a)
|
x |
f(x) |
xP(x) |
x2P(x) |
|
1.0000 |
1/3 |
0.333 |
0.333 |
|
2.0000 |
1/3 |
0.667 |
1.333 |
|
3.0000 |
1/3 |
1.000 |
3.000 |
|
|
total |
2.000 |
4.667 |
|
|
|
|
|
|
|
E(x) =?= |
?xP(x) = |
2.0000 |
|
E(x2) = |
?x2P(x) = |
4.6667 |
|
Var(x)=?2 = |
E(x2)-(E(x))2= |
0.6667 |
|
std
deviation= |
?= ??2 = |
0.8165 |
from above E(X)=2
b)SD(X)=0.8165
c)P(X=1)=P(x1=1 ; x2=1) =(1/3)*(!/3) =1/9
P(X=1.5)=P(x1=1 ; x2=2)+P(x1=2 ; x2=1)
=(1/3)*(!/3)+(!/3)*(!/3)=2/9
P(X=2)=P(x1=2 ; x2=2)+P(x1=3 ; x2=1)+P(x1=1 ; x2=3)
=(1/3)*(!/3)+ (1/3)*(!/3)+(!/3)*(!/3)=3/9=1/3
P(X=2.5)=P(x1=3 ; x2=2)+P(x1=2 ; x2=3)
=(1/3)*(!/3)+(!/3)*(!/3)=2/9
P(X=3)=P(x1=3 ; x2=3) =(1/3)*(!/3) =1/9
d)
|
x |
f(x) |
xP(x) |
x2P(x) |
|
1.0000 |
1/9 |
0.111 |
0.111 |
|
1.5000 |
2/9 |
0.333 |
0.500 |
|
2.0000 |
1/3 |
0.667 |
1.333 |
|
2.5000 |
2/9 |
0.556 |
1.389 |
|
3.0000 |
1/9 |
0.333 |
1.000 |
|
|
total |
2.000 |
4.333 |
|
|
|
|
|
|
|
E(x) =?= |
?xP(x) = |
2.0000 |
|
E(x2) = |
?x2P(x) = |
4.3333 |
|
Var(x)=?2 = |
E(x2)-(E(x))2= |
0.3333 |
|
std
deviation= |
?= ??2 = |
0.5774 |
from E(Xbar)=2.000
and SD(Xbar ) =0.5774
e)
as E(Xbar)=E(X)
as well SD(Xbar ) = ? /sqrt(n) =SD(X)/sqrt(2)
therefore answers are consistent
