November 2001 - Course 1 SOA Solutions

Course 1 Solutions
November 2001 Exams

1. A
For i = 1, 2, let
R = event that a red ball is drawn form urn i
i
B = event that a blue ball is drawn from urn i .
i
Then if x is the number of blue balls in urn 2,
( R1∩R2) ∪( B1∩B2) R1∩R2 [ B1∩B2]
[ R ] [ R ] + [ B ] [ B ]
0.44 = Pr[ ] = Pr[ ] + Pr
= Pr Pr Pr Pr
1 2 1 2
4 ⎛ 16 ⎞ 6 ⎛ x ⎞
= ⎜ ⎟+
⎜ ⎟
10 ⎝ x+ 16 ⎠ 10 ⎝ x+
16 ⎠
Therefore,
32 3x
3x+
32
2.2 = + =
x+ 16 x+ 16 x+
16
2.2x+ 35.2 = 3x+
32
0.8x
= 3.2
x = 4
2. C
We are given that
f x, y dA= 6 and dA=
2
∫∫ R
( )
∫∫ R
It follows that
∫∫⎡4 ( , ) 2 4 ( , ) 2
R ⎣ f xy− ⎤⎦dA= ∫∫f xydA−
dA
R ∫∫ R
= 4 6 − 2 2 = 20
( ) ( )
Course 1 Solutions 1 November 2001

3. A
Since
3 4 2 5
S = 175L A , we have
∂ S
1 4
=
2 5
262.5 L A > 0 for
∂L
L > 0 , A > 0
2
∂ S
− 1 4
2 5
= 131.25 L A > 0 for
2
∂L
L> 0 , A>
0
∂ S
3 −1 = 140 L 2
A 5 > 0 for
∂A
L > 0 , A > 0
2
∂ S
3 −6
2 5
=− 28 L A < 0 for
2
∂A
L> 0 , A>
0
It follows that S increases at an increasing rate as L increases, while S increases at a
decreasing rate as A increases.
4. B
Apply Baye’s Formula:
Pr ⎡⎣Seri. Surv. ⎤⎦
=
Pr ⎡⎣Surv. Seri. ⎤⎦Pr[ Seri. ]
Pr ⎡⎣Surv. Crit. ⎤ ⎦Pr[ Crit. ] + Pr ⎡⎣Surv. Seri. ⎤ ⎦Pr[ Seri. ] + Pr ⎡⎣Surv. Stab. ⎤⎦Pr[ Stab. ]
( 0.9)( 0.3)
0.29
( 0.6)( 0.1) + ( 0.9)( 0.3) + ( 0.99)( 0.6)
= =
Course 1 Solutions 2 November 2001

5. A
Let us first determine K. Observe that
⎛ 1 1 1 1 ⎞ ⎛60 + 30 + 20 + 15 + 12 ⎞ ⎛137
⎞
1= K⎜1+ + + + ⎟= K⎜ ⎟=
K⎜ ⎟
⎝ 2 3 4 5 ⎠ ⎝ 60 ⎠ ⎝ 60 ⎠
60
K =
137
It then follows that
Pr N = n = Pr ⎣⎡ N = n Insured Suffers a Loss⎤⎦Pr Insured Suffers a Loss
60 3
= ( 0.05 ) = , N = 1,...,5
137N
137N
P = E X where
[ ] [ ]
Now because of the deductible of 2, the net annual premium [ ]
⎧0 , if N ≤ 2
X = ⎨
⎩N
− 2 , if N > 2
Then,
P E [ X 5
] ( N 3 ⎛ 1 ⎞ ⎡ 3 ⎤ ⎡ 3 ⎤
= = ∑ − 2) = () 1 2 3 0.0314
N = 3
⎜ ⎟+ ⎢ ⎥+ ⎢ ⎥ =
137N
⎝137 ⎠ ⎣137( 4) ⎦ ⎣137( 5)
⎦
6. E
The line 5y− 4x= 3 has slope 4 5
4 dy dy dx 2t
= = =
5 dx dt dt 2t
+ 1
8t+ 4=
10t
2t
= 4
t = 2
. It follows that we need to find t such that
Course 1 Solutions 3 November 2001

