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Five people own and operate the Victrola Coffee Shop in Seattle. Each person is married, and the number of years each has been married is given below.

5, 3, 7, 2, 12

Suppose a random sample of two of the owners is selected with replacement. Let D be a statistic defined to be the absolute value of the difference in the number of years each has been married. For example, if 2 and 7 were selected, the value of D would be |2 − 7| = | −5| = 5. Find the sampling distribution of D.

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  • $\begingroup$ This is not a 'do my homework for free service'! $\endgroup$ Nov 9 '16 at 6:27
  • $\begingroup$ i literally have no clue what they are asking me to do $\endgroup$ Nov 9 '16 at 6:29
  • $\begingroup$ They are asking you to calculate all possible differences, and the probability of each one of them. $\endgroup$ Nov 9 '16 at 6:30
  • $\begingroup$ so as a permutation count $\endgroup$ Nov 9 '16 at 6:31
  • $\begingroup$ What??????????? $\endgroup$ Nov 9 '16 at 6:31
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Hints:

There are $5^2 = 25$ equally likely possibilities for choosing two numbers. Find the absolute difference for each. From there you can find the sampling distribution.

I have done a brief simulation in R with a million samples as specified, and tabled the resulting approximate distribution of $D.$ It should give you a rough idea what to expect. It seems possible values of $D$ are 0, 1, 2, 3, 4, 5, 7, 9, 10. What is the probability of each? (The simulation is accurate to only a few decimal places. Your probabilities should be of the form $i/25.$)

m = 10^6;  d = numeric(m)
pop = c(5, 3, 7, 2, 12)
for (i in 1:m) {
    d[i] = abs(diff(sample(pop, 2, repl=T))) }  
table(d)/m
## d
##        0        1        2        3        4 
## 0.199989 0.079764 0.160214 0.080052 0.080129 
##        5        7        9       10 
## 0.159982 0.079991 0.080073 0.079806 
mean(d);  sd(d)
## 3.839428
## 3.219872

The histogram below shows the approximate sampling distribution of $D$.

enter image description here

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  • $\begingroup$ You have some work to do. Good luck with it. (This is a little like finding the distribution of the sum when two fair dice are rolled, except that has 36 possibilities.) $\endgroup$
    – BruceET
    Nov 9 '16 at 6:51
  • $\begingroup$ yes...Thank you again $\endgroup$ Nov 9 '16 at 6:52
  • 1
    $\begingroup$ pop = c(5, 3, 7, 2, 12); table(abs(outer(pop, pop, "-")))/length(pop)^2 would give you the table without simulation $\endgroup$
    – Henry
    Nov 9 '16 at 15:00

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