1.
We have population values 2,5,6,8,10,12 population size N=6
μ=62+5+6+8+10+12=7.17
σ2=61((2−7.17)2+(5−7.17)2+(6−7.17)2+(8−7.17)2+(10−7.17)2+(12−7.17)2)=10.8
2.
The number of possible samples which can be drawn without replacement is
(nN)=(26)=15
SampleNo.123456789101112131415Samplevalues2,52,62,82,102,125,65,85,105,126,86,106,128,108,1210,12Sample mean(Xˉ)3.545675.56.57.58.578991011
3.
The sampling distribution of the sample means.
TotalXˉ3.5455.566.577.588.591011f111111211121115f(Xˉ)1/151/151/151/151/151/152/151/151/151/152/151/151/151Xˉf(Xˉ)0.230.260.330.360.40.430.930.50.530.561.20.660.737.17Xˉ2f(Xˉ)0.811.061.662.022.42.811.863.754.264.8210.86.668.155.68
4.
E(Xˉ)=∑Xˉf(Xˉ)=7.17
The mean of the sampling distribution of the sample means is equal to the mean of the population.
E(Xˉ)=μXˉ=7.17=μ
5.
Var(Xˉ)=∑Xˉ2f(Xˉ)−(∑Xˉf(Xˉ))2
=55.68−(7.17)2=4.27
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