7. A
Cov C , C = Cov X + Y, X + 1.2Y
( 1 2) ( )
= Cov ( X, X) + Cov ( Y, X) + Cov ( X,1.2Y) + Cov( Y,1.2Y)
= Var X + Cov ( X, Y) + 1.2Cov ( X, Y)
+ 1.2VarY
= Var X + 2.2Cov ( X, Y)
+ 1.2VarY
( ) ( ( ))
2
( ) ( E Y )
2 2
2
Var X = E X − E X = 27.4 − 5 = 2.4
VarY
2 2
= E Y − ( ) = 51.4 − 7 = 2.4
( X + Y) = X + Y + ( X Y)
Var Var Var 2Cov ,
1
Cov ( XY , ) = Var ( X+ Y)
−Var X−VarY
2
1
= ( 8 − 2.4 − 2.4 ) = 1.6
2
Cov , = 2.4 + 2.2 1.6 + 1.2 2.4 = 8.8
( C C ) ( ) ( )
1 2
( )
Course 1 Solutions 4 November 2001

7. Alternate solution:
We are given the following information:
C1
= X + Y
C2
= X + 1.2Y
E X = 5
[ ]
2
⎡ ⎤ =
E⎣X
⎦
EY = 7
[ ]
2
⎡ ⎤ =
27.4
E⎣Y
⎦ 51.4
Var[ X + Y]
= 8
Now we want to calculate
Cov C , C = Cov X + Y, X + 1.2Y
( 1 2) ( )
= E⎡⎣( X + Y)( X + 1.2Y) ⎤⎦− E[ X + Y] iE[ X + 1.2Y]
2 2
= ⎡
⎣ + + ⎤
⎦− [ ] + [ ] +
2 2
= E⎡ ⎣X ⎤
⎦+ 2.2E[ XY] + 1.2E⎡ ⎣Y
⎤
⎦− ( 5 + 7) 5 + ( 1.2)
7
= 27. 4 + 2.2E[ XY] + 1.2( 51.4) −( 12)( 13.4)
= 2.2E[ XY]
−71.72
Therefore, we need to calculate E [ XY ] first. To this end, observe
2
2
8= Var[ X + Y] = E⎡( X + Y) ⎤− ( E[ X + Y]
)
[ ]
E XY
( )( [ ] [ ])
( )
E X 2.2XY 1.2Y E X E Y E X 1.2E Y
⎣
⎦
2 2
2
= E ⎡
⎣X + 2XY + Y ⎤
⎦− ( E[ X] + E[ Y]
)
2 2
2
= E⎡ ⎣X ⎤
⎦+ 2E[ XY] + E⎡ ⎣Y
⎤
⎦− ( 5 + 7)
= 27.4 + 2E[ XY]
+ 51.4 −144
= 2E[ XY]
−65.2
( 8 65.2)
2 36.6
= + =
Finally,
Cov CC = 2.2 36.6 − 71.72 = 8.8
( 1, 2 ) ( )
Course 1 Solutions 5 November 2001

8. D
The function F() t , 1≤t
≤ 7 , achieves its minimum value at one of the endpoints of the
interval 1≤t
≤ 7 or at t such that
−t −t −t
0= F′
() t = e − te = e ( 1−t)
Since F′ () t < 0 for t > 1 , we see that F () t is a decreasing function on the interval
1< t ≤ 7 . We conclude that
F 7 = 7e − = 0.0064
( )
7
is the minimum value of F.
9. D
Let
C = event that patient visits a chiropractor
T = event that patient visits a physical therapist
We are given that
Pr[ C] = Pr[ T]
+ 0.14
Pr C∩T
= 0.22
( )
c c
( C ∩T
)
Pr = 0.12
Therefore,
c c
0.88 = 1− Pr ⎡
⎣C ∩T ⎤
⎦ = Pr C∪T = Pr C + Pr T −Pr
C∩T
= Pr[ T] + 0.14 + Pr[ T]
−0.22
= 2 Pr T −0.08
or
[ ]
Pr[ T ] = ( 0.88 + 0.08)
2 = 0.48
[ ] [ ] [ ] [ ]
Course 1 Solutions 6 November 2001

