A sample with size n = 3 is drawn from the set 5, 6, 8, 12, and 20. Construct the sampling distribution of the sample means.
The population size is "N=5" and the sample size is "n=3". The number of possible samples which can be drawn without replacement is
"\\binom{N}{n}=\\binom{5}{3}= 10"
The 10 combinations and their sample means are given below.
The sample mean is derived from the formula, "\\bar{x_i}={\\displaystyle\\sum^{10}_{i=1} x_i\\over 3}"
sample values Sample mean
(5,6,8) "{19\\over3}=6.33"
(5,6,12) "{23\\over3}=7.67"
(5,6,20) "{31\\over3}=10.33"
(5,8,12) "{25\\over3}=8.33"
(5,8,20) "{33\\over3}=11"
(6,8,12) "{26\\over3}=8.67"
(6,8,20) "{34\\over3}=11.33"
(8,12,20) "{40\\over3}=13.33"
(5,12,20) "{37\\over3}=12.33"
(6,12,20) "{38\\over3}=12.67"
The sampling distribution for the sample means is given as,
"\\bar{x_i}" 6.33 7.67 10.33 8.33 11 8.67 11.33 13.33 12.33 12.67
"p(\\bar{x_i})" "{1\\over10}" "{1\\over10}" "{1\\over10}" "{1\\over10}" "{1\\over10}" "{1\\over10}" "{1\\over10}" "{1\\over10}" "{1\\over10}" "{1\\over10}"
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