Question

Consider the experiment of tossing two dices. Let X denote the total of the upturned faces.

a. Calculate the pmf f(x) and cdf F(x) of X.

b. Calculate Q(p) such that p = 0.20, 0.50, and 0.75.

Answer 1

a) There are a total of 6*6 outcomes possible here.

The PMF for X is computed here as:

• 1 + 1 = 2, Prob = 1/36
• 1 + 2 or 2 + 1 = 3, Prob = 2/36
• 2 + 2 or 1 + 3 or 3 + 1, Prob = 3/36

Similarly for others the probabilities are computed to get:

P(X = 2) = 1/36
P(X = 3) = 2/36
P(X = 4) = 3/36
P(X = 5) = 4/36
P(X = 6) = 5/36
P(X = 7) = 6/36
P(X = 8) = 5/36
P(X = 9) = 4/36
P(X = 10) = 3/36
P(X = 11) = 2/36
P(X = 12) = 1/36

This is the required PMF for X.

The CDF is computed as:

P(X <= 2) = 1/36
P(X <= 3) = 1/36 + 2/36 = 3/36
P(X <= 4) = 3/36 + 3/36 = 6/36
P(X <= 5) = 6/36 + 4/36 = 10/36
P(X <= 6) = 10/36 + 5/36 = 15/36
P(X <= 7) = 15/36 + 6/36 = 21/36
P(X <= 8) = 21/36 + 5/36 = 26/36
P(X <= 9) = 26/36 + 4/36 = 30/36
P(X <= 10) = 30/36 + 3/36 = 33/36
P(X <= 11) = 33/36 + 2/36 = 35/36
P(X <= 12) = 1

b) The percentiles are now computed as: (from the CDF )

Q(0.2). We know that: 6/36 = 0.1667 and 10/36 = 0.2778 > 0.2, therefore the required percentile value here is:
Q(0.2) = 5

Similarly others are computed as:

P(X <= 6) = 15/36 < 0.5 but P(X<= 7) = 21/36 > 0.5

Therefore Q(0.5) = 7

30/36 = 0.833 > 0.75 but 26/36 = 0.722 < 0.75

Therefore Q(0.75) = 9