10. E
Note that the terms of ∑ ∞ ⎛ 1 ⎞
a
n=
1⎜
n
+ ⎟
⎝ n ⎠
exceed the terms of the harmonic series for
1 1
a = n
1 , a = n
, or an
2
n
= n
and will thus fail to converge.
Moreover, the terms of the series
n
∞
⎡( −1)
1⎤
∞ ⎛ 1 1 ⎞ ∞ 1
∑ ⎢ + ⎥ =
n= 1 ∑n= 1⎜ + ⎟=
∑n=
1
⎢⎣
n n⎥
⎦ ⎝2n 2n⎠
n
are seen to be identical to the terms of the divergent harmonic series. By contrast, the
series
∞ ⎡ 1⎤ ∞ ⎡1−
n 1⎤
∞ 1
∑ a
n= 1⎢ n
+
n= 1 2 n=
1 2
n⎥ = ∑ + =
⎣ ⎦
⎢
⎣ n n⎥
∑
⎦ n
is seen to converge by the integral test, since
∞
2 1
x −
dx ∞
∫ =−
1 = 1

12. E
Observe that
x
f′ x = f′′
t dt
For [ ]
( ) ( )
∫
0
= (area bounded by f′′
above the x-axis from 0 to x)
− (area bounded by f ′′ below the x-axis from 0 to x)
x ∈ 0,5 , f ′′ is drawn such that the area bounded by f ′′ above the x-axis from 0 to
x is strictly greater than the area bounded by f ′′ below the x-axis from 0 to x. Therefore,
f′ ( x) > 0 for 0 < x≤ 5 . We conclude that f is an increasing function on the interval
[ 0,5 ], and its maximum value occurs at x = 5 .
13. E
10 8
F y Y y X y ⎡ X ⎤ e −
⎣ ⎦ ⎢ 10 ⎥
⎣
⎦
( )
( ) [ ] ( ) 10 8
0.8 10
= Pr ≤ = Pr ⎡10 ≤ ⎤ = Pr ≤ Y = 1−
Therefore, ( ) ′( )
1 4
10
1 Y
f y = F y = ⎛ ⎞
⎜ ⎟ e −
8⎝10⎠
( Y ) 54
Y
14. A
L’Hôspital’s Rule may be applied since f and g are given differentiable and from the
diagram lim f( x) = lim g( x) = 0 .
x→0 x→0
f ( x)
f′
( x)
Therefore lim = lim
x→0 g ( x)
x→0
g′
( x)
Also from the diagram f′ ( x) f′
( )
This leads to
lim = 0 > 0
x→0
and g′ ( x) x→0
g′
( )
f ( x)
f′ ( x)
f′
( 0)
→0 g( x)
x→0
g′ ( x)
g′
( 0)
lim = 0 < 0
lim = lim = < 0
x
Course 1 Solutions 8 November 2001

15. E
Let X1,...,
X
100
denote the number of pensions that will be provided to each new recruit.
Now under the assumptions given,
⎧0 with probability 1− 0.4 = 0.6
⎪
X i = ⎨1 with probability ( 0.4)( 0.25)
= 0.1
⎪
⎩2 with probability ( 0.4)( 0.75)
= 0.3
for i = 1,...,100 . Therefore,
E X = 0 0.6 + 1 0.1 + 2 0.3 = 0.7 ,
[ i ] ( )( ) ( )( ) ( )( )
2 2 2 2
⎣ i ⎦ ( ) ( ) ( ) ( ) ( ) ( )
2
[ Xi] E⎡X ⎤
i { E[ Xi]
} ( )
E⎡X
⎤ = 0 0.6 + 1 0.1 + 2 0.3 = 1.3 , and
2 2
Var =
⎣ ⎦
− = 1.3 − 0.7 = 0.81
Since X1,...,
X
100
are assumed by the consulting actuary to be independent, the Central
Limit Theorem then implies that S = X1+ ... + X100
is approximately normally distributed
with mean
E S = E X + ... + E X = 100 0.7 = 70
[ ] [ ] [ ] ( )
1 100
and variance
Var S = Var X + ... + Var X = 100 0.81 = 81
[ ] [ ] [ ] ( )
1 100
Consequently,
⎡S
−70 90.5 −70⎤
Pr[ S ≤ 90.5]
= Pr
⎢
≤
⎣ 9 9 ⎥
⎦
= Pr[ Z ≤2.28]
= 0.99
16. C
Observe
Pr 4 < S < 8 = Pr ⎡ ⎣4 < S < 8 N = 1⎤ ⎦Pr N = 1 + Pr ⎡ ⎣4 < S < 8 N > 1⎤ ⎦Pr N > 1
1 − 8
( ) ( )
4 − 1 −
5 5 1 2 −1
= e − e + e −e
*
3 6
= 0.122
*Uses that if X has an exponential distribution with mean µ
[ ] [ ] [ ]
b
1 1
− −
−t
µ −t
µ µ
Pr ( a≤ X ≤ b) = Pr ( X ≥a) −Pr
( X ≥ b)
= ∫ e dt− e dt = e −e µ
µ ∫µ
∞
a
∞
b
a
Course 1 Solutions 9 November 2001

