The assets (in billions of dollars) of the four wealthiest people in a particular country are 47, 30, 28, 18. Assume that samples of size n=2 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.

The mean of a dataset is the average value of the dataset.

The dataset is given as:

47, 30, 28 and 18

From the above dataset, we can see that the data elements are 4

The sample size is given as:

n = 2

So, the number of different samples is:

The 16 samples are: {(47,47), (47,30), (47,28), (47,18),(30,47), (30,30), (30,28), (30,18),(28,47), (28,30), (28,28), (28,18),(18,47), (18,30), (18,28), (18,18)}

The mean values of the above samples are: {47, 38.5, 37.5 , 32.5 , 38.5 , 30 , 29 , 24 , 37.5 , 29 , 28 , 23 , 32.5 , 24 , 23 , 18}

Rearrange in ascending order

18, 23, 23, 24, 24, 28, 29,29, 30, 32.5, 32.5, 37.5, 37.5, 38.5, 38.5, 47

So, the table of the sample means is:

Mean             Frequency

18                       1

23                      2

24                      2

28                      1

29                      2

30                      1

32.5                   2

37.5                   2

38.5                   2

47                       1

Total                 16

Read more about sample mean at:

link

Step-by-step explanation:

The 16 possible permutations and their means are:

47, 47: μ = 47

47, 30: μ = 38.5

47, 28: μ = 37.5

47, 18: μ = 32.5

30, 47: μ = 38.5

30, 30: μ = 30

30, 28: μ = 29

30, 18: μ = 24

28, 47: μ = 37.5

28, 30: μ = 29

28, 28: μ = 28

28, 18: μ = 23

18, 47: μ = 32.5

18, 30: μ = 24

18, 28: μ = 23

18, 18: μ = 18

So the frequency distribution of the sample means is:

what is the answer?

e^-kt = n/n0 = 0.79  

-kt = ln 0.79  

t = - ln 0.79/0.0001  

= 2357 years

step-by-step explanation: