Answer to Question #218574 in Statistics and Probability for Marcus Maraka

Question

2 x 4

Answer 1

$f(x)=\begin{cases} kx & 2\leq x\leq 4 \\ 0 & otherwise \end{cases}$

$\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{2}^{4}kxdx=[\dfrac{kx^2}{2}]\begin{matrix} 4\\ 2 \end{matrix}$

$=\dfrac{16k}{2}-\dfrac{4k}{2}=6k=1=>k=\dfrac{1}{6}$

$E[X]=\displaystyle\int_{2}^{4}x(\dfrac{1}{6}x)dx=[\dfrac{x^3}{18}]\begin{matrix} 4\\ 2 \end{matrix}$

$=\dfrac{64}{18}-\dfrac{8}{18}=\dfrac{28}{9}$

$E[X^2]=\displaystyle\int_{2}^{4}x^2(\dfrac{1}{6}x)dx=[\dfrac{x^4}{24}]\begin{matrix} 4\\ 2 \end{matrix}$

$=\dfrac{256}{24}-\dfrac{16}{24}=10$

$Var(X)=\sigma^2=E[X^2]-(E[X])^2=10-(\dfrac{28}{9})^2=\dfrac{26}{81}$

$\sigma=\sqrt{\sigma^2}=\dfrac{\sqrt{26}}{9}\approx0.5666$