17. B
Note that
20
[ > ] = ( − )
∫
Pr X x 0.005 20 t dt
x
⎛ 1 2⎞
20
= 0.005⎜20t−
t ⎟ x
⎝ 2 ⎠
⎛
1 2 ⎞
= 0.005⎜400 −200 − 20x
+ x ⎟
⎝
2 ⎠
⎛ 1 2 ⎞
= 0.005⎜200 − 20x+
x ⎟
⎝
2 ⎠
where 0< x < 20 . Therefore,
[ X > ]
[ X ]
( ) 1
2( )
2
( ) 1 ( )
Pr 16 200 − 20 16 + 16 8 1
Pr ⎡ ⎣X
> 16 X > 8⎤ ⎦ = = = =
Pr > 8 200 − 20 8 + 8 72 9
2
2
18. D
Note that
⎧2 x for x< 0 and x>
1
⎪
f′ ( x)
= ⎨ 1
⎪
for 0 < x < 1
⎩2
x
Furthermore,
lim
lim
0 = x
− f′ ( x) ≠ ( )
0 x
+ f′
x =+∞
→
→0
and
lim
lim 1
2 = f ( x ) f ( x
+
′ ≠ −
′ ) =
x→1 x→1
2
so f is not differentiable at x = 0 or x = 1, although f is differentiable everywhere else. By
contrast, f is continuous everywhere since
lim
f ( x lim
) f ( x
+ = − ) = 0
x→0 x→0 and
lim
f x lim
= f x =
( ) ( )
+ − 1
x→1 x→1 Course 1 Solutions 10 November 2001

19. B
The amount of money the insurance company will have to pay is defined by the random
variable
⎧1000 x if x<
2
Y = ⎨
⎩2000 if x ≥ 2
where x is a Poisson random variable with mean 0.6 . The probability function for X is
−0.6
k
e ( 0.6)
p( x)
= k = 0,1, 2,3 $ and
k!
k
−0.6 −0.6
∞ 0.6
EY [ ] = 0 + 1000( 0.6)
e + 2000e
∑ k = 2
k !
k
−0.6 ⎛ −0.6 ∞ 0.6 −0.6 −0.6
⎞
= 1000( 0.6) e + 2000⎜e ∑ −e −( 0.6)
e
k = 0
⎟
⎝ k!
⎠
k
∞
−0.6 ( 0.6)
−0.6 −0.6 −0.6 −0.6
= 2000e ∑ −2000e − 1000( 0.6)
e = 2000 −2000e −600e
k = 0 k!
= 573
k
2 2 −0.6 2 −0.6
∞ 0.6
E⎡ ⎣Y ⎤
⎦ = ( 1000) ( 0.6) e + ( 2000)
e ∑ k = 2
k!
k
2 −0.6 ∞ 0.6
2 −0.6 2 2
−0.6
= ( 2000) e ∑ −( 2000) e −⎡( 2000) −( 1000) ⎤( 0.6)
e
k = 0
k!
⎣
⎦
2 2 −0.6
2 2
= ( 2000) −( 2000)
e −⎡
−0.6
( 2000) − ( 1000) ⎤( 0.6)
e
⎣
⎦
= 816,893
Var
2
[ Y] = E⎡
⎣Y ⎤
⎦−{ E[ Y]
}
2
= 816,893−( 573)
= 488,564
[ Y ]
Var = 699
2
20. B
The graph of E′ tells us that employment increases from April through August (because
E t
E′ t ≤ then).
′() > 0 then) and does not increase from August through April (since () 0
It follows that employment is a minimum in April.
Course 1 Solutions 11 November 2001

