Science:Math Exam Resources/Courses/MATH101 B/April 2024/Question 17 (d)/Solution 1

The expected value of $X^{2}$ is

$${\begin{aligned}\mathbb {E} (X^{2})&=\int _{-\infty }^{\infty }x^{2}f(x)\;\mathrm {d} x=\int _{-1/2}^{0}x^{2}\;\mathrm {d} x+\int _{0}^{1/4}2x^{2}\;\mathrm {d} x\\&=\left.\left[{\frac {x^{3}}{3}}\right]\right|_{-1/2}^{0}+\left.\left[{\frac {2x^{3}}{3}}\right]\right|_{0}^{1/4}={\frac {1}{3}}\cdot {\frac {1}{8}}+{\frac {2}{3}}\cdot {\frac {1}{64}}={\frac {1}{3}}\left({\frac {1}{8}}+{\frac {1}{32}}\right)={\frac {1}{3}}\cdot {\frac {5}{32}}={\frac {5}{3\cdot 2^{5}}}.\end{aligned}}$$

Recall from part (c) that $\mathbb {E} (X)=-{\frac {1}{16}}=-{\frac {1}{2^{4}}}$. Thus,

$${\begin{aligned}\mathrm {Var} (X)&=\mathbb {E} (X^{2})-(\mathbb {E} X)^{2}={\frac {5}{3\cdot 2^{5}}}-{\frac {1}{2^{8}}}={\frac {5\cdot 2^{3}-3}{3\cdot 2^{8}}}={\frac {37}{3\cdot 2^{8}}}.\end{aligned}}$$

Taking the square root of this, the standard deviation of $X$ is $\sigma (X)={\frac {\sqrt {37/3}}{16}}$. Therefore, the interval we are looking for is $[\mathbb {E} (X)-\sigma (X),\mathbb {E} (X)+\sigma (X)]=\left[-{\frac {1}{16}}-{\frac {\sqrt {37/3}}{16}},-{\frac {1}{16}}+{\frac {\sqrt {37/3}}{16}}\right]$.

Note that depending on how much or how little one simplified along the way, it is possible to get different (equivalent) answers, such as $\left[-{\frac {1}{16}}-{\sqrt {{\frac {5}{96}}-{\frac {1}{16^{2}}}}},-{\frac {1}{16}}+{\sqrt {{\frac {5}{96}}-{\frac {1}{16^{2}}}}}\right]$.