Download Math 408, Spring 2005 Final Exam Solutions

Math 408, Spring 2005
Final Exam Solutions
1. Assume A and B are independent events with P (A) = 0.2 and P (B) = 0.3. Let C
be the event that neither A nor B occurs, let D be the event that exactly one
of A or B occurs.
(a) Find P (C).
Solution. P (C) = P (A0 ∩ B 0 ) = P (A0 )P (B 0 ) = (1 − P (A))(1 − P (B)) =
0.8 · 0.7 = 0.56
0.56
(b) Find P (D).
Solution. P (D) = P (A ∪ B \ A ∩ B) = P (A ∪ B) − P (A)P (B) = P (A) +
P (B) − 2P (A ∩ B) = 0.38
0.38
(c) Find P (A|D).
Solution. P (A|D) = P (A ∩ D)/P (D) = P (A \ A ∩ B)/P (D) = (0.2 − 0.2 ·
0.3)/0.38 = 7/19
7
19
= 0.368
(d) Are C and D independent? Justify your answer!
Solution. C and D are not independent since P (C∩D) = 0, but P (C)P (D) =
0.56 · 0.38 6= 0.
NO
2. [2.4-9, Variant] Suppose A, B, and C are mutually independent events with probabilities P (A) = 0.5, P (B) = 0.8, and P (C) = 0.3. Find the probability that at
least one of these events occurs.
0.93
Solution. The probability in question is P (A ∪ B ∪ C). Using the formula for
the probability of a union of three events, and the independence of the three sets,
we get
P (A ∪ B ∪ C) = P (A) + P (B) + P (C) − P (A ∩ B) − P (A ∩ C) − P (B ∩ C)
+ P (A ∩ B ∩ C)
= 0.5 + 0.8 + 0.3 − (0.5 · 0.8) − (0.5 · 0.3) − (0.8 · 0.3)
+ (0.5 · 0.8 · 0.3) = 0.93 .
1
MATH 408
Final Exam
Spring 2005
3. How many ways are there to seat 10 people, consisting of 5 couples, in a row of
seats (10 seats wide) if all couples are to get adjacent seats?
Solution. 5!25 = 3840 (or 10 · 8 · 6 · 4 · 2 = 3840)
3840
4. [Actuarial Exam Problem #27] The probability that a randomly chosen male has
a circulation problem is 0.25. Males who have a circulation problem are twice as
likely to be smokers as those who do not have a circulation problem. What is the
conditional probability that a male has a circulation problem, given that he is a
smoker?
0.4
Solution. Let C denote the event “has a circulation problem” and S the event
“is a smoker”. We are given that P (C) = 0.25 and P (S|C) = 2P (S|C 0 ), and we
need to compute P (C|S). By Bayes’ Rule,
P (C|S) =
=
P (S|C)P (C)
P (S|C)P (C) + P (S|C 0 )P (C 0 )
0.25
2
P (S|C)0.25
=
=
= 0.4
P (S|C)0.25 + (1/2)P (S|C)(1 − 0.25)
0.25 + 0.375
5
5. [Similar to Actuarial Exam Problem #57] Suppose a random variable X has moment generating function
9
2 + et
M (t) =
.
3
Calculate the variance of X.
2
Solution. We use the formula Var X = E(X 2 ) − E(X)2 = M 00 (0) − M 0 (0)2 .
Computing the derivatives of M (t), we get
8 t
2 + et
e
M (t) = 9
· ,
3
3
8
7
2 + et
e2t
2 + et
00
M (t) = 3 · 8
+3
et ,
3
3
3
8
2 + e0
0
e0 = 3 ,
M (0) = 3
3
7
8
3
3
00
M (0) = 8
+3
= 11 ,
3
3
0
so Var(X) = 11 − 32 = 2 .
MATH 408
Final Exam
Spring 2005
6. [Actuarial Exam Problem #37] The lifetime of a printer costing 200 is exponentially distributed with mean 2 years. The manufacturer agrees to pay a full refund
to a buyer if the printer fails during the first year following its purchase, and a
one-half refund if it fails during the second year. If the manufacturer sells 100
printers, how much should it expect to pay in refunds?
10, 256
Solution. Let X denote the lifetime of a printer, and let Y denote the refund
paid by the manufacturer on a single printer. Now,


200 if X ≤ 1,
Y = 100 if 1 < X ≤ 2,


0
if X > 2,
so
E(Y ) = 200P (X ≤ 1) + 100P (1 < X ≤ 2)
= 200(1 − e−1/2 ) + 100((1 − e−2/2 ) − (1 − e−1/2 )) = 102.56,
since P (X ≤ x) = 1 − e−x/θ = 1 − e−x/2 by the given (exponential) distribution of
X. The expected refund on 100 such printers then is 100E(Y ) = 10, 256 .
7. [Actuarial Exam Problem #56] An insurance policy is written to cover a loss, X,
where X has uniform distribution on [0, 1000]. At what level must a deductible
be set in order for the expected payment to be 25% of what it would be with no
deductible?
