Enrichment Practice: Continuous Random Variables — Answers

Questions: Continuous Random Variables Drills.

(a)

$${\int }_0^{2}kxdx=1$$

gives $2k=1$, so $k=\frac{1}{2}$. [2]

(b)(i)

$$P(X>1)={\int }_1^2\frac{x}{2}dx=\frac{3}{4}.$$

[2]

(ii)

$$P(0.5<X<1.5)={\int }_{0.5}^{1.5}\frac{x}{2}dx=\frac{1}{2}.$$

[2]

(c)

$$F(x)=\{\begin{array}{ll}0,& x<0,\\ \frac{x^2}{4},& 0\le x\le 2,\\ 1,& x>2.\end{array}$$

[3]

(d)

$$E(X)={\int }_0^{2}x\left(\frac{x}{2}\right)dx=\frac{4}{3},$$

and

$$E(X^2)={\int }_0^2{x}^2\left(\frac{x}{2}\right)dx=2.$$

Therefore

$$\mathrm{Var}(X)=2-{\left(\frac{4}{3}\right)}^2=\frac{2}{9}.$$

[4]

(e) The median $m$ satisfies $F(m)=\frac{1}{2}$, so

$$\frac{m^2}{4}=\frac{1}{2}.$$

Since $0\le m\le 2$, $m=\sqrt{2}$. [2]