Possible samples of size 2.
(0,2), (0,4), (0,6), (0,8), (2,4), (2,6), (2,8), (4,6), (4,8), (6,8)
Sampling distribution
Possible samples of size 3,
(0,2,4), (0,2,6), (0,2,8), (0,4,6), (0,4,8), (0,6,8), (2,4,6), (2,4,8), (2,6,8), (4,6,8)
P ( X = 2 ) = P ( X = 2.67 ) = P ( X = 5.33 ) = P ( X = 6 ) = 0.1 P(X=2)=P(X=2.67)=P(X=5.33)=P(X=6)=0.1 P ( X = 2 ) = P ( X = 2.67 ) = P ( X = 5.33 ) = P ( X = 6 ) = 0.1
P ( X = 3.33 ) = P ( X = 4 ) = P ( X = 4.67 ) = 0.2 P(X=3.33)=P(X=4)=P(X=4.67)=0.2 P ( X = 3.33 ) = P ( X = 4 ) = P ( X = 4.67 ) = 0.2
Standard Error
μ = ∑ x f ( x ) = 4 \mu=∑xf(x)=4 μ = ∑ x f ( x ) = 4
σ = ∑ x 2 f ( x ) − 4 2 = 17.33 − 4 2 \sigma=\sqrt{∑x^2f(x)-4^2}=\sqrt{17.33-4^2} σ = ∑ x 2 f ( x ) − 4 2 = 17.33 − 4 2
= σ 2 = 1.33 =\sigma^2=1.33 = σ 2 = 1.33
S . E = σ n = 1.33 3 = 0.665 S.E=\frac{\sigma}{\sqrt{n}}=\frac{\sqrt{1.33}}{\sqrt{3}}=0.665 S . E = n σ = 3 1.33 = 0.665