Can you please help me with this statistics math problem? (All of them) Thank you !! The assets (in billions of dollars) of the four wealthiest people in a particular country are 45, 28, 21, 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.| \(\bar{x}\) | Probability ||--------------|-------------|| 45 | || 36.5 | || 33 | || 31.5 | || 28 | || 24.5 | || 23 | || 21 | || 19.5 | || 18 | |(Type integers or fractions.) b. **Compare the mean of the population to the mean of the sampling distribution of the sample mean.**The mean of the population, \(\mu\), is [textbox] [dropdown] the mean of the sample means, [textbox]. (Round to two decimal places as needed.)c. **Do the sample means target the value of the population mean? In general, do sample means make good estimates of population means? Why or why not?**The sample means [dropdown] the population mean. In general, sample means [dropdown] make good estimates of population means because the mean is [textbox] estimator.

Given that: Assets (in billion of dollars) of 4 wealthiest people: 45, 28, 21, 18Since n=2 are randomly selected with replacement from this population of four values. So, the number of samples are 42=16. (a) There are 16 different possible samples: 45,45,45,28,45,21,45,1828,45,28,28,28,21,28,1821,45,21,28,21,21,21,1818,45,18,28,18,21,18,18 Their sample means are computed by adding the values of pairs and then divide it by 2. The 16 sample means are: 45,36.5,33,31.536.5,28,24.5,2333,24.5,21,19.531.5,23,19.5,18 Construct a table representing the sampling distribution of the sample mean: x¯ Probability x¯ Probability 45 116=0.0625 24.5 216=0.125 36.5 216=0.125 23 216=0.125 33 216=0.125 21 116=0.0625 31.5 216=0.125 19.5 216=0.125 28 116=0.0625 18 116=0.0625 (b) The population mean should be compared to the sample mean's sampling distribution mean. Obtain the population mean by adding all the values and then divide it by 4: 45+28+21+184=1124=28 Similarly, obtain the mean of sample means by adding all the 16 means as obtained in part(a) of the problem and then divide it by 16: 45+36.5+33+31.5+36.5+28+24.5+23+33+24.5+21+19.5+31.5+23+19.5+1816=44816=28 Thus, the mean of the population, 28 is equal to the mean of the sampling distribution of the sample mean, 28.(c) The answer is Yes. From part(b) of the problem, it is clear that the population mean and sample means are equal. The sample means target the population mean. In general, sample means do make good estimates of population means because the mean is an unbiased estimator.