Question

The probability density function of a ramdom variable X is given by

$f_x\left(x\right)=\{\begin{array}{cc}\lambda (x-2)(x^2+4);& 2\le x<4\\ 0& otherwise\end{array}$

The value of the constant $\lambda$ is

• A. $\frac{3}{92}$
• B. $\frac{3}{74}$
• C. $\frac{3}{86}$
• D. $\frac{7}{92}$

Answer 1

The correct option is A $\frac{3}{92}$

${\int }_2^4\lambda (x-2)(x^2+4)dx=1$

${\int }_2^4\lambda (x^3+4x-2x^2-8)dx=1$

$\lambda {[\frac{x^4}{4}+\frac{4x^2}{2}-\frac{2x^3}{3}-8x]}_2^4=1$

$\lambda [\frac{4^4-2^4}{4}+\frac{4}{2}(4^2-2^2)-\frac{2}{3}(4^3-2^3)-8(4-2\left)\right]=1$

$\Rightarrow \lambda [60+24-\frac{2}{3}\times 56-16]=1$

$\Rightarrow \lambda \left[\frac{204-112}{3}\right]=1$

$\Rightarrow \lambda =\frac{3}{92}$