A population consists of the numbers 2, 4, 5, 9, 10. List all possible sample size of 3 from this population without replacement and determine the mean of each sample.
The number of possible samples which can be selected without replacement is
"\\begin{pmatrix}\n N \\\\\n n\n\\end{pmatrix}=\\cfrac{N! } {n! \\cdot(N-n)! }=\\cfrac{5! } {3! \\cdot2! }=10."
"\\textnormal{Sample\\qquad\\qquad Mean}\\\\ \n\\ \\\\\n2,4,5\\qquad \\cfrac{2+4+5}{3}=3\\cfrac23\\\\\n2,4,9\\qquad \\cfrac{2+4+9}{3}=5\\\\\n2,4,10\\qquad \\cfrac{2+4+10}{3}=5\\cfrac13\\\\\n2,5,9\\qquad \\cfrac{2+5+9}{3}=5\\cfrac13\\\\\n2,5,10\\qquad \\cfrac{2+5+10}{3}=5\\cfrac23\\\\\n2,9,10\\qquad \\cfrac{2+9+10}{3}=7\\\\\n4,5,9\\qquad \\cfrac{4+5+9}{3}=6\\\\\n4,5,10\\qquad \\cfrac{4+5+10}{3}=6\\cfrac13\\\\\n4,9,10\\qquad \\cfrac{4+9+10}{3}=7\\cfrac23\\\\\n5,9,10\\qquad \\cfrac{5+9+10}{3}=8"
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