Question

Let X be a random variable that records the outcome of a dice toss. The probability distribution for X is as follows:
Outcome: 1, 2, 3, 4, 5, 6
Probability: 1/6, 1/6, 1/6, 1/6, 1/6, 1/6

Suppose now that we toss two dice. X1 records the outcome of the first toss, and X2 records the outcome of the second toss. We are interested in the sum of the two dice tosses,
Y= X1+ X2.
What is the mean value of Y, μY?

Answer 1

To find the mean value of Y, μY, we need to sum up the values of Y multiplied by their respective probabilities.

The possible outcomes for Y are:

2: (1, 1)
3: (1, 2), (2, 1)
4: (1, 3), (2, 2), (3, 1)
5: (1, 4), (2, 3), (3, 2), (4, 1)
6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
9: (3, 6), (4, 5), (5, 4), (6, 3)
10: (4, 6), (5, 5), (6, 4)
11: (5, 6), (6, 5)
12: (6, 6)

We can calculate the value of Y for each outcome and also calculate the probability for each outcome by multiplying the probabilities of each individual dice toss. Then, we multiply Y by its probability and sum up these products to find μY.

μY = (2 * P(2)) + (3 * P(3)) + (4 * P(4)) + (5 * P(5)) + (6 * P(6)) + (7 * P(7)) + (8 * P(8)) + (9 * P(9)) + (10 * P(10)) + (11 * P(11)) + (12 * P(12))

Using the given probability distribution for X, we can calculate μY as follows:

μY = (2 * 1/36) + (3 * 2/36) + (4 * 3/36) + (5 * 4/36) + (6 * 5/36) + (7 * 6/36) + (8 * 5/36) + (9 * 4/36) + (10 * 3/36) + (11 * 2/36) + (12 * 1/36)

Simplifying this equation:

μY = 2/36 + 6/36 + 12/36 + 20/36 + 30/36 + 42/36 + 40/36 + 36/36 + 30/36 + 22/36 + 12/36

μY = 244/36

μY ≈ 6.7778

Therefore, the mean value of Y, μY, is approximately 6.7778.