Fall 2002 - Course 3 Solutions

Question #39
Answer: D
Let A x and a x be calculated with µ x t
* *
b g and δ = 006 .
b g
Let Ax
and ax
be the corresponding values with µ x t
increased by 0.03 and δ decreased by 0.03
a
a
x
*
x
1 Ax
= − 0.
4
= = 6.
667
δ 0.
06
= a
x
L
N
M
t
*
¥ - µ x s + 0. 03 ds -0.
03t
x
0
0
Proof: a = e e dt
=
=
z
z
z
= a
* *
x x x
A = 1− 003 . a = 1−
0.
03a
0
x
¥
¥
0
t
x
0
e e e dt
-
z
-z
t
z0
µ
d
µ
= 1−
003 . 6667 .
= 08 .
x
b g
b g
b g
s ds
s ds
-0. 03t
-0.
03t
-0.
06t
e e dt
b gb g
i