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.

2. Originally Posted by 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

I Really Don't Understand The Answer. What Is 2^5 For Example. Sorry I Wish I Did Understand It More. I Also Recall That When My Teacher Solves Problems Such As That We Don't Include The Zero. Why Is The Zero Included I Don't Understand? I Know It Probably Works Because It All Add's Up To 1 In The End But I Just Not Clear As To Why It Works.

4. Originally Posted by hello there
I Really Don't Understand The Answer. What Is 2^5 For Example. Sorry I Wish I Did Understand It More. I Also Recall That When My Teacher Solves Problems Such As That We Don't Include The Zero. Why Is The Zero Included I Don't Understand? I Know It Probably Works Because It All Add's Up To 1 In The End But I Just Not Clear As To Why It Works.
2^5 is a standard method of writing 2 to the power 5 using ASCII or text
charaters only.

As I said in my most the number of "4"s has a binomial distribution so
if k is the number of "4"s, then:

p(k,5-k) = b(k,5,0.3)

where b(k, N, p) is the probability of k successes in N trials where the
probability of sucess in a single trial is p.

The table in my post is of the mean of k "4"s and (5-k) "8"s against
b(k,5,0.3) which is the probability of that outcome.

RonL

5. ## thanks

that clears it up a little more for me...see the ^ is what confused me because how did you get .7^5 y=8 i don't see how that matches.

6. Originally Posted by hello there
that clears it up a little more for me...see the ^ is what confused me because how did you get .7^5 y=8 i don't see how that matches.
This is p(0,5) = b(0, 5, 0.3) = 5!/[0! (5-0)!] 0.3^0 0.7^5 = 0.7^5.
This has an average of 8 as all five of the instances have the value 8
(that is there are 0 "4"'s)

RonL

7. ## okay

Yes thanks it's clear now that helped alot i just wanted to know the reasoning behind the y='s

8. ## Sorry

I just have one more question why are you doing the permutation on only
p(2,3) and p( 3,2)

9. Originally Posted by hello there
I just have one more question why are you doing the permutation on only
p(2,3) and p( 3,2)
I'm not sure what you mean, I have:

p(2,3) = b(2, 5, 0.3) = 5!/[2! (5-2)!] 0.3^2 0.7^3 = 5!/[2! 3!] 0.3^2 0.7^3

which is just the binomial probability for 2 "4"s., and the equivalent for p(3,2).

RonL