right )\geqslant 0 \textup{ and } \int_{a}^{\infty }f\left ( x \right )dx=1$. Here

$\int_{-a}^{a }f\left ( x \right )dx=\int_{-a}^{a }1.5x^2dx\\ \int_{-a}^{a }1.5x^2dx=\left [ 0.5x^3 \right ]_{-a}^{a }\\ a^3=1\Rightarrow {\color{Blue} a=1}$