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A random variable takes the values 1, 2, and 3, each with equal probability. List all possible samples of size two that may be chosen, without replacement, from this population and hence construct the sampling distribution of the sample mean.

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    $\begingroup$ What have you tried so far? What part or parts of the question are you having difficulty with? $\endgroup$ – heropup Jan 19 '14 at 1:18
  • $\begingroup$ I listed down all the possibilities already. But I don't know how to 'construct the sampling distribution of sample mean'. $\endgroup$ – Jollie Jan 19 '14 at 1:21
  • $\begingroup$ Easy start: Determine the probability of selecting a first number and then a second number. For sure, you should get all equal probabilities. $\endgroup$ – NasuSama Jan 19 '14 at 1:22
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    $\begingroup$ So, for each of the possible samples of size two, find the mean of the sample. Then find the probability of having observed that sample. This becomes your sampling distribution. For example, if you get a sample $\{1, 3\}$, the mean of that sample is $2$, and the probability of observing $\{1, 3\}$ is...? Then $\Pr[\bar X = 2] = \ldots$. $\endgroup$ – heropup Jan 19 '14 at 1:23
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    $\begingroup$ @Jollie : Questions posted here should not be phrased in a manner in which it is appropriate for an instructor to assign homework problems. You should be asking a question, not passing on to us a question written by someone other than you, with no indication that you've thought about it. $\endgroup$ – Michael Hardy Jan 19 '14 at 1:24
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Without replacement: \begin{align} 1+2 & = 3 \\ 1+3 & = 4 \\ \\ 2+1 & = 3 \\ 2+3 & = 5 \\ \\ 3+1 & = 4 \\ 3+2 & = 5 \end{align}

Sorting the sums into increasing order, you have: $\quad 3,3, \quad 4,4, \quad 5,5$

With replacement: \begin{align} 1+1 & = 2 \\ 1+2 & = 3 \\ 1+3 & = 4 \\ \\ 2+1 & = 3 \\ 2+2 & = 4 \\ 2+3 & = 5 \\ \\ 3+1 & = 4 \\ 3+2 & = 5 \\ 3+3 & = 6 \end{align}

Sorting the sums into increasing order, you have: $\quad 2, \quad 3,3, \quad 4,4,4 \quad 5,5, \quad 6$

The sample mean for a sample of size $2$ is just half the sum of the observations in a sample.

(Of course, you should write the solution that you turn in the way I wrote what you see above, but this should tell you how to solve the problem.)

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  • $\begingroup$ The sampling occurs without replacement. $\endgroup$ – heropup Jan 19 '14 at 1:32
  • $\begingroup$ I missed that. Now I've edited my answer to include both. $\endgroup$ – Michael Hardy Jan 19 '14 at 1:41

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