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ASM Study Manual for Exam P, Third Edition
By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA ([email protected])
Errata
Effective July 5, 2013, only the latest edition of this manual will have its errata
updated. You can find the errata for all latest editions of my books at:
http://smartURL.it/errata
Posted April 5, 2013
The second to last formula in the solution of Problem 12 in Practice Examination 5
should be:
2 2−x
x=2
2
3
3
1 ⎞
⎛3
⎛ 3 1⎞ 1
x dy dx = ∫ x ( 2 − x ) dx = ⎜ x 2 − x 3 ⎟
= ( 3 − 2) − ⎜ − ⎟ = .
∫
⎝
⎠
⎝ 4 4⎠ 2
4
4
4
4
x=1
1 0
1
The calculation shown was correct but there was a typo in limits of integral
calculation.
Pr ( X > 1) = ∫
Posted April 2, 2013
The last formula in the solution of Problem 10 in Practice Examination 4 should be
2
+∞
2
+∞
⎛ 2.5 ⋅ 0.6 2.5 ⎞
⎛ 2.5 ⋅ 0.6 2.5 ⎞
2.5 ⋅ 0.6 2.5
5 ⋅ 0.6 2.5
E (Y ) = ∫ x ⋅ ⎜
dx
+
2
⋅
dx
=
dx
+
dx =
3.5
3.5
2.5
3.5
∫
∫
∫
⎟⎠
⎜⎝
⎟⎠
x
x
x
x
⎝
0.6
2
0.6
2
x=2
x→∞
2.5 ⋅ 0.6 2.5
2 ⋅ 0.6 2.5
2.5 ⋅ 0.6 2.5 2.5 ⋅ 0.6
0.6 2.5
=−
−
=
−
+
−
0
+
≈ 0.934273.
1.5x1.5 x=0.6
x 2.5 x=2
1.5 ⋅ 21.5
1.5
21.5
The calculation shown was correct, but a factor in the numerator of the last fraction
was mistyped as 2.5, when it should have been 2.
Posted March 6, 2013
In the solution of Problem 6 in Practice Examination 2, the expression
rectangle [ 0,1] × [ 0, 2 ],
should be
rectangle [ 0,1] × [ 0, 2 ],
This expression appears twice in the solution.
Posted April 23, 2012
The fourth formula in the solution of Problem 4 in Practice Examination 7 should
be:
2 ± 4 − 4 ⋅ 0.0005 ⋅1850 2 ± 0.5477226 ⎧⎪ 2547.72,
M=
≈
≈⎨
2 ⋅ 0.0005
0.001
⎩⎪ 1452.28.
intead of
M=
2 ± 4 − 4 ⋅ 0.0005 ⋅1850 2 ± 0.5477226 ⎧⎪ 2547.72,
≈
≈⎨
0.0001
0.001
⎪⎩ 1452.28.
Posted April 20, 2012
The first formula in the last line of the solution of Problem 9 in Practice
1
1
Examination 7 should be 2 ⋅ Pr ( X ≥ 800 ) ≤ instead of 2 ⋅ Pr ( X ≥ 800 ) ≥ .
4
4
Posted March 10, 2011
The second sentence of the solution of Problem 10 in Practice Examination 1 should
be:
As the policy has a deductible of 1 (thousand), the claim payment is
⎧ 0,
when there is no damage, with probability 0.94,
⎪
Y = ⎨ max ( 0, X − 1) , when 0 < X < 15, with probability 0.04,
⎪ 14,
in the case of total loss, with probability 0.02.
⎪⎩
Posted January 15, 2011
In Problem 16 in Practice Examination 6, the calculation of the second moment of X
should be:
1
3
3
E X 2 = ⋅ 0 2 + ⋅ (1 + 1) = .

4
4 Second moment of T 2
( )
instead of
E(X2 ) = E(X) =
1 2 3
3
⋅ 0 + ⋅ (1+ 1) = .
4
4 Second
2
moment of T
Posted July 24, 2010
In the solution of Problem 21 in Practice Examination 6, the statement under the
first expression on the right-hand side of the third to last formula should be:
number of ways to
pick ordered samples
of size 2 from
population of size n
instead of
number of ways to
pick ordered samples
of size n − 2 from
population of size n
Posted July 3, 2010
In the solution of Problem 1, Practice Examination 5, at the end of the first part of
the fourth sentence of the solution, 5/6 is a typo, it should be 5/36, as used in the
formula for Pr (Y = 6).
Posted January 5, 2010
In the description of the gamma distribution in Section 2, the condition for the
range of its MGF should be t < β , not 0 < t < β .
Posted June 18, 2009
The last formula in the solution of Problem 19 of Practice Examination 5 should be
1
1
1
1
instead of Pr ( X ≥ 10 ) < 2 = .