500
Solution. Let Y be the insurance payment and D the deductible. Then
(
0
if X ≤ D,
Y =
X − D if D ≤ X ≤ 1000.
Since X has density f (x) = 1/1000 on the interval [0, 10000], we get
Z 1000
1
(1000 − D)2
E(Y ) =
(x − D)
dx =
.
1000
2000
D
Now the expected payment with no deductible is E(Y ) = 500 (obtained by setting
D = 0 above, or by symmetry arguments), so we need to choose D such that
E(Y ) = 125. This leads to the equation (1000 − D)2 = 125 · 2000, and solving for
D we get D = 500 .
MATH 408
Final Exam
Spring 2005
8. [Actuarial Exam Problem #46] A device that continuously measures and records
seismic activity is placed in a remote region. The time, T , to failure of this
device is exponentially distributed with mean 3 years. Since the device will not be
monitored during its first two years of service, the time to discovery of its failure
is X = max(T, 2). Determine E(X).
3.54
Solution. First note that since T has exponential distribution with mean 3, the
density of T is f (t) = (1/3)e−t/3 for t ≥ 0. Since X = max(T, 2) equals 2 if T ≤ 2,
and T if T ≥, it follows that (using integration by parts for the second integral)
Z 2
Z ∞
1 −t/3
1
E(X) =
2· e
dt +
t · e−t/3 dt
3
3
0
2
Z ∞
−t/3 2 −t/3 ∞
= 2 −e
+ −e
t 2 +
e−t/3 dt
0
2
1 −2/3
e
= 2 + 3e−2/3 = 3.54 .
= 2(1 − e−2/3 ) − (−e−2/3 2) +
1/3
9. [Actuarial Exam Problem #76] Claim amounts for wind damage to insured homes
are independent random variables with common density function f (x) = 3x−4 for
x > 1, and f (x) = 0 otherwise, where x is the amount of a claim in thousands.
Suppose 3 such claims are made. What is the expected value of the largest of the
three claims?
2025
Solution. Let X1 , X2 , X3 denote the three claims, and Y = max(X1 , X2 , X3 )
the largest of these. We need to compute E(Y ). This requires first computing the
c.d.f. F (y) and the p.d.f. f (y) = F 0 (y) of Y . By the maximum trick,
F (y) = P (max(X1 , X2 , X3 ) ≤ y) = P (X1 ≤ y)P (X2 ≤ y)P (X3 ≤ y)
3
Z y
3 1
−4
=
3x dx = 1 − 3 ,
y
1
so
1
f (y) = F (y) = 3 1 − 3
y
0
2
3
9
=
y4
y4
1
1− 3
y
2
for y ≥ 1. Therefore
2
9
1
E(Y ) =
yf (y)dy =
1 − 3 dy
y3
y
1
1
Z ∞
1
2
1
=9
− 6 + 9 dy
3
y
y
y
1
81
1 2 1
=9
− +
=
= 2.025
2 5 8
40
Z
∞
Z
∞
Since the units are thousands of dollars, the answer is 2025 .
MATH 408
Final Exam
Spring 2005
10. [5.3-15] Let X have uniform distribution on the interval [0, 2], and given X = x,
let Y have uniform distribution on the interval [0, x2 ].
(a) Find the joint density f (x, y) of X and Y . (Be sure to specify the range!)
Solution. We are given that fX (x) = 1/2, 0 ≤ x ≤ 2 (uniform distribution
on [0, 2]), and g(y|x) = 1/x2 , 0 ≤ y ≤ x2 (uniform distribution on [0, x2 ]).
Therefore
(a)
f (x, y) = g(y|x)fX (x) =
=
1
,
2x2
1 1
·
x2 2
0 ≤ x ≤ 2, 0 ≤ y ≤ x2 ,
(b) Find the marginal density fY (y) of Y . (Be sure to specify the range!)
Solution.
x=2
Z
Z 2
1
1
(b)
fY (y) = f (x, y)dx =
dx = −
2
√
2x x=√y
x= y 2x
1
1
= − + √ , 0≤y≤4
4 2 y
(c) Find E(XY ).
Solution. We have
Z
Z
E(XY ) = xyf (x, y)dydx =
2
Z
x=0
=
1
2
Z
2
x=0
2 2
1 (x )
1
·
dx =
x
2
4
Z
0
x2
(xy)
y=0
2
3
1
dydx
2x2
x dx =
1 1 4
· 2 = 1.
4 4
MATH 408
Final Exam
Spring 2005
11. [From Double Integral handout, variant of Actuarial Exam Problem #91] Let X
and Y be random variables with joint density
f (x, y) = x − y + 1,
0 ≤ x, y ≤ 1.
Find the probability that X + Y ≥ 0.5.