21. D
Let N1 and N
2
denote the number of claims during weeks one and two, respectively.
Then since N1 and N
2
are independent,
7
[ N1+ N2 = ] = ∑ [ N1 = n] [ N2
= −n]
n=
0
Pr 7 Pr Pr 7
7 ⎛ 1 ⎞⎛ 1 ⎞
= ∑n=
0⎜ n+ 1⎟⎜ 8−n
⎟
⎝2 ⎠⎝2
⎠
7 1
= ∑n=
0 9
2
8 1 1
= = =
9 6
2 2 64
22. D
The amount of the drug present peaks at the instants an injection is administered. If A(n)
is the amount of the drug remaining in the patient’s body at the time of the nth injection,
then
A 1 = 250
()
−16
( 2) = 250( 1+
e )
A
%
n−1 −k
6
An ( ) = 250∑
e
k = 0
and the least upper bound is
∞ −k
6 250
lim An ( ) = 250∑
e = = 1628 (noting the
n→∞
k = 0
−16
1−
e
−16
series is geometric with ratio e )
Course 1 Solutions 12 November 2001

23. C
Observe that
sin ( θ + π) + 3 cos ( θ + π)
=− ⎡
⎣
sinθ + 3 cosθ⎤
⎦
r, θ and − r,
θ + π define the same point, and we may restrict attention to
Therefore, ( ) ( )
π
θ such that 0 ≤ θ < π . Now for 0 ≤θ
≤ , 0
2 r > and the graph of r is in the first
π
π
quadrant (i.e., to the right of θ = ). On the other hand, for < θ < π, r > 0 when
2
2
sinθ
+ 3 cosθ
> 0
sinθ
> 3 cosθ
sinθ
> 3
cosθ
tanθ
> 3
⎡π
⎤
This is true in the interval
⎢
, π
⎣ 2 ⎥
⎦ when π 2π
< θ < . Consequently, the area bounded
2 3
π
2π
3
2
by the graph of r to the left of θ = is given by rd
2
∫ θ .
π 2
[It is not needed to do the problem, but the graph of the curve is a circle of radius 1
3 1
centered at x= , y = ].
2 2
Course 1 Solutions 13 November 2001

24. D
1 1
4 2
100 x y must be maximized subject to x+ y = 150,000
Since y = 150,000 −x
, this reduces to maximizing
( ) ( )
1 1
4 2
S x = 100x 150,000 −x , 0 ≤ x≤150,000
−3 1
4 4
S ( x) x ( x) x ( x)
( 150,000 −x)
− 2x=
0
1 2 −1
2
′ = 25 150,000 − −50 150,000 − = 0
3x
= 150,000
x = 50, 000
(This value of x is a maximum since S ( x) 0
50,000 < x < 150,000 ).
′ > for 0 x 50,000
1 1
4 2
Maximum sales are then ( ) ( )
Alternate solution using Lagrange Multipliers
Solve x+ y− 150,000 = 0
∂ 1 1 ∂
4 2
100x y = λ x+ y−150,000
∂x
∂x
∂ 1 1 ∂
4 2
100x y = λ x+ y−150,000
∂y
∂y
From the last two equations
25x
50x
Eliminating λ
−3 1
4 2
1 4
y
y
−1 2
′ < for
< < and S ( x) 0
100 50,000 150,000 − 50,000 = 472,871
= λ
= λ
−3 1 1 −1
4 2 4 2
25x y = 50x y
25y
= 50x
y = 2x
Using the first equation
x+ 2x− 150,000 = 0
x = 50,000
y = 100,000
( )
( )
1 1
4 2
( ) ( )
The extreme value (which must be a maximum) is 100 50,000 100,000 = 472,871
Course 1 Solutions 14 November 2001

25. E
3 f
Pr ⎣⎡ 1< Y < 3 X = 2⎤ ⎦ = ∫ dy
1
f
1
( − )
( 2, y)
( 2)
2 ( ) 1
f ( 2, y)
= y = y
4 2 1 2
∞
− 4−1 2−1 −3
∞
−3 −2
1 1 1
fx
( 2)
= ∫ y dy =− y =
2 4 4
x
1
1
3
−3
y dy
1 2
3
2
− 1 8
1 1
9 9
Finally, Pr ⎡ ⎣1 < Y < 3 X = 2⎤ ⎦ = =− y = 1− =
∫
4
26. C
The increase in sales from year two to year four that is attributable to the advertising
campaign is given by
4⎡⎛ 4
2 1⎞ ⎛ 5⎞⎤
2
∫
( 2)
2 ⎢⎜t + ⎟− ⎜t + ⎟
2 2
⎥dt = ∫ t −t − dt
2
⎣⎝ ⎠ ⎝ ⎠⎦
3 2
⎛t
t ⎞ 4
= ⎜ − −2t
⎟ 2
⎝ 3 2 ⎠
64 8
= −8 −8 − + 2 + 4
3 3
56 26
= − 10 =
3 3
Course 1 Solutions 15 November 2001

