A.
We have population values 23, 26, 29, 32, 35, population size N=5 and sample size n=3.
Mean of population (μ) = 523+26+29+32+35=29
Variance of population
σ2=NΣ(xi−xˉ)2=536+9+0+9+36=18
σ=σ2=18=32≈4.24264
The number of possible samples which can be drawn without replacement is NCn=5C3=10.
no12345678910Sample23,26,2923,26,3223,26,3523,29,3223,29,3523,32,3526,29,3226,29,3526,32,3529,32,35Samplemean (xˉ)26272828293029303132
Xˉ26272829303132f(Xˉ)1/101/102/102/102/101/101/10Xˉf(Xˉ)26/1027/1056/1058/1060/1031/1032/10Xˉ2f(Xˉ)676/10729/101568/101682/101800/10961/101024/10
Mean of sampling distribution
μXˉ=E(Xˉ)=∑Xˉif(Xˉi)=29=μ
The variance of sampling distribution
Var(Xˉ)=σXˉ2=∑Xˉi2f(Xˉi)−[∑Xˉif(Xˉi)]2
=108440−(29)2=3=nσ2(N−1N−n)
σXˉ=3≈1.73205
B.
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