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If a r.v. X has p.d.f f(x) = {cx, 1<x<3,c>00, otherwise Find c, E(X), and Var(X). Also Find F(x). - Mathematics and Statistics

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Sum

If a r.v. X has p.d.f f(x) = `{("c"/x","  1 < x < 3"," "c" > 0),(0","  "otherwise"):}` 
Find c, E(X), and Var(X). Also Find F(x).

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Solution

a. Given that f(x) represents p.d.f. of r.v. X

∴ `int_1^3 f(x)*"d"x` = 1

∴ `int_1^3 "c"/x*"d"x` = 1

∴ `"c" int_1^3 (1)/x*"d"x` = 1

∴ `"c"[logx]_1^3` = 1

∴ c [log 3 – log 1] = 1

∴ c [log 3 – 0] = 1

∴ c = `(1)/log3`

b. E(X) = `int_(-oo)^(oo) xf(x)`

= `int_1^3 xf(x)*"d"x`

= `int_1^3 x "c"/x*"d"x`

= `"c" int_1^3 1*"d"x`

= `(1)/log3 [x]_1^3`

= `(1)/log3[3 - 1]`

= `(2)/log3`.

c. E(X2) = `int_(-oo)^(oo) x^2f(x)`

= `int_1^3 x^2f(x)*"d"x`

= `-int_1^3 x^2. "c"/x*"d"x`

= `"c" int_1^3x*"d"x`

= `(1)/(2log3)[x^2]_1^3`

= `(1)/(2log3) [9 - 1]`

= `8/(2log3)`

= `(4)/(log3)`

∴ Var(X) = E(X2) – [E(x)]2

= `(4)/log3 -(2/log3)^2`

= `(4)/((log3)) -  4/(log3)^2`

= `(4log3 - 4)/(log3)^2`

= `(4(log3 - 1))/(log3)^2`

F(x) = `int_1^x f(x)*"d"x`

= `int_1^x "c"/x*"d"x`

= `"c" int_1^x (1)/x*"d"x`

= `"c"[logx]_1^x`

= c[log x – log 1]

= c log x

= `log x/log 3`

Concept: Probability Distribution of a Continuous Random Variable
  Is there an error in this question or solution?
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