N=5
n=3
Without replacement
1. Number of outcomes =N!n!(N−n)!= \frac{N!}{n!(N-n)!}=n!(N−n)!N!
=5!3!(5−3)!=4×52=10= \frac{5!}{3!(5-3)!} = \frac{4 \times 5}{2} = 10=3!(5−3)!5!=24×5=10
2.
The sampling distribution of sample mean
3. P(Mean=4) =110=0.1= \frac{1}{10}=0.1=101=0.1
4. P(Mean = 2) =110=0.1= \frac{1}{10}=0.1=101=0.1
5. P(Mean=3.33) =210=0.2= \frac{2}{10}=0.2=102=0.2
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