Question #208473

7. Two fair cubical dice are thrown: one is red and one is blue. The random variable M represents the score on the red die minus the score on the blue die. (a) Find the distribution of M. (b) Write down E(M). (c) Find Var(M).


1
Expert's answer
2021-06-22T09:00:25-0400
SampleRed,BlueM11,1021,2−131,3−241,4−351,5−461,6−572,1182,2092,3−1102,4−2112,5−3122,6−4133,12143,21153,30163,4−1173,5−2183,6−3194,13204,22214,31224,40234,5−1244,6−2255,14265,22275,32285,41295,50305,6−1316,15326,24336,33346,42356,51366,60\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Red,Blue & M \\ \hline 1 & 1,1 & 0 \\ \hdashline 2 & 1,2 & -1 \\ \hdashline 3 & 1,3 & -2 \\ \hdashline 4 & 1,4 & -3\\ \hdashline 5 & 1,5 & -4 \\ \hdashline 6 & 1,6 & -5 \\ \hdashline 7 & 2,1 & 1 \\ \hdashline 8 & 2,2 & 0 \\ \hdashline 9 & 2,3 & -1 \\ \hdashline 10 & 2,4 & -2\\ \hdashline 11 & 2,5 & -3 \\ \hdashline 12 & 2,6 & -4 \\ \hdashline 13 & 3,1 & 2 \\ \hdashline 14 & 3,2 & 1 \\ \hdashline 15 & 3,3 & 0 \\ \hdashline 16 & 3,4 & -1\\ \hdashline 17 & 3,5 & -2 \\ \hdashline 18 & 3,6 & -3 \\ \hdashline 19 & 4,1 & 3 \\ \hdashline 20 & 4,2 & 2 \\ \hdashline 21 & 4,3 & 1 \\ \hdashline 22 & 4,4 & 0\\ \hdashline 23 & 4,5 & -1 \\ \hdashline 24 & 4,6 & -2 \\ \hdashline 25 & 5,1 & 4 \\ \hdashline 26 & 5,2 & 2 \\ \hdashline 27 & 5,3 & 2 \\ \hdashline 28 & 5,4 & 1\\ \hdashline 29 & 5,5 & 0 \\ \hdashline 30 & 5,6 & -1 \\ \hdashline 31 & 6,1 & 5 \\ \hdashline 32 & 6,2 & 4 \\ \hdashline 33 & 6,3 & 3 \\ \hdashline 34 & 6,4 & 2\\ \hdashline 35 & 6,5 & 1 \\ \hdashline 36 & 6,6 & 0 \\ \hline \end{array}




Mff(M)Mf(M)M2f(M)−511/36−5/3625/36−422/36−8/3632/36−333/36−9/3627/36−244/36−8/3616/36−155/36−5/365/36066/3600155/365/365/36244/368/3616/36333/369/3627/36422/368/3632/36511/365/3625/36Total361035/6\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} M & f & f(M) & Mf(M)& M^2f(M) \\ \hline -5 & 1 & 1/36 & -5/36 & 25/36 \\ \hdashline -4 & 2 & 2/36 & -8/36 & 32/36 \\ \hdashline -3 & 3 & 3/36 & -9/36 & 27/36 \\ \hdashline -2 & 4 & 4/36 & -8/36 & 16/36 \\ \hdashline -1 & 5 & 5/36 & -5/36 & 5/36 \\ \hdashline 0 & 6 & 6/36 & 0 & 0 \\ \hdashline 1 & 5 & 5/36 & 5/36 & 5/36 \\ \hdashline 2 & 4 & 4/36 & 8/36 & 16/36 \\ \hdashline 3 & 3 & 3/36 & 9/36 & 27/36 \\ \hdashline 4 & 2 & 2/36 & 8/36 & 32/36 \\ \hdashline 5 & 1 & 1/36 & 5/36 & 25/36 \\ \hdashline Total & 36 & 1 & 0 & 35/6 \\ \hline \end{array}


1.


Mp(M)−51/36−42/36−33/36−24/36−15/3606/3615/3624/3633/3642/3651/36\begin{matrix} M & p(M) \\ \hline -5 & 1/36 \\ -4 & 2/36 \\ -3 & 3/36 \\ -2 & 4/36 \\ -1 & 5/36 \\ 0 & 6/36 \\ 1 & 5/36 \\ 2 & 4/36 \\ 3 & 3/36 \\ 4 & 2/36 \\ 5 & 1/36 \\ \end{matrix}


2.


E(M)=∑Mp(M)=0E(M)=\sum Mp(M)=0

The mean of the sampling distribution of the sample means is equal to the

the mean of the population.


E(Xˉ)=0=μE(\bar{X})=0=\mu



3.


Var(M)=∑M2p(M)−(∑Mp(M))2Var(M)=\sum M^2p(M)-(\sum Mp(M))^2




=356−(0)2=356=\dfrac{35}{6}-(0)^2=\dfrac{35}{6}




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