Question

A random variable X has a probability density function
$f\left(x\right)=(\begin{array}{cc}{\mathrm{ke}}^n{e}^{-x};& x\ge 0\\ 0;& \mathrm{otherwise}\end{array}$ (n is an integer)
with mean 3. The values of {k, n} are

• A. $\{\frac{1}{2},1\}$
• B. $\{\frac{1}{4},2\}$
• C. $\{\frac{1}{2},2\}$
• D. {1, 2}

Answer 1

The correct option is C $\{\frac{1}{2},2\}$

We know that
${\int }_{-\infty }^{\infty }f\left(x\right)dx=1$

Let n is 2 then
${\int }_0^{\infty }k{x}^2{e}^{-x}dx=1$

or $x^2(-e^{-x}){]}_0^{\infty }-{\int }_0^{\infty }2x.(-e^{-x})dx=\frac{1}{k}$

or $0+2{\int }_0^{\infty }x{e}^{-x}dx=\frac{1}{k}$

or $0+{\int }_0^{\infty }{e}^{-x}dx=\frac{1}{2k}$

or $-e^{-x}{]}_0^{\infty }=\frac{1}{2k}$

or $-[0-1]=\frac{1}{2k}$

or $k=\frac{1}{2}$

Hence, option (c) is correct