Solution
Population size N=5
1.(a) mean
μ=N∑Xi
μ=56+8+10+12+13
μ=9.8
(b) Variance
Xi68101213Xi−Xˉ−3.8−1.80.22.23.2(Xi−Xˉ)214.443.240.044.8410.24
σ2=N∑(Xi−Xˉ)=532.8
σ2=6.56
2. Samples of size 4 and their means
Sample size n=4
Possible samples
=C54=(45)=5 samples
Sample mean9.009.259.7510.2510.75frequency11111Probability1/51/51/51/51/5no.12345Samples6,8,10,126,8,10,136,8,12,136,10,12,138,10,12,13means9.009.259.7510.2510.75
3. Sampling distribution of the sample means
mean9.009.259.7510.2510.75freq11111Prob1/51/51/51/51/5
4. The mean of the sampling distribution
Xˉ=X1P(1)+X2P(2)+X3P(3)+X4P(4)+X5P(5)
Xˉ=9.00(1/5)+9.25(1/5)+9.75(1/5)+10.25(1/5)+10.75(1/5)
Xˉ=9.8
The mean of the sampling distribution of the sample means is equal to the mean of the population.
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