Question

Can I ask that how to solve this question? Thanks so much!

Question 3: Suppose that you toss a pair of fair dice and write down the value of the top face from each die. That is, the value of the top face is a random variable. (a) What is the population distribution for one die? Write it out in a table. (b) Determine the sampling distribution of the sample means obtained by tossing two dice. Write it out in a table. (c) Calculate the variance of the sample mean, using the population distribution for one die (Show your work!). (d) Calculate the variance of the sample mean, using the distribution of the sam- ple mean (Show your work!). Comment.

Answer 1

Let X = random variable denoting the value of the top face and since die is fair so P(X = z) = 1/6, I = 1, 2, 3, 4, 5, 6. Hence the distribution of X X = 1 1 2 3 4 5 6 P(X = ) 1/6 1/6 1/6 1/6 1/6 1/6 (6) Let X = sample mean obtained by tossing two dice then P(X = 1) = P({(1,1)}) = 1/36, P(X = 1.5) = P({(1,2), (2,1)) = 2/36, P(X = 2) = P({(2, 2), (1,3), (3,1)}) = 3/36, P(X = 2.5) = P({(2,3), (3, 2), (1,4), (4,1))) = 4/36, P(X = 3) = P({(3, 3), (1,5), (5, 1), (2, 4), (4,2)}) = 5/36, P(X = 3.5) = P({(1,6), (2,5), (3, 4), (4,3), (5,2), (6,1)}) = 6/36, P(X = 4) = P({(2,6), (3,5), (4,4), (5,3), (6,2)}) = 5/36, P(X = 4.5) = P({(3,6), (4,5), (5,4), (6,3)}) = 4/36, P(X = 5) = P({(4, 6), (5,5), (6,4)}) = 3/36, P(X = 5.5) = P({(5,6), (6,5)}) = 2/36, P(X = 6) = P({(6,6)}) = 1/36

Hence the distribution of X :

I= X	1	1.5	2	2.5	3	3.5	4	4.5	5	5.5	6
P(X = 1)	1/36	2/36	3/36	4/36	5/36	6/36	5/36	4/36	3/36	2/36	1/36

(C) Var(X) = į Var(83) (since X1 and X2 are independent where, Xi = value of top face of ith die, i = 1,2) Again X1 and X2 are identical so Var(X) = Var(11) = + (E(X4) – E²(X1)) = (15.1667 – 3.52)/2 = 1.4584 where, E(X) = (1 + 2 + 32 +42 +52 + 6)/6 = 15.1667 E(X) = (1+2+3+4+5+6)/6 = 3.5 (d) E(X) = 1*(1/36) +1.5*(2/36) +2* (3/36) + 2.5*(4/36) +3* (5/36) + 3.5*(6/36) + 4* (5/36) + 4.5*(4/36) + 5* (3/36) +5.5+ (2/36) +6 *(1/36) = 3.5 E(X) = 12 *(1/36)+1.52* (2/36) +22 * (3/36) +2.52* (4/36) +32 * (5/36) + 3.52 * (6/36)+42*(5/36)+4.52*(4/36)+52* (3/36)+5.52 * (2/36)+62 *(1/36) = 13.7083 Var(X) = F(X) - E?(X) = 13.7083 – 3,52 = 1.4584 Hence we get same value as (c).