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

given a probability   function   f(x)= (1/3), 1≤x≤4 and f(x)=0  in other x. find the value   of σ^(2 )  ?

$${given}\:{a}\:{probability}\: \\ $$$${function}\: \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{1}\leqslant{x}\leqslant\mathrm{4}\:{and}\:{f}\left({x}\right)=\mathrm{0} \\ $$$${in}\:{other}\:{x}.\:{find}\:{the}\:{value}\: \\ $$$${of}\:\sigma^{\mathrm{2}\:} \:? \\ $$

Commented by jagoll last updated on 10/Feb/20

my way σ^2  = ∫_1 ^( 4) (x−μ)^2 f(x)dx  where μ= ∫ _1^4  xf(x)dx  it correct?

$${my}\:{way}\:\sigma^{\mathrm{2}} \:=\:\int_{\mathrm{1}} ^{\:\mathrm{4}} \left({x}−\mu\right)^{\mathrm{2}} {f}\left({x}\right){dx} \\ $$$${where}\:\mu=\:\int\:_{\mathrm{1}} ^{\mathrm{4}} \:{xf}\left({x}\right){dx} \\ $$$${it}\:{correct}? \\ $$

Commented by Joel578 last updated on 10/Feb/20

yes

$${yes} \\ $$

Commented by Joel578 last updated on 10/Feb/20

another formula  σ_X ^2  = ∫_(−∞) ^∞  (x − μ)^2  f(x) dx         = ∫_(−∞) ^∞  (x^2  − 2xμ + μ^2 )f(x) dx         = ∫_(−∞) ^∞ x^2  f(x) dx − 2μ ∫_(−∞) ^∞ x f(x) dx + μ^2  ∫_(−∞) ^∞ f(x) dx         = ∫_(−∞) ^∞ x^2  f(x) dx − 2μ . μ + μ^2  . 1         = ∫_(−∞) ^∞  x^2  f(x) dx − μ^2          = E[X^2 ] − μ^2

$${another}\:{formula} \\ $$$$\sigma_{{X}} ^{\mathrm{2}} \:=\:\underset{−\infty} {\overset{\infty} {\int}}\:\left({x}\:−\:\mu\right)^{\mathrm{2}} \:{f}\left({x}\right)\:{dx} \\ $$$$\:\:\:\:\:\:\:=\:\underset{−\infty} {\overset{\infty} {\int}}\:\left({x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\mu\:+\:\mu^{\mathrm{2}} \right){f}\left({x}\right)\:{dx} \\ $$$$\:\:\:\:\:\:\:=\:\underset{−\infty} {\overset{\infty} {\int}}{x}^{\mathrm{2}} \:{f}\left({x}\right)\:{dx}\:−\:\mathrm{2}\mu\:\underset{−\infty} {\overset{\infty} {\int}}{x}\:{f}\left({x}\right)\:{dx}\:+\:\mu^{\mathrm{2}} \:\underset{−\infty} {\overset{\infty} {\int}}{f}\left({x}\right)\:{dx} \\ $$$$\:\:\:\:\:\:\:=\:\underset{−\infty} {\overset{\infty} {\int}}{x}^{\mathrm{2}} \:{f}\left({x}\right)\:{dx}\:−\:\mathrm{2}\mu\:.\:\mu\:+\:\mu^{\mathrm{2}} \:.\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:=\:\underset{−\infty} {\overset{\infty} {\int}}\:{x}^{\mathrm{2}} \:{f}\left({x}\right)\:{dx}\:−\:\mu^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\:{E}\left[{X}^{\mathrm{2}} \right]\:−\:\mu^{\mathrm{2}} \\ $$

Commented by jagoll last updated on 10/Feb/20

o yes thank you

$${o}\:{yes}\:{thank}\:{you} \\ $$

Answered by Joel578 last updated on 10/Feb/20

f_X (x) =  { (((1/3)   , 1 ≤ x ≤ 4)),(( 0     , elsewhere)) :}    E[X] = μ =  ∫_(−∞) ^∞  x f(x) dx = ∫_1 ^4  x((1/3)) dx = (5/2)  E[X^2 ] = ∫_(−∞) ^∞ x^2  f(x) dx = ∫_1 ^4  x^2 ((1/3)) dx = 7    σ_X ^2  = E[X^2 ] − μ^2  = 7 − ((25)/4) = (3/4)

$${f}_{{X}} \left({x}\right)\:=\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{3}}\:\:\:,\:\mathrm{1}\:\leqslant\:{x}\:\leqslant\:\mathrm{4}}\\{\:\mathrm{0}\:\:\:\:\:,\:\mathrm{elsewhere}}\end{cases} \\ $$$$ \\ $$$${E}\left[{X}\right]\:=\:\mu\:=\:\:\underset{−\infty} {\overset{\infty} {\int}}\:{x}\:{f}\left({x}\right)\:{dx}\:=\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:{x}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:{dx}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${E}\left[{X}^{\mathrm{2}} \right]\:=\:\underset{−\infty} {\overset{\infty} {\int}}{x}^{\mathrm{2}} \:{f}\left({x}\right)\:{dx}\:=\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:{dx}\:=\:\mathrm{7} \\ $$$$ \\ $$$$\sigma_{{X}} ^{\mathrm{2}} \:=\:{E}\left[{X}^{\mathrm{2}} \right]\:−\:\mu^{\mathrm{2}} \:=\:\mathrm{7}\:−\:\frac{\mathrm{25}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\: \\ $$

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