27. D
Let X denote the number of employees that achieve the high performance level. Then X
follows a binomial distribution with parameters n= 20 and p = 0.02 . Now we want to
determine x such that
Pr X > x ≤0.01
[ ]
or, equivalently,
20
0.99 Pr[ ] x ( )( 0.02) k ( 0.98) 20 −
≤ X ≤ x =∑ k
k = 0 k
The following table summarizes the selection process for x:
x Pr X = x Pr X ≤ x
[ ] [ ]
20
( ) =
19
( )( ) =
2 18
( ) ( ) = 53 0.993
0 0.98 0.668 0.668
1 20 0.02 0.98 0.272 0.940
2 190 0.02 0.98 0.0
Consequently, there is less than a 1% chance that more than two employees will achieve
the high performance level. We conclude that we should choose the payment amount C
such that
2C = 120,000
or
C = 60,000
Course 1 Solutions 16 November 2001

28. B
Let X and Y denote the two bids. Then the graph below illustrates the region over which
X and Y differ by less than 20:
Based on the graph and the uniform distribution:
1
− ⋅
Shaded Region Area
Pr 20
2
⎣⎡ X − Y < ⎤ ⎦ = =
2 2
200
( 2200 − 2000)
( )
2
200 2 180
2
180
2
= 1− = 1− 2
( 0.9)
= 0.19
200
More formally (still using symmetry)
Pr ⎡⎣ X − Y < 20⎤ ⎦ = 1−Pr ⎡⎣ X −Y ≥20⎤ ⎦ = 1−2Pr[ X −Y
≥20]
2200 x−20 1 2200 1 x−20
= 1− 2∫ dydx 1 2 y
2 2 2000
dx
2020 ∫ = −
2000
200
∫2020
200
2 1
= 1− x−20 − 2000 dx= 1− x−2020
2 2020
2
200
∫
200
⎛180
⎞
= 1− ⎜ ⎟ = 0.19
⎝200
⎠
( ) ( )
2200 2 2200
2020
2
2
Course 1 Solutions 17 November 2001

29. B
Let X and Y denote repair cost and insurance payment, respectively, in the event the auto
is damaged. Then
⎧0 if x ≤ 250
Y = ⎨
⎩x
− 250 if x > 250
and
2
1500 1 1 2
[ ] ( ) ( )
1500 1250
EY = ∫ x− 250 dx= x− 250
250
= = 521
250
1500 3000 3000
3
1500
2 1 2 1 3 1500 1250
E⎡ ⎣Y ⎤
⎦ = ∫ ( x− 250) dx= ( x− 250)
250
= = 434,028
250
1500 4500 4500
2
2 2
Var[ Y] = E⎡
⎣Y ⎤
⎦− { E[ Y]
} = 434,028 −( 521)
Var Y = 403
[ ]
Course 1 Solutions 18 November 2001

30. C
Let ( , )
below:
and
f t1 t
2
denote the joint density function of
1 2
T
T . The domain of f is pictured
Now the area of this domain is given by
A = 6 −
2
6− 4 = 36− 2=
34
Consequently,
⎧ 1
⎪ , 0< t1 < 6 , 0< t2 < 6 , t1+ t2
< 10
f ( t1, t2)
= ⎨34
⎪⎩ 0 elsewhere
and
2 1 ( ) 2
[ + ] = [ ] + [ ] = [ ]
ET1 T2 ET1 ET2 2 ET1
(due to symmetry)
⎧ 4 6 1 6 10−t1
1 ⎫
= 2⎨∫ t1 2 1 1 2 1
0 ∫ dt dt +
0
34
∫ t
4 ∫ dt dt ⎬
0
⎩
34 ⎭
⎧ 4 ⎡t
6
2 6 ⎤ ⎡t2
10−t
⎤ ⎫
1
= 2⎨∫
t1 0 1 1 0 1
0 ⎢34 ⎥
dt + ∫ t
4 ⎢34
⎥
dt ⎬
⎩ ⎣ ⎦ ⎣ ⎦ ⎭
⎧ 43t 6
1
1
2 ⎫
= 2⎨∫
dt
1+ ( 10 t
1 t
1 ) dt
1
0
17
∫ − ⎬
4
⎩
34
⎭
2
⎧3t1
4 1 ⎛ 2 1 3⎞
6
⎫
= 2⎨
0+ ⎜5t1 − t1
⎟ 4 ⎬
⎩34 34 ⎝ 3 ⎠ ⎭
⎧24 1 ⎡
64⎤⎫
= 2 ⎨ + 180 72 80
17 34 ⎢
− − + ⎬
3 ⎥
⎩ ⎣
⎦⎭
= 5.7
Course 1 Solutions 19 November 2001

