Answer to Question #269332 in Statistics and Probability for sarah

Question

Two 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.

Answer 1

The random variable $X$ may take on the following difference between the values of the two cubes.

$x=0,1,2,3,4,5$ and its probability distribution is,

$x$ 0 1 2 3 4 5

$p(x)$ 1/6 5/18 2/9 1/6 1/9 1/18

a.

The mean of the probability distribution above is,

$E(x)=\sum x*p(x)=(0*1/6)+(1*5/18)+(2*2/9)+(3*1/6)+(4*1/9)+(5*1/18)=70/36= 1.944444$

b.

To find the variance of the random variable $X,$ we need to find the value of $E(X^2)$ given as,

$E(x^2)=\sum x^2*p(x)=(1*5/18)+(4*2/9)+(9*1/6)+(16*1/9)+(25*1/18)= 5.833333$

We now use the formula below to find the variance,

$V(X)=E(x^2)-(E(X))^2=5.833333-(1.944444)^2=2.052471$

The standard deviation is given as,

$SD(X)=\sqrt{V(X)}=\sqrt{2.052471}= 1.432645$

Therefore the variance and the standard deviation of $X$ are 5.833333 and 1.432645 respectively.

c.

Given that

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

$E(A)=E(X-10)=E(X)-10=1.944444-10= -8.055556$ and

$E(B)=E(1/2*X-5)=1/2*E(X)-5=(1/2*1.944444)-5= -4.027778$

Therefore, $E(A)$ and $E(B)$ are -8.055556 and -4.027778 respectively.

d.

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

$V(A)=V(X-10)=V(X)-V(10)=V(X)=2.052471$ since the variance of a constant is zero.

$V(B)=V(1/2*X-5)=(1/2)^2V(X)-V(5)=1/4*2.052471-0=1/4*2.052471= 0.5131178$

Therefore, $V(A)$ and $V(B)$ are 2.052471 and 0.5131178 respectively.

e.

When Arnold records his score using the random variable A and Brian uses the random variable B, then the random variable $A\sim Binom(n,p)$ where $n$ is the number of times Arnold rolls the pair of fair cubes and $p$ is the probability of success at each of the values that the random variable $A$ can take. Also, the random variable $B\sim Binom(n,p)$ where $n$ is the number of times Brian rolls the pair of fair cubes and $p$ is the probability of success at each of the values that the random variable $B$ can take. Therefore, the probability distribution for rolling fair cubes for both will have the same Binomial distribution with parameters $n$ and $p.$