Decision Analysis

Q.5 Solution
Bill Denomination (X)
Number of Bills (f)
X*f
X2*f
$1
520
520
520
$5
260
1,300
6500
$10
120
1,200
12000
$20
70
1,400
28000
$50
29
1,450
72500
$100
1
100
10000
Total
1000
5,970
129520

a.

b.
c. The probability that a bar containing $50 or $100 bill is purchased is 30 / 1000 = 0.03
Hence customer have to buy 100 bars of soap, so that he or she has purchased three bars containing a $50 or $100 bill.

d. Given n=1000, E(X) = 5.97, SD(x) = 9.68912
The probability that a soap contains a bill of $20 or above is 100 / 1000 = 0.0.1
Binomial with n =8, p = 0.1

Hence the probability that at least one of these bar contains a bill of $20 or larger is 0.5595

Q.6 Answer
Number of Children
Frequency

1996
2006
1
3671455
3971276
2
100750
137085
3
5298
6118
4
560
355
5
81
67
a.
X
1
2
3
4
5
P (x)
0.97125
0.02665
0.00140
0.00015
0.00002
b. x 1
2
3
4
5
Total
P(x)
0.97125
0.02665
0.00140
0.00015
0.00002
1.00 x*P (x)
0.97125
0.0533
0.0042
0.00059
0.00011
1.03 x2*P (x)
0.97125
0.10661
0.01261
0.00237
0.00054
1.09

Expected value = E(x) = ∑ x*P (x) = 1.03
Variance = E(x2) – [E(x)]2 = 1.09-1.03*1.03 = 0.0291
c.
y
1
2
3
4
5
Total
P(y)
0.96463
0.03330
0.00149
0.00009
0.00002
1.00

d. y 1
2
3
4
5
Total
P(y)
0.96463
0.03330
0.00149
0.00009
0.00002
1.00 y*P (y)
0.96463
0.06660
0.00446
0.00034
0.00008
1.04 y2*P (y)
0.96463
0.13319
0.01337
0.00138
0.00041
1.11

Expected value = E(x) = ∑ x*P (x) = 1.04
Variance = E(x2) – [E(x)]2 = 1.11-1.04*1.04 = 0.0284
e. The expected value of the number of children born in a single pregnancy in 1996 was 1.03 and in 2006 it was 1.04. hence we do not support the conclusion that increased use of fertility drugs by older women has generated an upward trend in multiple births.
Q.7 Answer

Unit demand (x)
300
400
500
600
Total
p(x)
0.2
0.3
0.35