Question

A population consists of the numbers 1, 2, 3, and 4 with the sample size of 2 (without replacement). Which mean that has a probability of 2/6 or 1/3 or 0.33?

Answer 1

The number of possible samples which can be selected without replacement is

$\begin{pmatrix} N \\ n \end{pmatrix}=\cfrac{N! } {n! \cdot(N-n)! }=\cfrac{4! } {2! \cdot2! }=6.$

$\textnormal{Sample\qquad\qquad Mean}\\ \ \\ 1,2\qquad \cfrac{1+2}{2}=1.5\\ 1,3\qquad \cfrac{1+3}{2}=2\\ 1,4\qquad \cfrac{1+4}{2}=2.5\\ 2,3\qquad \cfrac{2+3}{2.5}=2.5\\ 2,4\qquad \cfrac{2+4}{2}=3\\ 3,4\qquad \cfrac{3+4}{2}=3.5.$

We see that the mean = 2.5 met twice, its probability is $\cfrac{2}{6}=\cfrac{1}{3}.$

All the other means met once, their probabilities are $\cfrac{1}{6}.$