0.875
Solution. The probability is given by the double integral over the above density
function over the part of the unit square on which x + y ≥ 0.5:
Z 0.5 Z 1
Z 1 Z 1
(x − y + 1)dydx +
(x − y + 1)dydx
x=0 y=0.5−x
x=0.5 y=0
Z 1 Z 0.5 1 2
1 2
1
1
=
xy − y + y dx +
xy − y + y dx
2
2
y=0.5−x
y=0
x=0
x=0.5
Z 0.5 1
1
1
1
=
x(1 − + x) − (1 − ( − x)2 ) + (1 − + x) dx
2
2
2
2
0
Z 1
1
+
x+
dx
2
0.5
Z 0.5 1
3 2
1 2 1 1
+ x + x dx +
x + x =
8
2
2
2
0.5
0
=
0.77
1 1
1
1
3 3 1
7
· +
+
· + + =
= 0.875
2
3
2 8 2·2
3·2 2 8 4
8
12. [Actuarial Exam Problem #87, Variant] A computer generates 48 random real
numbers, rounds each number to the nearest integer and then computes the average
of these 48 rounded values. Assume that the numbers generated are independent of
each other and that the rounding errors are distributed uniformly on the interval
[−0.5, 0.5]. Find the approximate probability that the average of the rounded
values is within 0.05 of the average of the exact numbers.
P
Solution. Let X1 , . . . , X48 denote the 48 rounding errors, and X = (1/48) 48
i=1 Xi
their average. We need to compute P (|X| ≤ 0.05). Since a rounding error
is uniformly
on [−0.5, 0.5], its mean is µ = 0 and its variance is
R 0.5 2 distributed
3
0.5
σ = −0.5 x dx = [x /3]−0.5 = 1/12. By the Central Limit Theorem, X has approximate distribution N (µ, σ 2 /n) = N (0, (1/12)/48) = N (0, 1/242 ). Thus 24X
is approximately standard normal, so
P (|X| ≤ 0.05) ≈ P (24 · (−0.05) ≤ 24X ≤ 24 · 0.05)
= Φ(1.2) − Φ(−1.2) = 2Φ(1.2) − 1 = 0.77 .
MATH 408
Final Exam
Spring 2005
13. [Actuarial Exam Problem #83] A company manufactures a brand of light bulb
with a lifetime in months that is normally distributed with mean 3 and variance
1. A consumer buys a number of these bulbs with the intention of replacing them
successively as they burn out. The light bulbs have independent lifetimes. What
is the smallest number of bulbs to be purchased so that the succession of light
bulbs produces light for at least 40 months with probability 0.9772?
16
Solution.
Let n be the (unknown) numberPof light bulbs to be purchased,
X1 , . . . , Xn their respective lifetimes, and S = ni=1 Xi the total lifetime of all n
bulbs. We need to choose n minimal so that P (S ≥ 40) ≥ 0.9772. Now, by the
CLT,
S − 3n
40 − 3n
40 − 3n
√ ≥
√
√
P (S ≥ 40) ≈ P
≈1−Φ
1 n
1 n
1 n
√
√
This is equal√
to 0.9772 when (40−3n)/ n = −2, or equivalently, 3n−2 n−40 = 0.
Setting x = n, the latter equation becomes 3x2 − 2x − 40 = 0. Solving (ignoring
√
the negative solution) gives x = (1/6)(2 + 4 + 480 = 4, so n = x2 = 16 is the
number sought.
14. [6.3-12] Assume the math scores on the SAT test are normally distributed with
mean 500 and standard deviation 60, and the verbal scores are normally distributed
with mean 450 and standard deviation 80. If two students who took both tests
are chosen at random, what is the probability that the first student’s math score
exceeds the second student’s verbal score?
0.6915
Solution. Let X and Y denote the scores of the two students. Then X − Y is
N (500 − 450, 602 + 802 ) = N (50, 1002 ), so
X − Y − 50
0 − 50
P (X > Y ) = P (X − Y > 0) = P
>
= P (Z > −1/2)
100
100
≈ 1 − Φ(−1/2) = Φ(1/2) = 0.6915
15. [Example 6.3-2] Let X1 , X2 , X3 , X4 be a random sample of size 4 from the normal distribution N (76.4, 383), and let X be the sample mean and S 2 the sample
variance. Determine a such that P (S 2 ≤ a) = 0.90.
a = 798
Solution.
We know that S 2 (n − 1)/σ 2 = (3/383)S 2 = 0.0078S 2 has χ2 (3)
distribution, and from the chi-square table the 90-th percentile of a χ2 (3) is 6.251.
Thus,
0.90 = P ((3/383)S 2 ≤ 6.251) = P (S 2 ≤ 6.251(383/3)) = P (S 2 ≤ 798).
so a = 798 .
MATH 408
Final Exam
Spring 2005
16. [6.1-1(b)] Suppose X and Y are independent, each having Poisson distribution
with means 2 and 3, respectively. Let Z = X + Y . Find P (X + Y = 1).
0.0336
Solution.
P (X + Y = 1) = P (X = 0, Y = 1) + P (X = 1, Y = 0)
= (e−2 20 /0!)(e−3 31 /1!) + (e−3 30 /0!)(e−2 21 /1!)
= 5e−5 = 0.0336 .