Question

If X is a continuous random variable whose probability density function is given by
$f\left(x\right)=(\begin{array}{cc}k(5x-2x^2),& 0\le x\le 2\\ 0,& \mathrm{otherwise}\end{array}$
Then P(x > 1) is

• A. 3/14
• B. 4/5
• C. 14/17
• D. 17/28

Answer 1

The correct option is D 17/28

If f(x) is p.d.f then
${\int }_{-\infty }^{\infty }f\left(x\right)dx=1\&P(a<x<b)={\int }_a^{b}f\left(x\right)dx$

So ${\int }_0^{2}k(5x-2x^2)dx=1\Rightarrow k=\frac{3}{14}$

& $P(X>1)={\int }_1^{\infty }f\left(x\right)dx={\int }_1^2\frac{3}{14}(5x-2x^2)dx=\frac{17}{28}$