Question

Let X be a random variable with probability density function
$f\left(x\right)=(\begin{array}{cc}0.2,& \mathrm{for}\left|x\right|\le 1\\ 0.1,& \mathrm{for}1<\left|x\right|\le 4\\ 0,& \mathrm{otherwise}\end{array}$

The probability P(0.5 < X < 5) is .

• A. 0.45

Answer 1

The correct option is A 0.45

f(x)=0.2,for |x|≤10.1,for 1<|x|≤40,otherwise

$=(\begin{array}{cc}0.2,& -1\le x\le 1\\ 0.1,& x<-1\&x>1\\ 0.1,& -4\le x\le 4\\ 0,& \mathrm{otherwise}\end{array}$

$=(\begin{array}{cc}0.2,& -1\le x\le 1\\ 0.1,& x\end{array}$∈[-4, -1)∪(1, 4]0,otherwise

$P\left(\frac{1}{2}<x<5\right)={\int }_{1/2}^{5}f\left(x\right)\mathrm{dx}$

$={\int }_{1/2}^1(0.2)\mathrm{dx}+{\int }_1^4(0.1)\mathrm{dx}+{\int }_4^{5}0\mathrm{dx}$

$=(0.2)\left(1-\frac{1}{2}\right)+(0.1)(4-1)$

= 0.1 + 0.3 = 0.4