Question #269333

wo fair cubes are rolled. The random variable X represents the difference between the values of the two cubes.






a) Find the mean of this probability distribution. (i.e. Find E[X] )



b) Find the variance and standard deviation of this probability distribution.



(i.e. Find V[X] and SD[X])



The random variables A and B are defined as follows:



A = X-10 and B = [(1/2)X]-5



c) Show that E[A] and E[B].



d) Find V[A] and V[B].



e) Arnold and Brian play a game using two fair cubes. The cubes are rolled, and Arnold records his score using the random variable A and Brian uses the random variable B. They repeat this for a large number of times and compare their scores. Comment on any likely differences or similarities of their scores.




1
Expert's answer
2021-11-22T14:30:24-0500

probability distribution:


values on cubes difference probability

1,1 0 6/36=1/6

1,2 1 10/36=5/18

2,1 1 10/36=5/18

2,2 0 6/36=1/6

2,3 1 10/36=5/18

3,2 1 10/36=5/18

3,3 0 6/36=1/6

3,4 1 10/36=5/18

4,3 1 10/36=5/18

4,4 0 6/36=1/6

4,5 1 10/36=5/18

5,4 1 10/36=5/18

5,5 0 6/36=1/6

5,6 1 10/36=5/18

6,5 1 10/36=5/18

6,6 0 6/36=1/6

1,3 2 8/36=2/9

3,1 2 8/36=2/9

4,2 2 8/36=2/9

2,4 2 8/36=2/9

5,3 2 8/36=2/9

3,5 2 8/36=2/9

4,6 2 8/36=2/9

6,4 2 8/36=2/9

1,4 3 6/36=1/6

4,1 3 6/36=1/6

5,2 3 6/36=1/6

2,5 3 6/36=1/6

3,6 3 6/36=1/6

6,3 3 6/36=1/6

5,1 4 4/36=1/9

1,5 4 4/36=1/9

2,6 4 4/36=1/9

6,2 4 4/36=1/9

1,6 5 2/36=1/18

6,1 5 2/36=1/18


a)

μ=E(X)=xipi=\mu=E(X)=\sum x_ip_i=

=105/18+282/9+361/6+441/9+521/18=11.67=10\cdot5/18+2\cdot8\cdot2/9+3\cdot6\cdot1/6+4\cdot4\cdot1/9+5\cdot2\cdot1/18=11.67


b)

V(X)=pi(xiμ)2=11.672+10510.672/18+829.672/9+V(X)=\sum p_i (x_i-\mu)^2=11.67^2+10\cdot5\cdot10.67^2/18+8\cdot2\cdot9.67^2/9+

+8.672+47.672/9+26.672/18=724.93+8.67^2+4\cdot7.67^2/9+2\cdot6.67^2/18=724.93


SD(X)=V(X)=724.93=26.92SD(X)=\sqrt{V(X)}=\sqrt{724.93}=26.92


c)

A = X-10 and B = [(1/2)X]-5

E[A]=E[X]10E[A]=E[X]-10

E[B]=E[X]/25E[B]=E[X]/2-5

E[A]=2E[B]E[A]=2E[B]


d)

V[A]=E[A2](E[A])2=E[(X10)2](E[X])2+20E[X]100=V[A]=E[A^2]-(E[A])^2=E[(X-10)^2]-(E[X])^2+20E[X]-100=

=E[X2]20E[X]+100(E[X])2+20E[X]100=E[X2](E[X])2=V[X]=E[X^2]-20E[X]+100-(E[X])^2+20E[X]-100=E[X^2]-(E[X])^2=V[X]


V[B]=E[B2](E[B])2=E[A2/4](E[A])2/4=V[X]/4V[B]=E[B^2]-(E[B])^2=E[A^2/4]-(E[A])^2/4=V[X]/4


e)

In probability theory, the law of large numbers  is a theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value and will tend to become closer to the expected value as more trials are performed.

So, for a large number of times scores of Arnold and Brian will be the same.


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