Answer to Question #206384 in Statistics and Probability for Chilu

Question

Suppose that a pair of fair dice are to be tossed, and let the random variable X denote the  of the points.  a) obtain the probability distribution for X. b)  find the cumulative distribution function F(x) for the random variable X     c) graph this cumulative distribution function. 

Answer 1

a) probability distribution

The random variable X is the sum of the points after two dices are tossed.

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Thus for (3, 2) we have X = 5. Using the fact that all 36 sample points are equally probable, so that each sample point has probability 1/36.

b) the cumulative distribution function

Given from the above f(2) = 1/36, f(3) = 2/36, f(3) = 3/36 etc

F(2) = f(2) = 1/36

F(3) = f(2) + f(3) = 1/36 + 2/36 = 3/36

F(4) = f(2) + f(3) + f(4) = 1/36 + 2/36 + 3/36 = 6/36

F(5) = f(2) + f(3) + f(4) + f(5) = 1/36 + 2/36 + 3/36 + 4/36 = 10/36

F(6) = f(2) + f(3) + f(4) + f(5) + f(6) = 1/36 + 2/36 + 3/36 + 4/36 + 5/36 = 15/36

F(7) = f(2) + f(3) + f(4) + f(5) + f(6) + f(7) = 1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36 = 21/36

F(8) = f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) = 1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36 + 5/36 = 26/36

F(9) = f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) + f(9) = 1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36 + 5/36 + 4/36 = 30/36

F(10) = f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) + f(9) + f(10) = 1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36 + 5/36 + 4/36 + 3/36 = 33/36

F(11) = f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) + f(9) + f(10) + f(11) = 1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36 + 5/36 + 4/36 + 3/36 + 2/36 = 35/36

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

Hence the distribution function is given by

$$F(X)=\begin{cases} 0 \ \ \ \ \ \ \ \ \ \ \ \ for \ x< 2\\ \frac{1}{36} \ \ \ \ \ \ \ \ \ \ for \ 2\le x< 3 \\ \frac{3}{36}=\frac{1}{12} \ for \ 3\le x< 4 \\ \frac{6}{36}=\frac{1}{6} \ \ \ for \ 4\le x< 5 \\ \frac{10}{36}=\frac{5}{18} \ for \ 5\le x< 6 \\ \frac{15}{36}=\frac{5}{12} \ for \ 6\le x< 7 \\ \frac{21}{36}=\frac{7}{12} \ for \ 7\le x< 8 \\ \frac{26}{36}=\frac{13}{18} \ for \ 8\le x< 9 \\ \frac{30}{36}=\frac{5}{6} \ \ for \ 9\le x< 10 \\ \frac{33}{36}=\frac{11}{12} \ for \ 10\le x< 11 \\ \frac{35}{36} \ \ \ \ \ \ \ \ \ \ for \ 11\le x< 12 \\ \frac{36}{36}=1 \ \ for \ x \ge 12 \\ \end{cases}$$

c)