http://oi48.tinypic.com/207l2tl.jpg . . . . [Mean and standard deviation] Suppos

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[Mean and standard deviation] Suppose three fair dice are rolled independently, so the sample space is Ohm = {(i, j, k) : 1 i, j, l 6} and all outcomes are equally likely. Let X be the number showing on the first die, X(i, j, l) = i, and let Y be the random variable defined by Y(i, j, k) = min{i, j, l}. Derive the pmf of X and sketch it. Find the mean E[X] and standard deviation, sigma X, of X. Correct numerical answers are fine, but show your work. Derive the pmf of Y and sketch it. Find the mean E[Y] and standard deviation, sigma Y, of Y. Which is larger, sigma X or sigma Y? Is that consistent with your sketches of the pmfs?

Explanation / Answer

a) Here X can take the values 1,2,...,6 and all the values are equally likely. Hence the pmf of X is as follows:

i:--------1-------2-------3-------4-------5-------6
p(i):----1/6----1/6-----1/6----1/6----1/6----1/6

b) E(X)= summation i from 1 to 6 [ i*p(i)]
= summation i from 1 to 6 [ i/6]
= 1/6*(6*(6+1)/2)= (6+1)/2= 3.5

E(X^2)= summation i from 1 to 6 [ i^2*p(i)]
= summation i from 1 to 6 [ i^2/6]
=1/6*[6(6+1)(2*6+1)/6] = 7*13/6 = 91/6 = 15.166667

hence Var(X)= E(X^2)-(E(X)^2)= 15.166667-(3.5)^2 = 2.91667

c) Y is min{1,j,k}.

here Y can take values 1,2,3,4, 5 and 6.

P( Y=i)= 3*1/6*((6-i)/6)^2+ 3*1/36*(6-i)/6+ 1/6^3= 1/216*[3*(6-i)^2+ 3*(6-i)+1] as out of i,j and k one, two or three can be i. If only one of them is i, then that can happen in 3 ways and for any of the 3 ways, the probability = probability that one outcome is i times probability that out of the remaining 2 outcomes, each is greater than i . This accounts for the first term. If two of them are i, then that can happen in 3C2=3 ways and for any of the 3 ways, the probability = probability that two outcomes are i times probability that the remaining outcome is greater than i . This accounts for the second term. And probability that all 3 are i is 1/6^3 accounting for the 3rd term.

P(Y=1)= 1/216[ 3* 25+ 3* 5 +1]= 91/216

P(Y=2)= 1/216[ 3* 16+ 3* 4 +1]=61/216

P(Y=3)= 1/216[ 3* 9+ 3* 3 +1]= 37/216

P(Y=4)=1/216[ 3* 4+ 3* 2 +1]=19/216

P(Y=5)=1/216[ 3* 1+ 3* 1 +1]=7/216

P(Y=6)= 1/216

c) E(Y)= summation i from 1 to 6 [i*P(Y=i)]
= 91/216+ 61*2/216+ 37*3/216+ 19*4/216+ 7*5/216+ 1*6/216= 2.041666667

E(Y^2)= summation i from 1 to 6 [i^2*P(Y=i)]
=91/216+ 61*4/216+ 37*9/216+ 19*16/216+ 7*25/216+ 1*36/216
=5.47685

hence Var(Y)= E(Y^2)-(E(Y)^2)= 1.30845

d) clearly sigmaX is more than sigmaY.

that is consistent with the sketches of the pmf's as for X, the values 1,2..6 are equally likely and hence the chances of variation is more but for Y, the values close to the eman occur with more probability than values far away from it.. hence variance is less.

PLEASE RATE AND REWARD..:)

REGARDS ABHISHEK..

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