A population is given by 5 8 10 15 16 20 draw all possible samples of size 2 without replacement. Show that E(x ̅ )=μ
N=6
n=2
μ=5+8+10+15+16+206=12.33\mu = \frac{5+ 8+ 10+ 15+ 16+ 20}{6} = 12.33μ=65+8+10+15+16+20=12.33
Number of samples =6!2!(6−2)!=5×62=15= \frac{6!}{2!(6-2)!} = \frac{5 \times 6}{2} = 15=2!(6−2)!6!=25×6=15
E(xˉ)=6.5+7.5+...+17.5+1815=12.33E(\bar{x}) = \frac{6.5+7.5+...+17.5+18}{15}=12.33E(xˉ)=156.5+7.5+...+17.5+18=12.33
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