Question

In a certain population, the random variable x can only assume the values 4 and 8. The probability distribution for x is given below. Construct the sampling distribution for the sample mean x bar if the random samples of size n=3 are drawn from the population.

x|4|8
p(x)|.2|.8

Answer 1

if your sample size is 3, it means you can either get (4,4,8) or you can get(8,8,4), or (8,8,8) or (4,4,4) or (8,4,4) or (4,8,8) or (4,8,4) or (8,4,8). the mean for the first one would be (4+4+8)/3=5.3333 and the next is 6.666667 and the next is 8 and the next is 4 and the next is 5.333 and the next is 6.6666 and the next is 5.333 and the last is 6.6666. now you draw the table with
mean(1)=5.333 probability of mean(1)= 0.032*3
mean(2)=6.666 probability of mean(2)=0.128*3

Answer 2

sorry hadn't finished..
mean(3)= 8 probability of mean(3)=0.512
mean(4)=4 probaility of mean (4) = 0.008
if you add up these probabilities you will get 1.

now for the second part of the question, mean will be summation of (x*p(x)) and variance will be (summation of x^2)- (mean)^2
tell me if you don't understand any step.

Answer 3

So would it look like this?

Answer 4

yes that's it.

Answer 5

Thank you!