Pr ( X ≥ 10 ) ≤ 2 =
4
16
4
16
Posted June 17, 2009
In the statement of Problem 9 in Practice Examination 7, Pr ( X > 800 ) should be
Pr ( X ≥ 800 ) , and in the solution, all inequalities should be changed accordingly.
Posted April 6, 2009
In the text of Problem 7 in Practice Examination 3, the word “whwther” should be
“whether”.
Posted March 5, 2009
The first sentence of Problem 16 in Practice Examination 6 should end with t < 1,
instead of t > 1.
Posted March 2, 2009
The properties of the cumulant moment-generating function should be:
The cumulant generating function has the following properties: ψ X ( 0 ) = 0,
( )
d
ln E etX
dt
d2
ψX
dt 2
3
=
(
E XetX
( )
d E ( Xe )
(t ) =
dt E ( e )
E e
t =0
)
= E ( X ),
tX
t =0
tX
=
tX
t=0
E ( X 2 etX ) E ( etX ) − E ( XetX ) E ( XetX )
(
t=0
(
)
E ( etX )
)
3
d
ψ X ( t ) = E ( X − E ( X )) ,
3
dt
t=0
but for k > 3,
k
dk
ψ X ( t ) = ψ (Xk ) ( 0 ) < E ( X − E ( X )) .
k
dt
t=0
Also, if X and Y are independent (we will discuss this concept later),
ψ aX+b ( t ) = ψ X ( at ) + bt,
(
)
= Var ( X ) ,
2
t=0
and
ψ X+Y ( t ) = ψ X ( t ) + ψ Y ( t ) .
Posted January 13, 2009
In Practice Examination 8, Problem 24, the calculation of the expected value had a
typo, an extra, unnecessary p in the second line, and it instead should be:
+∞
+∞
E ( X ) = ∑ k ⋅ Pr ( X = k ) = 1⋅ Pr ( X = 1) + ∑ k ⋅ Pr ( X = k ) =
k=1
k=2
+∞
+∞
+∞
k=1
k=1
k=1
= p + ∑ ( k + 1) ⋅ Pr ( X = k + 1) = p + ∑ k ⋅ Pr ( X = k + 1) + ∑ Pr ( X = k + 1) =
⎛
⎞
⎜ ⎛ +∞
⎟
⎞
= p + ∑ k ⋅ (1− p ) ⋅ Pr ( X = k ) + ⎜ ⎜ ∑ Pr ( X = k + 1)⎟ − Pr ( X = 0 + 1)⎟ =


 ⎝ k=0
⎠
k=1
⎜ 
⎟


=Pr( X=k+1)
⎝ this sum is equal to 1
⎠
+∞
+∞
= p + (1− p ) ⋅ ∑ k ⋅ Pr ( X = k ) + (1− Pr ( X = 1)) =
k=1
= p + (1− p ) ⋅ E ( X ) + (1− p ) = 1+ (1− p ) ⋅ E ( X ) .
Posted September 1, 2008
In the solution of Problem 16 in Practice Examination 2, the left-hand side of the
second formula should be E X 2 , not E ( X ) .
( )
Posted July 24, 2008
In the solution of Problem 6 of Practice Examination 7, the phrase
Therefore, recalling that for a discrete random variable, whose only possible values are
possible integers,
should be
Therefore, recalling that for a discrete random variable, whose only possible values are
positive integers,
Posted July 23, 2008
In the solution of Problem 5 in Practice Examination 3 the expression
Var ( X2 ) = 40,000 should be Var ( X2 ) = 250,000.
Posted February 7, 2008
The discussion of the lack of memory property of the geometric distribution should
have the formula
Pr ( X = n + k X ≥ n ) = Pr ( X = k )
corrected to:
Pr ( X = n + k X > n ) = Pr ( X = k ) .
Posted November 13, 2007
In the solution of Problem No. 26 of Practice Examination No. 5, the sentence:
Of the five numbers, 1 can never be the median.
should be
Of the five numbers, neither 1 nor 5 can ever be the median.
Posted October 5, 2007
In the solution of Problem 1 in Practice Examination 3, the random variable Y
should refer to class B.
Posted September 13, 2007
The beginning of the third sentence in the solution of Problem 21 in Practice
Examination 2 should be
The mean error of the 48 rounded ages,
instead of
The mean of the 48 rounded ages.
Posted May 15, 2007
In the solution of Problem 1, Practice Examination 7, both events were mistakenly
labeled E1 . In order to correct that, the expression
Define the following events:
E1 = { X1 = 0} ∩ { X2 = 1} ,
E1 = { X1 = 1} ∩ { X2 = 2} .
should be replaced by:
Define the following events:
E1 = { X1 = 0} ∩ { X2 = 1} ,
E2 = { X1 = 1} ∩ { X2 = 2} .
Posted May 9, 2007
In the solution of Problem No. 15, Practice Examination No. 3, the derivative is
missing a minus, and it should
2
dX
1
−
= − (8 − Y ) 3 .
dY
3
The rest of the solution is unaffected.
