Question

Discrete Probability Distributions: part 1 1. Consider the experiment of rolling two dice. a.) Define a random variable that takes on all possible values for the minimum value of the two dice face showing when the dice come to rest after the roll. b.) Before doing anything at all (do not write out the distribution yet), think about this experiment and the random variable. Tell me what you think the shape of the distribution might look like AND WHY you think this. No points off for incorrect answers for this one. Just do your best to form a mental picture. c.)Now in the space below, write out the probability distribution for the random variable that you defined in part a. Was your answer in part b correct? If not, tell me why (now that you have written out the distribution) you know your answer in partb is not correct d.) Find the mean, variance and the standard deviation of the distribution in part c. Please use proper notation!!!!

Please answer all parts of question. Thank you!

Answer 1

a)

When we roll a fair die possible outcomes are 1, 2, 3, 4, 5, and 6.

Let X is a random variable shows the minimum of two dice. That is X can take value 1, 2, 3, 4, 5 or 6.

b)

As the value of X increases, it probability decrease. Since X shows minimum of two values so chance to get X=1 must be higher than chances to get X=6. So distribution must be skewed to right. It is not symmetric.

c)

Following is the sample space of the rolling of two fair dice and minimum of two:

$S=\begin{Bmatrix} (1,1)=1 &(1,2)=1 &(1,3)=1 & (1,4)=1 &(1,5)=1 & (1,6)=1\\ (2,1)=1 &(2,2)=2 &(2,3)=2 & (2,4)=2 &(2,5)=2 & (2,6)=2 \\ (3,1)=1 &(3,2)=2 &(3,3)=3 & (3,4)=3 &(3,5)=3 & (3,6)=3\\ (4,1)=1 &(4,2)=2 &(4,3)=3 & (4,4)=4 &(4,5)=4 & (4,6)=4\\ (5,1)=1 &(5,2)=2 &(5,3)=3 & (5,4)=4 &(5,5)=5 & (5,6)=5\\ (6,1)=1 &(6,2)=2 &(6,3)=3 & (6,4)=4 &(6,5)=5 & (6,6)=6 \end{Bmatrix}$

Here in each outcome, first number shows outcome of first die and second number of shows outcome of second die. After equal to sign minimum of two numbers is shown.

The pdf of X is:

X	P(X=x)	P(X=x)
1	11/36	0.3056
2	9/36	0.25
3	7/36	0.1944
4	5/36	0.1389
5	3/36	0.0833
6	1/36	0.0278
Total	1	1

Following is the graph:

0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 2 х Ln (x=x)d

As expected graph is skewed to right. It is not symmetric.

(c)

Following table shows the calculations:

X	P(X=x)	xP(X=x)	x^2*P(X=x)
1	0.3056	0.3056	0.3056
2	0.25	0.5	1
3	0.1944	0.5832	1.7496
4	0.1389	0.5556	2.2224
5	0.0833	0.4165	2.0825
6	0.0278	0.1668	1.0008
Total		2.5277	8.3609

The mean is

E(X) P(X )2.5277

The variance is

ΣΡΙ -ΣΡ- rP(X) Var(X) = x) 8.3609 2.52772 1.9716

The standard deviation is

2 P(X) P(X r) 1.4041