Q UIZ 2
1. You are given:

0.04 if 0 < x < 40,
µ(x) =
0.05 if x > 40.
◦
Calculate e25:25 .
(A) 15.6
(B) 17.6
(C) 19.6
(D) 21.6
(E) 23.6
Solution. Let s denote the survival function of a newborn; then

Z x
 −0.04x
e
if x ≤ 40,
s(x) = exp −
µ(t) dt =
e−0.04·40 e−0.05·(x−40) if x ≥ 40,
0
and

−0.04t


e
if x + t ≤ 40,
s(x + t)
= something we do not need if x ≤ 40, x + t ≥ 40,
t px =

s(x)

 −0.05t
e
if x ≥ 40.
Finally,
◦
◦
◦
e25:25 = e25:15 + 15 p25 e40:10
Z 15
Z
−0.04t
−0.04·15
=
e
dt + e
0
10
e−0.05t dt
0
1 − e−0.05·10
1−e
+ e−0.04·15
0.04
0.05
−0.6
−0.5
1−e
1−e
=
+ e−0.6
≈ 15.5985.
0.04
0.05
−0.04·15
=
1
2
2. For T , the future lifetime random variable for (0):
(a) ω > 70;
(b) 40 p0 = 0.6;
(c) E(T ) = 62;
(d) E(min(T, t)) = t − 0.005 t2 , 0 < t < 60.
Calculate the complete expectation of life at 40.
(A) 10
(B) 20
(C) 30
(D) 40
(E) 50
Solution. Since
◦
◦
◦
e0 = e0:40 + 40 p0 e40 ,
we have
◦
◦
e40
◦
e0 − e0:40
=
40 p0
E(T ) − E(min(T, 40))
=
◦
e40:0
=
62 − (40 − 0.005 · 402 )
= 50.
0.6
3. Scientists are searching for a vaccine for a disease. You are given:
(i) 100, 000 lives age x are exposed to the disease.
(ii) Future lifetimes are independent, except that the vaccine, if available, will be given to all at the end of year 1.
(iii) The probability that the vaccine will be available is 0.2.
(iv) For each life during year 1, qx = 0.02.
(v) For each life during year 2, qx+1 = 0.01 if the vaccine has been
given, and qx+1 = 0.02 if it has not been given.
Calculate the standard deviation of the number of survivors at the
end of year 2.
(A) 100
(B) 200
(C) 300
(D) 400
(E) 500
Solution. Let S be the number of survivors, and
A = { Vaccine is available. }
P(A = N o) = 1 − 0.2 = 0.8
2 p0 given No = (0.98 during year 1)(0.98 during year 2) = 0.9604.
3
Since E(S|N o),Var(S|N o) and E(S2|N o) are just binomial, with
n = 100, 000, and p(success) = 0.9604, we calculate E(S|N o) =
np = 96, 040, Var(S|N o) = npq = 3, 803.
Similarly, 2 p0 given Yes = (0.98 during year 1)(0.99 during year 2)
= 0.9702. E(S|Y es) = np = 97, 020, Var(S|N o) = npq = 2, 891.
By the conditional variance formula:
Var(S) = Var[E(S|A)] + E[V ar(S|A)]
= 0.2(0.8)(97, 020 − 96, 040)2 + 0.2(2, 891) + 0.8(3, 803)
= 153, 664 + 3, 621 = 157, 285
Hence the standard deviation of S is 397.
4. You are given the survival function:
x 31
S0 (x) = 1 −
, where 0 ≤ x ≤ 60.
60
Calculate 1000µ35 .
(A) 5.6
(B) 6.7
(C) 13.3
(D) 16.7
Solution.
d
1 d
x
ln S0 (x) = −
ln 1 −
dx
3 dx
60
1
.
=
3(60 − x)
µx = −
Therefore, µ35 =
1
75
= 0.0133.
(E) 20.1