Science:Math Exam Resources/Courses/MATH101 C/April 2025/Question 03 (b)/Solution 1

$\sigma (X)=\sqrt{\mathbb{𝔼}(X^2)-\mathbb{𝔼}(X{)}^2}$, so we compute $$\begin{array}{r}\mathbb{𝔼}(X^2)=\underset{0}{\overset{1}{\int }}{x}^2\cdot 2xdx=\underset{0}{\overset{1}{\int }}2x^{3}dx=\frac{2}{4}{x}^4{|}_0^1=\frac{1}{2}.\end{array}$$ Hence, $\text{Var}(X)=\mathbb{𝔼}(X^2)-\mathbb{𝔼}(X{)}^2=\frac{1}{2}-(\frac{2}{3}{)}^2=\frac{1}{18}$, and so $\sigma (X)=\frac{1}{\sqrt{18}}=\frac{1}{3\sqrt{2}}$.

Alternatively, we could compute the variance directly, $$\begin{array}{r}\text{Var}(X)=\underset{0}{\overset{1}{\int }}(x-\frac{2}{3}{)}^{2}2xdx=\underset{0}{\overset{1}{\int }}2x(x^2-\frac{4}{3}x+\frac{4}{9})dx=\underset{0}{\overset{1}{\int }}(2x^3-\frac{8}{3}{x}^2+\frac{8}{9}x)dx\\ [\frac{2}{4}{x}^4-\frac{8}{9}{x}^3+\frac{4}{9}{x}^2]{|}_0^1=\frac{1}{2}-\frac{8}{9}+\frac{4}{9}=\frac{1}{18}.\end{array}$$ Therefore we have $\sigma (X)=\sqrt{\text{Var}(X)}=\sqrt{\frac{1}{18}}=\frac{1}{3\sqrt{2}}$.