Asked by sharik
two dices are tossed once. let the random variable be t he sum of the up faces on the dice. A). find and graph the probability distribution of the random variable. and b) calculate the mean (or expectation) of this distribution
Answers
Answered by
Mgraph
X=2=1+1 (1st face + 2nd face)P=1/36
X=3=1+2=2+1 P=2/36
X=4=1+3=3+1=2+2 P=3/36
X=5=1+4=4+1=2+3=3+2 P=4/36
X=6=1+5=5+1=2+4=4+2=3+3 P=5/36
X=7=1+6=6+1=2+5=5+2=3+4=4+3 P=6/36
X=8=2+6=6+2=3+5=5+3=4+4 P=5/36
X=9=3+6=6+3=4+5=5+4 P=4/36
X=10=4+6=6+4=5+5 P=3/36
X=11=5+6=6+5 P=2/36
X=12=6+6 P=1/36
Note 1/36=(1/6)*(1/6),
1/36+2/36+3/36+4/36+5/36+6/36+
5/36+4/36+3/36+2/36+1/36=36/36=1
B) m(X)=2*(1/36)+3*(2/36)+4*(3/36)+
5*(4/36)+6*(5/36)+7*(6/36)+8*(5/36)+
9*(4/36)+10*(3/36)+11*(2/36)+12*(1/36)=
(2+6+12+20+30+42+40+36+30+22+12)/36=
252/36=7
We can also use the independence of numbers on two dice. The mean of the sum
= the sum of two means: if X=X1+X2 and X1, X2 are independent then m(X1+X2)=m(X1)+m(X2)
7=3.5+3.5
X=3=1+2=2+1 P=2/36
X=4=1+3=3+1=2+2 P=3/36
X=5=1+4=4+1=2+3=3+2 P=4/36
X=6=1+5=5+1=2+4=4+2=3+3 P=5/36
X=7=1+6=6+1=2+5=5+2=3+4=4+3 P=6/36
X=8=2+6=6+2=3+5=5+3=4+4 P=5/36
X=9=3+6=6+3=4+5=5+4 P=4/36
X=10=4+6=6+4=5+5 P=3/36
X=11=5+6=6+5 P=2/36
X=12=6+6 P=1/36
Note 1/36=(1/6)*(1/6),
1/36+2/36+3/36+4/36+5/36+6/36+
5/36+4/36+3/36+2/36+1/36=36/36=1
B) m(X)=2*(1/36)+3*(2/36)+4*(3/36)+
5*(4/36)+6*(5/36)+7*(6/36)+8*(5/36)+
9*(4/36)+10*(3/36)+11*(2/36)+12*(1/36)=
(2+6+12+20+30+42+40+36+30+22+12)/36=
252/36=7
We can also use the independence of numbers on two dice. The mean of the sum
= the sum of two means: if X=X1+X2 and X1, X2 are independent then m(X1+X2)=m(X1)+m(X2)
7=3.5+3.5