A population consists of the four numbers 3, 4, 2, 5. Consider all possible
distinct samples (without replacement) of size two and verify that the population mean
is equal to the mean of sample means
μ=(3+4+2+5)/4=3.5\mu=(3+4+2+5)/4=3.5μ=(3+4+2+5)/4=3.5
μ(3,5)=4\mu(3,5)=4μ(3,5)=4
μ(3,2)=2.5\mu(3,2)=2.5μ(3,2)=2.5
μ(3,4)=3.5\mu(3,4)=3.5μ(3,4)=3.5
μ(4,2)=3\mu(4,2)=3μ(4,2)=3
μ(4,5)=4.5\mu(4,5)=4.5μ(4,5)=4.5
μ(2,5)=3.5\mu(2,5) =3.5μ(2,5)=3.5
f(4)=1f(4)=1f(4)=1/6
f(2.5)=1/6
f(3.5)=2/6
f(3)=1/6
f(4.5)=1/6
E(x)=∑f(x)x=4/6+(2.5)/6+7/6+0.5+(4.5)/6=3.5E(x)=\sum f(x)x=4/6+(2.5)/6+7/6+0.5+(4.5)/6=3.5E(x)=∑f(x)x=4/6+(2.5)/6+7/6+0.5+(4.5)/6=3.5
E(x)=μE(x)=\muE(x)=μ
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments