Z ∼ B i n ( 4 , 0.5 ) Z ∈ { 0 , 1 , 2 , 3 , 4 } P ( Z = 0 ) = 0. 5 4 = 0.0625 P ( Z = 1 ) = C 4 1 ⋅ 0.5 ⋅ 0. 5 3 = 0.25 P ( Z = 2 ) = C 4 2 ⋅ 0. 5 2 ⋅ 0. 5 2 = 0.375 P ( Z = 3 ) = C 4 3 ⋅ 0.5 ⋅ 0. 5 3 = 0.25 P ( Z = 4 ) = 0. 5 4 = 0.0625 Z\sim Bin\left( 4,0.5 \right) \\Z\in \left\{ 0,1,2,3,4 \right\} \\P\left( Z=0 \right) =0.5^4=0.0625\\P\left( Z=1 \right) =C_{4}^{1}\cdot 0.5\cdot 0.5^3=0.25\\P\left( Z=2 \right) =C_{4}^{2}\cdot 0.5^2\cdot 0.5^2=0.375\\P\left( Z=3 \right) =C_{4}^{3}\cdot 0.5\cdot 0.5^3=0.25\\P\left( Z=4 \right) =0.5^4=0.0625 Z ∼ B in ( 4 , 0.5 ) Z ∈ { 0 , 1 , 2 , 3 , 4 } P ( Z = 0 ) = 0. 5 4 = 0.0625 P ( Z = 1 ) = C 4 1 ⋅ 0.5 ⋅ 0. 5 3 = 0.25 P ( Z = 2 ) = C 4 2 ⋅ 0. 5 2 ⋅ 0. 5 2 = 0.375 P ( Z = 3 ) = C 4 3 ⋅ 0.5 ⋅ 0. 5 3 = 0.25 P ( Z = 4 ) = 0. 5 4 = 0.0625
E Z = 0 ⋅ 0625 + 1 ⋅ 0.25 + 2 ⋅ 0.375 + 3 ⋅ 0.25 + 4 ⋅ 0.0625 = 2 E Z 2 = 0 2 ⋅ 0625 + 1 2 ⋅ 0.25 + 2 2 ⋅ 0.375 + 3 2 ⋅ 0.25 + 4 2 ⋅ 0.0625 = 5 D Z = E Z 2 − ( E Z ) 2 = 5 − 2 2 = 1 σ Z = D Z = 1 EZ=0\cdot 0625+1\cdot 0.25+2\cdot 0.375+3\cdot 0.25+4\cdot 0.0625=2\\EZ^2=0^2\cdot 0625+1^2\cdot 0.25+2^2\cdot 0.375+3^2\cdot 0.25+4^2\cdot 0.0625=5\\DZ=EZ^2-\left( EZ \right) ^2=5-2^2=1\\\sigma Z=\sqrt{DZ}=1 EZ = 0 ⋅ 0625 + 1 ⋅ 0.25 + 2 ⋅ 0.375 + 3 ⋅ 0.25 + 4 ⋅ 0.0625 = 2 E Z 2 = 0 2 ⋅ 0625 + 1 2 ⋅ 0.25 + 2 2 ⋅ 0.375 + 3 2 ⋅ 0.25 + 4 2 ⋅ 0.0625 = 5 D Z = E Z 2 − ( EZ ) 2 = 5 − 2 2 = 1 σ Z = D Z = 1
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