Fall 2002 - Course 3 Solutions

Question # 16
Answer: C
Let N = number of prescriptions then S
b g
E N
E
E
∞
∑
= 4 = 1−
F j
j=
0
c
b gh
n f n
0
1
2
3
bS − 80g = 40 × E N − 2 +
+
= N × 40
Nb g FNbng 1− FNbng
0.2000
0.1600
0.1280
0.1024
∞
b g = 40× ∑c1
− Fb jgh
j=
2
L ∞
1
O
∑
NM
c Fb jgh ∑c
Fb jgh
j=
0
j=
0 QP
b . g . .
= 40 × 1 − − 1 −
= 40 4 − 144 = 40 × 2 56 = 102 40
∞
bS −120g = 40× E N − 3 = 40× 1−
F j
+
b g+
∑c
b gh
j=
3
L ∞
2
O
∑
NM
c Fb jgh ∑c
Fb jgh
j=
0
j=
0 QP
b . g . .
= 40 × 1 − − 1 −
= 40 4 − 1952 = 40 × 2 048 = 8192
Since no values of S between 80 and 120 are possible,
0.2000
0.3600
0.4880
0.5904
0.8000
0.6400
0.5120
0.4096
E
b
g
S − 100 =
+
b120− 100g × E bS − 80g + 100− 80 × E S −120
+
b g b g
+
= 9216 .
120
Alternatively,
∞
b g+
∑b g Nb g Nb g Nb g Nb g
j=
0
E S − 100 = 40j − 100 f j + 100 f 0 + 60 f 1 + 20 f 2
(The correction terms are needed because b40 j −100g would be negative for j = 0, 1, 2; we need
to add back the amount those terms would be negative)
¥
¥
Nb g Nb g b gb g b gb g b gb g
j=
0 j=
0
= 40 j · f j - 100 f j + 100 0. 200 + 016 . 60 + 0128 . 20
b g . .
= 40 E N − 100+ 20+ 9 6+
256
= 160 − 67. 84 = 9216 .