given-a-probability-function-f-x-1-3-1-x-4-and-f-x-0-in-other-x-find-the-value-of-2-

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

g i v e n a p r o b a b i l i t y

f u n c t i o n

f (x) = \frac{1}{3}, 1 ⩽ x ⩽ 4 a n d f (x) = 0

i n o t h e r x . f i n d t h e v a l u e

o f σ^{2} ?

Commented by jagoll last updated on 10/Feb/20

m y w a y σ^{2} = \int_{1}^{4} {(x - μ)}^{2} f (x) d x

w h e r e μ = \int_{1}^{4} x f (x) d x

i t c o r r e c t ?

Commented by Joel578 last updated on 10/Feb/20

y e s

Commented by Joel578 last updated on 10/Feb/20

a n o t h e r f o r m u l a

σ_{X}^{2} = \underset{- \infty}{\int^{\infty}} {(x - μ)}^{2} f (x) d x

= \underset{- \infty}{\int^{\infty}} (x^{2} - 2 x μ + μ^{2}) f (x) d x

= \underset{- \infty}{\int^{\infty}} x^{2} f (x) d x - 2 μ \underset{- \infty}{\int^{\infty}} x f (x) d x + μ^{2} \underset{- \infty}{\int^{\infty}} f (x) d x

= \underset{- \infty}{\int^{\infty}} x^{2} f (x) d x - 2 μ . μ + μ^{2} . 1

= \underset{- \infty}{\int^{\infty}} x^{2} f (x) d x - μ^{2}

= E [X^{2}] - μ^{2}

Commented by jagoll last updated on 10/Feb/20

o y e s t h a n k y o u

Answered by Joel578 last updated on 10/Feb/20

f_{X} (x) = {\begin{cases} \frac{1}{3}, 1 ⩽ x ⩽ 4 \\ 0, elsewhere \end{cases}

E [X] = μ = \underset{- \infty}{\int^{\infty}} x f (x) d x = \underset{1}{\int^{4}} x (\frac{1}{3}) d x = \frac{5}{2}

E [X^{2}] = \underset{- \infty}{\int^{\infty}} x^{2} f (x) d x = \underset{1}{\int^{4}} x^{2} (\frac{1}{3}) d x = 7

σ_{X}^{2} = E [X^{2}] - μ^{2} = 7 - \frac{25}{4} = \frac{3}{4}

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