Question

consider a set of cards that has four cards labeled 2, 4, 6, and 8. suppose you pick two cards, without replacement, to obtain the mean of the two numbers that are drawn from the set. which of the following tables shows the sampling distribution?

Explanation:

Step1: Calculate number of samples

The number of ways to choose 2 cards out of 4 without replacement is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 4$ and $r=2$. So $C(4,2)=\frac{4!}{2!(4 - 2)!}=\frac{4\times3\times2!}{2!\times2!}=\frac{4\times3}{2\times 1}=6$ samples.

Step2: List all possible samples and their means

• Sample (2,4): Mean $\frac{2 + 4}{2}=3$
• Sample (2,6): Mean $\frac{2+6}{2}=4$
• Sample (2,8): Mean $\frac{2 + 8}{2}=5$
• Sample (4,6): Mean $\frac{4+6}{2}=5$
• Sample (4,8): Mean $\frac{4 + 8}{2}=6$
• Sample (6,8): Mean $\frac{6+8}{2}=7$

Step3: Determine the sampling - distribution

Count the frequency of each mean value:

• Mean = 3: Frequency = 1
• Mean = 4: Frequency = 1
• Mean = 5: Frequency = 2
• Mean = 6: Frequency = 1
• Mean = 7: Frequency = 1

The sampling - distribution table should have columns for the sample - mean values and their corresponding probabilities (frequency divided by total number of samples, which is 6 in this case).

Answer:

The table with the following rows:

Sample Mean	Probability
4	$\frac{1}{6}$
5	$\frac{2}{6}=\frac{1}{3}$
6	$\frac{1}{6}$
7	$\frac{1}{6}$