Question #206384

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. 


1
Expert's answer
2021-11-15T17:40:50-0500

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)={0            for x<2136          for 2x<3336=112 for 3x<4636=16   for 4x<51036=518 for 5x<61536=512 for 6x<72136=712 for 7x<82636=1318 for 8x<93036=56  for 9x<103336=1112 for 10x<113536          for 11x<123636=1  for x12F(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)

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