HW 4 Solutions
H&G 2.5
(a)
x
|
f(x)
|
1
|
1/6
|
2
|
1/6
|
3
|
1/6
|
4
|
1/6
|
5
|
1/6
|
6
|
1/6
|
b) P(X=4) = f(4) = 1/6
P(X = 4 or 5) = P(X = 4) + P(X = 5) = f(4) + f(6) = 2/6
c) E(X) = (1/6) * 1 + (1/6)*2 + .... + (1/6) *6 = 3.5 using Equation
2.3.1
Meaning: If we were to toss the die infinitely many times, the average
would be 3.5.
d) E(X2) = (1/6)*12 + (1/6)*22 + (1/6)*32
+ (1/6)*42 + (1/6)*52 + (1/6)*62 = 15.1667
using Equation 2.3.2a
e) There are two ways. The longer way is to use the definition given in class:
Var(X) = E(X - E(X))2 = (1-3.5)2 * (1/6) + (2 - 3.5)2
* (1/6) + .... + (6-3.5)2 * (1/6)
The easier way is to use 2.3.4
Var(X) = E(X2) - (E(X))2 = 15.1667 - 3.52
= 2.9167
p. 226, #14
The probabilities have to add to 1. So p(0) + p(3) + p(7) + p(11) +
p(21) + p(2100) + p(free ticket) = 1
but we're ignoring p(free ticket), and so
p(0) + p(3) + .. + p(2100) = .9
To make all of the probabilities add to 1, we divide both sides of the equation
by .9. And the table of probabilities is then:
prize
|
0
|
3
|
7
|
11
|
21
|
2100
|
prob
|
?
|
.03703704
|
.011111
|
.011111
|
.0074
|
.00000370
|
X = payoff
The missing probability is 1 - (.0111111*2 + .03703704 + + .0074 +
.0000037) = .933337
E(X) = 3*.03703704 + 7*.011111 + 11*.0111111 + 21*.0074 + 2100*.0000037 =
$.474
Var(X) = (0 - .474)^2 * .03703704 + (7 - .474)^2* .0111111 + (11-.474)^2
* .011111 + (21-.474)^2*.0074 + (2100-.474)^2 * .00000370 = 21.13997
So SD(X) = sqrt(21.13997) = 4.60