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Question Number 81219 by jagoll last updated on 10/Feb/20

$$givenaprobability$$$$function$$$$f\left(x\right)=\frac{1}{3},1⩽x⩽4andf\left(x\right)=0$$$$inotherx.findthevalue$$$$of{\sigma }^2?$$

Commented by jagoll last updated on 10/Feb/20

$$myway{\sigma }^2={\int }_1^4{(x-\mu )}^{2}f\left(x\right)dx$$$$where\mu =\int {}_1^{4}xf\left(x\right)dx$$$$itcorrect?$$

Commented by Joel578 last updated on 10/Feb/20

$$yes$$

Commented by Joel578 last updated on 10/Feb/20

$$anotherformula$$$${\sigma }_X^2=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}{(x-\mu )}^{2}f\left(x\right)dx$$$$=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}(x^2-2x\mu +{\mu }^2)f\left(x\right)dx$$$$=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}{x}^{2}f\left(x\right)dx-2\mu \underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}xf\left(x\right)dx+{\mu }^2\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}f\left(x\right)dx$$$$=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}{x}^{2}f\left(x\right)dx-2\mu .\mu +{\mu }^2.1$$$$=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}{x}^{2}f\left(x\right)dx-{\mu }^2$$$$=E\left[X^2\right]-{\mu }^2$$

Commented by jagoll last updated on 10/Feb/20

$$oyesthankyou$$

Answered by Joel578 last updated on 10/Feb/20

$$f_X\left(x\right)=\{\begin{array}{l}\frac{1}{3},1⩽x⩽4\\ 0,\mathrm{elsewhere}\end{array}$$$$$$$$E\left[X\right]=\mu =\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}xf\left(x\right)dx=\underset{1}{\stackrel{4}{\int }}x\left(\frac{1}{3}\right)dx=\frac{5}{2}$$$$E\left[X^2\right]=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}{x}^{2}f\left(x\right)dx=\underset{1}{\stackrel{4}{\int }}{x}^2\left(\frac{1}{3}\right)dx=7$$$$$$$${\sigma }_X^2=E\left[X^2\right]-{\mu }^2=7-\frac{25}{4}=\frac{3}{4}$$

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