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प्रश्न
If random variable X has probability distribution function.
f(x) = `c/x`, 1 < x < 3, c > 0, find c, E(x) and Var(X)
उत्तर १
∴ X has probability distribution function
∴ `int_1^3 c/x dx = 1`
`[ c log x]_1^3` = 1
c [ log3 - log 1 ] = 1
c log 3 = 1
c = `1/log 3`
E(X) = `int_1^3 xf(x)`
= `int_1^3 x . c/x dx`
= `c int_1^3 1 dx`
= `c | x |_1^3`
= c | 3 - 1|
= 2c
E(X) = `2/log 3`
V(X) = `E( "X"^2 ) - E( "X"^2 )`
V(X) = `int_1^3 x^2f(x) dx - [ int_1^3 x x f(x) dx ]^2`
= `int_1^3 x^2 c/x dx - [ 2/log 3 ]^2`
= `int_1^3 cx dx - [ 2/log 3 ]^2`
= `c[ x^2/2 ]_1^3 - [ 2/log 3 ]^2`
= `1/log 3 xx [ (9 - 1)/2] - 4/[(log 3)^2]`
V(X) = = `4/log 3- 4/[(log 3)^2]`
उत्तर २
∴ X has probability distribution function
∴ `int_1^3 c/x dx = 1`
`[ c log x]_1^3` = 1
c [ log3 - log 1 ] = 1
c log 3 = 1
c = `1/log 3`
E(X) = `int_1^3 xf(x)`
= `int_1^3 x . c/x dx`
= `c int_1^3 1 dx`
= `c | x |_1^3`
= c | 3 - 1|
= 2c
E(X) = `2/log 3`
V(X) = `E( "X"^2 ) - E( "X"^2 )`
V(X) = `int_1^3 x^2f(x) dx - [ int_1^3 x x f(x) dx ]^2`
= `int_1^3 x^2 c/x dx - [ 2/log 3 ]^2`
= `int_1^3 cx dx - [ 2/log 3 ]^2`
= `c[ x^2/2 ]_1^3 - [ 2/log 3 ]^2`
= `1/log 3 xx [ (9 - 1)/2] - 4/[(log 3)^2]`
V(X) = = `4/log 3- 4/[(log 3)^2]`
उत्तर ३
∴ X has probability distribution function
∴ `int_1^3 c/x dx = 1`
`[ c log x]_1^3` = 1
c [ log3 - log 1 ] = 1
c log 3 = 1
c = `1/log 3`
E(X) = `int_1^3 xf(x)`
= `int_1^3 x . c/x dx`
= `c int_1^3 1 dx`
= `c | x |_1^3`
= c | 3 - 1|
= 2c
E(X) = `2/log 3`
V(X) = `E( "X"^2 ) - E( "X"^2 )`
V(X) = `int_1^3 x^2f(x) dx - [ int_1^3 x x f(x) dx ]^2`
= `int_1^3 x^2 c/x dx - [ 2/log 3 ]^2`
= `int_1^3 cx dx - [ 2/log 3 ]^2`
= `c[ x^2/2 ]_1^3 - [ 2/log 3 ]^2`
= `1/log 3 xx [ (9 - 1)/2] - 4/[(log 3)^2]`
V(X) = = `4/log 3- 4/[(log 3)^2]`
उत्तर ४
∴ X has probability distribution function
∴ `int_1^3 c/x dx = 1`
`[ c log x]_1^3` = 1
c [ log3 - log 1 ] = 1
c log 3 = 1
c = `1/log 3`
E(X) = `int_1^3 xf(x)`
= `int_1^3 x . c/x dx`
= `c int_1^3 1 dx`
= `c | x |_1^3`
= c | 3 - 1|
= 2c
E(X) = `2/log 3`
V(X) = `E( "X"^2 ) - E( "X"^2 )`
V(X) = `int_1^3 x^2f(x) dx - [ int_1^3 x x f(x) dx ]^2`
= `int_1^3 x^2 c/x dx - [ 2/log 3 ]^2`
= `int_1^3 cx dx - [ 2/log 3 ]^2`
= `c[ x^2/2 ]_1^3 - [ 2/log 3 ]^2`
= `1/log 3 xx [ (9 - 1)/2] - 4/[(log 3)^2]`
V(X) = = `4/log 3- 4/[(log 3)^2]`
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