Answer:
[tex]\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}[/tex]
Explanation:
Given
[tex]Cards = \{1,2,3,4\}[/tex]
Required
The sampling distribution
The possible selection of 2 cards without replacement is as follows:
[tex]S = \{(1,2) (1,3) (1,4) (2,1) (2,3) (2,4) (3,1) (3,2) (3,4) (4,1) (4,2) (4,3)\}[/tex]
Calculate the mean
[tex]\begin{array}{cccccccccccc}{Selection} & {(1,2)} & {(1,3)} & {(1,4)} & {(2,1)} & {(2,3)} & {(2,4)}& {(3,1)} & {(3,2)} & {(3,4)} & {(4,1)} & {(4,2)} & {(4,3)} \ \\ {Mean} & {1.5} & {2} & {2.5} & {1.5} & {2.5} & {3} & {2} & {2.5} & {3.5} & {2.5} & {3} & {3.5}\ \end{array}[/tex]
List out the mean and the respective frequency
[tex]1.5 \to 2[/tex]
[tex]2 \to 2[/tex]
[tex]2.5 \to 4[/tex]
[tex]3 \to 2[/tex]
[tex]3.5\to 2[/tex]
[tex]Total \to 12[/tex]
Calculate the probability of each mean
[tex]P(1.5) \to \frac{2}{12} \to \frac{1}{6}\\[/tex]
[tex]P(2) \to \frac{2}{12} \to \frac{1}{6}\\[/tex]
[tex]P(2.5) \to \frac{4}{12} \to \frac{1}{3}\\[/tex]
[tex]P(3) \to \frac{2}{12} \to \frac{1}{6}\\[/tex]
[tex]P(3.5) \to \frac{2}{12} \to \frac{1}{6}[/tex]
So, the table of sampling distribution is:
[tex]\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}[/tex]
