a)
here as probabiolity of an outcome on a die p=1/6
hence
| Y1 |
F(Y1) |
Y1P(Y1) |
Y12P(Y1) |
| 1 |
1/6 |
0.167 |
0.167 |
| 2 |
1/6 |
0.333 |
0.667 |
| 3 |
1/6 |
0.500 |
1.500 |
| 4 |
1/6 |
0.667 |
2.667 |
| 5 |
1/6 |
0.833 |
4.167 |
| 6 |
1/6 |
1.000 |
6.000 |
|
total |
3.500 |
15.167 |
|
|
|
|
|
E(Y1) =μ= |
ΣY1P(Y1) = |
3.5000 |
|
E(Y1^2) = |
ΣY12P(Y1) = |
15.1667 |
|
Var(Y1)=σ2 = |
E(Y12)-(E(Y1))2= |
2.9167 |
|
std deviation=Â Â |
     σ= √σ2 = |
1.7078 |
mean of Y1 =E(Y1)=3.5
Std deviaiton of Y1 SD(Y1)=1.7078
b)
for P(Y2=0)=P(0 heads)=(1/2)*(1/2)=1/4
P(Y2=1)=P(1st toss head and secnd tail)+P(1st tail and 2nd
head)=(1/2)*(1/2)+(1/2)*(1/2)=1/2
P(Y2=2)=P(both heads)=(1/2)*(1/2)=1/4
| Y2 |
F(Y2) |
Y2P(Y2) |
Y22P(Y2) |
| 0 |
1/4 |
0.000 |
0.000 |
| 1 |
1/2 |
0.500 |
0.500 |
| 2 |
1/4 |
0.500 |
1.000 |
|
total |
1.000 |
1.500 |
|
|
|
|
|
E(Y2) =μ= |
ΣY2P(Y2) = |
1.0000 |
|
E(Y2^2) = |
ΣY22P(Y2) = |
1.5000 |
|
Var(Y2)=σ2 = |
E(Y22)-(E(Y2))2= |
0.5000 |
|
std deviation=Â Â |
     σ= √σ2 = |
0.7071 |
Here mean of Y2 =E(Y2)=1
std deviation of Y2 =0.7071
c)
mean of Y1+Y2 =E(Y1)+E(Y2)=3.5+1 =4.5
std deviaiton of Y1+Y2
=sqrt(Var(Y1)+Var(Y2))=sqrt(2.9167+0.5)=1.8484
d)
mean of 5+Y1 =5+E(Y1)=5+3.5=8.5
std deviaiton of 5+Y1 =SD(Y1)=1.7078
e)
mean of 5Y1 =5*E(Y1)=5*3.5=17.5
std deviation of 5Y1 =5*SD(Y1)=5*1.7078=8.539
