3. Dice Sum Distributions: General Properties Consider the sum of the dots showi
ID: 3202204 • Letter: 3
Question
3. Dice Sum Distributions: General Properties Consider the sum of the dots showing on a pair of six-sided dice. Verify that the (discrete) probability distribution describing this sum is: 13 -v 36 for V (7) 8 9, 10, 11.12 an for 36 Sketch this distribution, verify that it is properly normalized, and that it has a mean of 7 Verify that the variance is equal to 35/6 in two separate ways Direct: Calculate according to the definition of variance for this discrete distribution. Indirect: Use the following (important theorem. (Note: You do not need to prove the theorem.) The variance of the sum of independent random variables is equal to the sum of their variances. Note: This probability distribution is used in Experiment 1.Explanation / Answer
Solution
Back-up Theory
If X is a discrete random variable with p(x) = P(X = x),
Mean X = E(X) = sum{x.p(x)}, sum over all possible values of X ………. (1)
E(X2) = sum{x2.p(x)}, sum over all possible values of X ………. …………(2)
Variance of X = V(X) = E(X2) – { E(X)}2 ……………………………………. (3)
Now, to work out the answer,
Let S = sum of dots on the two dice.
Then, S = 2, 3, ….., 12 and
p(2) = p(12) = 1/36; p(3) = p(11) = 2/36; p(4) = p(10) = 3/36; p(5) = p(9) = 4/36; p(6) = p(8) = 5/36; and p(7) = 6/36. [these probabilities are obtained by actual enumeration the possibilities for each case]
E(S) by (1), = 42/6 = 7………………………………………………………………….(4)
E(S2) by (2), = 1974/36 = 545/6
So, V(S), by (3) = 210/36 = 5.83 ANSWER 1
Now, to get V(S) by application of theorem,
Let S1 = number of dots on one die and S2 = number of dots on the other die.
Then, S = S1 + S2 and hence by the theorem, V(S) = V(S1) + V(S2) …………… (5)
S1 and S2 are clearly independent and also have the same probability distribution given by
p(x) = 1/6 for all x = 1,2,3,4,5,6. And hence V(S2) = V(S2)
So, V(S) = 2V(S1)
Now, E(S1) = 21/6; E(S22) = 91/6 and hence V(S1) = 105/6
So, V(S) = 2V(S1) = 210/6 which is the same as Answer 1. ANSWER 2
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