3) from given pmf:
| x | P(x) | xP(x) | x2P(x) |
| 1 | 0.1 | 0.1 | 0.1 |
| 2 | 0.2 | 0.4 | 0.8 |
| 3 | 0.2 | 0.6 | 1.8 |
| 4 | 0.1 | 0.4 | 1.6 |
| 5 | 0.25 | 1.25 | 6.25 |
| 6 | 0.15 | 0.9 | 5.4 |
| total | 1 | 3.65 | 15.95 |
a)E(X) =
xP(x) =3.65
b) E(X2) =
x2P(x)
=15.95
hence Var(X) = E(X2)-(E(X))2 =15.95-3.652 =2.6275
c) from Y=2/X ;
below is distribution of Y:
| y | P(y) | yP(y) | y2P(y) |
| 2.000 | 0.100 | 0.200 | 0.400 |
| 1.000 | 0.200 | 0.200 | 0.200 |
| 0.667 | 0.200 | 0.133 | 0.089 |
| 0.500 | 0.100 | 0.050 | 0.025 |
| 0.400 | 0.250 | 0.100 | 0.040 |
| 0.333 | 0.150 | 0.050 | 0.017 |
| total | 1 | 0.7333 | 0.7706 |
i) E(Y) =0.7333
ii) Var(Y) =0.7706-0.73332 =0.2328
4)
E(Y) =E((X-
x)/
x)
=(1/
x)*(E(X)-E(
x))
=(1/
x)*(
x-
x)
=0
Var(Y) =Var((X-
x)/
x)
=(1/
x)*(Var(X)+Var(
x))
=(1/
x)*
x
=1
3. Let X be a discrete random variable with the following PMF: 0.1 for x 0.2...
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