Question

The assets​ (in billions of​ dollars) of the four wealthiest people in a particular country are 46 comma 27 comma 21 comma 13. Assume that samples of size nequals2 are randomly selected with replacement from this population of four values. a. After identifying the 16 different possible samples and finding the mean of each​ sample, construct a table representing the sampling distribution of the sample mean. In the​ table, values of the sample mean that are the same have been combined. x overbar Probability x overbar Probability 46 nothing 24 nothing 36.5 nothing 21 nothing 33.5 nothing 20 nothing 29.5 nothing 17 nothing 27 nothing 13 nothing ​(Type integers or​ fractions.)

Answer 1

4 wealthy person :46, 27, 21, 13

Since sampling is happening with replacement, the idea is even if I have chosen the wealthy person 46, I place the person back in the population (replacement) and can then pick one again from the population to complete my sample size of 2. Applying simple permutation and combination, you will find that the total sample sizes possible is 16 (4x4)... I am mentioning below an easy way of finding the sample, without missing out any possibility:

Sample- finding all possible samples	46	27	21	13
46	(46,46)	(46,27)	(46,21)	(46,13)
27	(27,46)	(27,27)	(27,21)	(27,13)
21	(21,46)	(21,27)	(21,21)	(21,13)
13	(13,46)	(13,27)	(13,21)	(13,13)

Now sample mean is nothing but  (obseravtion 1 + observation 2)/2.... we can generate a table below using this formula and the samples indentified above. Note we also calculate the probability of occurence of each sample size and the corresponding mean by the formula occurrence of that particular sample / total number of observations

Samples	Sample mean is (obseravtion 1 + observation 2)/2	Probability of occurrence
(46,46)	46	1/16
(27,46)	36.5	1/16
(21,46)	33.5	1/16
(13,46)	29.5	1/16
(46,27)	36.5	1/16
(27,27)	27	1/16
(21,27)	24	1/16
(13,27)	20	1/16
(46,21)	33.5	1/16
(27,21)	24	1/16
(21,21)	21	1/16
(13,21)	17	1/16
(46,13)	29.5	1/16
(27,13)	20	1/16
(21,13)	17	1/16
(13,13)	13	1/16

Probability distribution table of sample means. Now since the question says values of the sample mean that are the same have been combined, we find the probability distribution of sample means by adding the probability for those which have common sample means. For eg both samples (27,46) and (46,27) have the sample mean 36.5. So we simply add their probabilities and say that the probability of having a sample with sample mean 36.5 is 2/16.... basis this logic the below table can be easily created:

X bar	Probability
46	1/16
36.5	2/16
33.5	2/16
29.5	2/16
27	1/16
24	2/16
20	2/16
21	1/16
17	2/16
13	1/16