N=5
n=3
Without replacement
1. Number of outcomes "= \\frac{N!}{n!(N-n)!}"
"= \\frac{5!}{3!(5-3)!} = \\frac{4 \\times 5}{2} = 10"
2.
The sampling distribution of sample mean
3. P(Mean=4) "= \\frac{1}{10}=0.1"
4. P(Mean = 2) "= \\frac{1}{10}=0.1"
5. P(Mean=3.33) "= \\frac{2}{10}=0.2"
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