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

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

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