distribusi normal (menentukan varian)

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DISTRIBUSI NORMAL Variansi Var (x ) = −∞ xf ( x ) dx , < x < Var (x)= E (( µ x ) 2 ) = π σ 2 1 . ) ( 2 µ x dx e x 2 ] / ) ).[( 2 1 ( σ µ Misal z = σ µ x dx = σ dz & x - µ = σ z Var (x) = 2 2 2 2 2 z e z π σ dz Misal u = z du = dz dv = z . 2 ) 2 1 ( z e dz v = 2 ) 2 1 ( . z e z dz, misal s = - 2 2 z ds = z dz v = ds e s = -e -s = e −Z 2 2 Sehingga, Var (x ) = u × v v du Var(x) = + dz e z e z z 2 2 2 2 2 2 π σ = + dz e z e z z 2 2 2 2 2 2 1 . 2 1 π π σ = ) 2 1 0 ( 2 2 2 dz e z + π σ Karena dz e z 2 2 2 1 π adalah luas di bawah urva n!rmal dengan 0 = µ dan 1 = σ , maa nilain"a = #. Var (x) = 2 σ ( $ % # ) = 2 σ

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DISTRIBUSI NORMALVariansi

Var (x) =E (()2)

=

Misal z = dx = dz & x - = z

Var (x)=dz

Misal u = z du = dz

dv = z . dz

v = dz, misal s = -ds = z dz

v = = -e-s = Sehingga,

Var(x)=

=

=

Karena adalah luas di bawah kurva normal dengan dan , maka nilainya = 1.

Var (x)=( 0 + 1 )

=