31. E
The general solution of the differential equation may be determined as follows:
1
∫ dy = ( k1 k2)
dt
y
∫ −
ln y = k − k t+
C
(C is a constant)
( ) ( 1 2)
c ( k1 k2)
t
() = e e −
y t
When t = 0 ,
c
y( 0)
= e , so
() ( ) ( k ) 1 k2
t
y t = y 0 e −
1
Now we are given that if k1
= 0, y( 8) = y( 0)
.
2
Therefore,
1
−8k2
y( 0) = y( 8) = y( 0)
e
2
1 −8k2
= e
2
8k2
= ln( 2)
1
k2
= ln ( 2 )
8
24( k1−k2)
[Note the problem also gives y( 24) = 2y( 0) = y( 0)
e , but that information is not
needed to determine k
2
].
32. B
Observe
[ N ]
[ N ]
Pr 1 ≤ ≤4 ⎡1 1 1 1 ⎤ ⎡1 1 1 1 1 ⎤
Pr ⎡⎣N
≥1 N ≤4⎤ ⎦ = = + + + + + + +
Pr ≤ 4 ⎢
⎣6 12 20 30⎥ ⎦
⎢
⎣2 6 12 20 30⎥
⎦
10 + 5 + 3 + 2 20 2
= = =
30 + 10 + 5 + 3 + 2 50 5
Course 1 Solutions 20 November 2001

33. E
Let X and Y denote the year the device fails and the benefit amount, respectively. Then
the density function of X is given by
x−1
f ( x) = ( 0.6) ( 0.4 ) , x=
1,2,3...
and
⎧⎪ 1000( 5 − x)
if x=
1,2,3, 4
⎪⎩
It follows that
y = ⎨
0 if x > 4
2 3
[ ] 4000( 0.4) 3000( 0.6)( 0.4) 2000( 0.6) ( 0.4) 1000( 0.6) ( 0.4)
EY = + + +
= 2694
34. D
The diagram below illustrates the domain of the joint density ( )
f xy , of Xand
Y.
We are told that the marginal density function of X is f ( x) = 1,0< x<
1
while f ( yx) = 1, x< y< x+
1
yx
It follows that f ( x,
y) f ( x) f ( y x)
Therefore,
⎧1 if 0< x < 1, x< y< x+
1
= x yx
=⎨
⎩ 0 otherwise
[ ] [ ]
1 1
2 2
0 ∫ x
Pr Y > 0.5 = 1−Pr Y ≤ 0.5 = 1−
dydx
∫
1 1
1
2 2
2⎛1
⎞
= 1− ∫ y
x
dx= 1− 0 ∫0
⎜ −x⎟dx
⎝2
⎠
⎡1 1 1
2 ⎤ 1 1 7
2
= 1− ⎢
x− x
0
= 1− + =
⎣2 2 ⎥
⎦ 4 8 8
[Note since the density is constant over the shaded parallelogram in the figure the
solution is also obtained as the ratio of the area of the portion of the parallelogram above
y = 0.5 to the entire shaded area.]
x
Course 1 Solutions 21 November 2001

