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**hello there** 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 ( used by x bar) if random samples of size n=5 are drawn from the population.

for x=4 the probability is p(x)= .3

for x= 8 the probability is p(x)= .7

Note: Each possible sample is not likely to be chosen.

The number of 4's in a sample of size 5 has a binomial distribution B(5,0.5=3).

Denote the possible samples by (a,b), where a is the number of 4's and

b the number of 8's, and let Y denote the random variable equal to the

mean of a sample of size 5.

Then

p(0,5) = 0.7^5, y=8

p(1,4) = 5 (0.3)(0.7^4), y=7.2

p(2,3) = [5!/(3!*2!)] (0.3^2)(0.7^3), y=6.4

p(3,2) = [5!/(2!*3!)] (0.3^3)(0.7^2), y=5.6

p(4,1) = 5 (0.3^4)(0.7), y=4.8

p(5,0) = 0.3^5, y=4

So we have the distribution for y:

Code:

y ~p(y)
4.0 0.00243
4.8 0.02835
5.6 0.1323
6.4 0.3087
7.2 0.36015
8.0 0.16807

RonL