Question

For a random variable x having the PDF shown in the figure given below

the mean and the variance are, respectively

• A. 0.5 and 0.66
• B. 2.0 and 1.33
• C. 1.0 and 0.66
• D. 1.0 and 1.33

Answer 1

The correct option is D 1.0 and 1.33

Given,


So, $f_x\left(x\right)=\frac{1}{4}in-1\le x\le 3$

Mean $E\left[x\right]={\int }_{-\infty }^{\infty }x{f}_x\left(x\right)dx$

$={\int }_{-1}^{3}x.\frac{1}{4}dx$

$=\frac{1}{4}{\left[\frac{x^2}{2}\right]}_{-1}^3$

$=\frac{1}{4}\left[\frac{9-1}{2}\right]=\frac{8}{8}=1$

Variance; $Var\left[x\right]=E\left[x^2\right]-{\left\{E\right[x\left]\right\}}^2$

$E\left[x^2\right]={\int }_{-\infty }^{\infty }{x}^2{f}_x\left(x\right)dx$

$={\int }_1^3{x}^2\frac{1}{4}dx=\frac{1}{4}{\left[\frac{x^3}{3}\right]}_{-1}^3$

$=\frac{1}{4}\left[\frac{27-(-1)}{3}\right]=\frac{28}{12}=\frac{7}{3}$

$Var\left[x\right]=E\left[x^2\right]-{\left\{E\right[x\left]\right\}}^2$

$=\frac{7}{3}-(1{)}^2=\frac{7}{3}-1=\frac{4}{3}=1.33$