35. E
If X is the random variable representing claim amounts, the probability that X exceeds the
deductible is
∞
[ ] ∫
1
−x −x ∞ −x
Pr X > 1 = xe dx=− xe + e dx (integration by parts)
1 1
−1 −x
∞ −1 −1 −1
1
2
= e − e = e + e = e
= 0.736
It follows that the company expects ( 100)( 0.736)
= 74 claims.
∫
∞
36. C
Let
Then
x = price in excess of 60 that the company charges,
p(x) = price per policy that the company charges,
n(x) = number of policies the company sells per month, and
R(x) = revenue per month the company collects
( ) = 60 + x
( ) = 80 −x
( ) = ( ) ( ) = ( 80 − )( 60 + )
′( ) =− ( 60 + ) + ( 80 − ) = 20 −2
′′( x)
=− 2<
0
p x
n x
R x n x p x x x
R x x x x
R
It follows that R(x) is a maximum when R ( x) 0
20 − 2x
= 0
x = 10
R 10 = 80 − 10 60 + 10 = 4900
and ( ) ( )( )
when R(x) is a maximum.
′ = . We conclude that
Course 1 Solutions 22 November 2001

37. A
The joint density of T1 and T
2
is given by
−t
−t
f t , t = e e , t > 0 , t > 0
( )
1 2
1 2 1 2
Therefore,
Pr X ≤ x = Pr 2T + T ≤ x
[ ] [ ]
1 2
⎡
= ∫∫
= −
⎢⎣
1
1
x ( x−t2
) x
( x−t2
)
2 −t1 −t2 −t2 −t1
2
e e dt1dt2 e e dt2
0 0 ∫ ⎢ ⎥
0
0
⎡ ⎤ ⎛ ⎞
= ∫ − = −
⎣ ⎦ ⎝ ⎠
1 1 1 1
x
− x+ t2 x
− x − t
−t
2
2 2 2 −t2
2 2
e ⎢1
e ⎥dt2 ⎜e e e ⎟dt2
0 ∫0
⎡
⎤
= ⎢− e + e e ⎥ =− e + e e + − e
⎣
⎦
1 1 1 1 1
− x − t2
− x − x − x
−t2
2 2 x − x 2 2 2
2
0
2 1 2
1 1
− x
− x
− x −x
2 2 −x
= 1− e + 2e − 2e = 1− 2 e + e , x>
0
It follows that the density of X is given by
1
d ⎡ − x
2 − x
⎤
g ( x)
= ⎢1− 2e + e ⎥
dx ⎣
⎦
1
− x
2
− x
= e − e , x>
0
⎤
⎥⎦
38. E
X , X , and X denote annual loss due to storm, fire, and theft, respectively. In
Let
1 2 3
addition, let Y Max( X , X , X )
Then
= .
1 2 3
[ Y > ] = − [ Y ≤ ] = − [ X ≤ ] [ X ≤ ] [ X ≤ ]
Pr 3 1 Pr 3 1 Pr 3 Pr 3 Pr 3
(
−
e )( −3 e )( −3
e )
(
−
e 3 )(
−
e )( −5
e )
= 1− 1− 1− 1−
= 1− 1− 1− 1−
1 2 3
= 0.414
* Uses that if X has an exponential distribution with mean µ
1
Pr ( X ≤ x) = 1−Pr ( X ≥ x) = 1− ∫ e dt = 1−( − e ) x
= 1−e
µ
∞
x
*
−t µ −t µ ∞ −x
µ
Course 1 Solutions 23 November 2001

39. D
We are given that
P x =− x + 50x−25
( )
2
before the translation occurs. The revised profit function P ( x)
follows:
P*
x = P x−2 −3
( ) ( )
2
( x ) ( x )
=− − 2 + 50 −2 −25−3
2
=− x + x− + x− −
2
=− x + x−
4 4 50 100 28
54 132
* may be determined as
40. B
Observe that
E X + Y = E X + E Y = 50 + 20 = 70
[ ] [ ] [ ]
[ ] [ ] [ ] [ ]
Var X + Y = Var X + Var Y + 2 Cov X , Y = 50 + 30 + 20 = 100
for a randomly selected person. It then follows from the Central Limit Theorem that T is
approximately normal with mean
ET = 100 70 = 7000
[ ] ( )
and variance
Var T = 100 100 = 100
[ ] ( )
2
Therefore,
⎡T
−7000 7100 −7000⎤
Pr[ T < 7100]
= Pr
⎢
<
⎣ 100 100 ⎥
⎦
= Pr[ Z < 1]
= 0.8413
where Z is a standard normal random variable.
Course 1 Solutions 24 November 2001