Posted May 9, 2007
Answer C in Problem No. 7 in Practice Examination No. 6 should be 0.2636, not
0.2626. It is the correct answer.
Posted May 9, 2007
In Problem No. 10, Practice Examination No. 7, in the first part of the Practice
Examination (not in the Solutions part), the expression E ( X ) = µY should be
E ( X ) = µX .
Posted May 8, 2007
Problem No. 24 in Practice Examination No. 5 should be (it had a typo in the list of
possible values of x)
May 1992 Course 110 Examination, Problem No. 35
Ten percent of all new businesses fail within the first year. The records of new
businesses are examined until a business that failed within the first year is found. Let X
be the total number of businesses examined prior to finding a business that failed within
the first year. What is the probability function for X?
A. 0.1⋅ 0.9 x , for x = 0,1,2, 3,...
B. 0.9 ⋅ 0.1x , for x = 0,1,2, 3,...
C. 0.1x ⋅ 0.9 x , for x = 1,2, 3,...
D. 0.9x ⋅ 0.1x , for x = 1,2, 3,...
E. 0.1⋅ ( x − 1) ⋅ 0.9 x , for x = 2, 3, 4,...
Solution.
Let a failure of a business be a success in a Bernoulli Trial, and a success of a business be
a failure in the same Bernoulli Trial. Then X has the geometric distribution with p = 0.1,
and therefore f X ( x ) = 0.1⋅ 0.9 x for x = 0, 1, 2, 3, … .
Answer A.
Posted May 5, 2007
Problem No. 10 in Practice Examination No. 3 should be (it had a typo in one of the
integrals and I decided to reproduce the whole problem to provide a better
explanation):
Sample Course 1 Examination, Problem No. 35
Suppose the remaining lifetimes of a husband and a wife are independent and uniformly
distributed on the interval (0, 40). An insurance company offers two products to married
couples:
• One which pays when the husband dies; and
• One which pays when both the husband and wife have died.
Calculate the covariance of the two payment times.
A. 0.0
B. 44.4
C. 66.7
D. 200.0
E. 466.7
Solution.
Let H be the random time to death of the husband, W be the time to death of the wife, and
X be the time to the second death of the two. Clearly,
X = max ( H ,W ) .
1
We have fH ( h ) = fW ( w ) =
for 0 ≤ h ≤ 40, and 0 ≤ w ≤ 40. Thus
40
E ( H ) = E (W ) = 20.
Furthermore,
FX ( x ) = Pr ( X ≤ x ) = Pr ( max ( H ,W ) ≤ x ) =
= Pr ({ H ≤ x} ∩ {W ≤ x}) = Pr ( H ≤ x ) ⋅ Pr (W ≤ x ) =
x x
x2
⋅
=
.
40 40 1600
This implies that
x2
sX ( x ) = 1 −
1600
for 0 ≤ x ≤ 40, and
40
⎛
x2 ⎞
40 3 120 40 80
E ( X ) = ∫ ⎜1 −
dx
=
40
−
=
−
=
.
1600 ⎟⎠
4800
3
3
3
0 ⎝
w to find covariance, we also need to find
In order
40
max
w )(=Hw,W )) .
E ( XH ) =here
E ( H( h,
max
We separate the double integral into two parts: one based on the region where the wife
lives longer and one based on the region where the husband lives longer, as illustrated in
the graph below
max ( h, w ) = h
here
40
h
E ( H max ( H ,W )) =
+∞ +∞
∫ ∫ h max ( h, w ) ⋅ f ( h ) ⋅ f ( w ) ⋅ dwdh =
H
W
−∞ −∞
h = 40 w = 40
⎛ w=h 2 1 1
⎞
⎛
⎞
1 1
= ∫ ⎜ ∫ h ⋅ ⋅ dw ⎟ dh + ∫ ⎜ ∫ hw ⋅ ⋅ dw ⎟ dh =
40 40 ⎠
40 40 ⎠
h=0 ⎝ w=0
h=0 ⎝ w=h
h = 40
⎛ h2w
= ∫⎜
1600
0 ⎝
40
40 ⎛
⎞
hw 2
dh
+
⎟
∫0 ⎜⎝ 3200
w=0 ⎠
w=h
w = 40
w=h
40
40
⎞
h3
1600h − h 3
dh
=
dh
+
dh =
⎟
∫
∫
1600
3200
⎠
0
0
40
1
1
⎛1
⎞
⎛1
⎞
= ∫⎜ h+
h 3 ⎟ dh = ⎜ h 2 +
h4 ⎟
⎝2
⎝4
3200 ⎠
12800 ⎠
0
h = 40
=
h=0
1
1
⋅ 40 2 +
⋅ 40 4 = 600.
4
12800
Finally,
Cov ( X, H ) = E ( XH ) − E ( X ) E ( H ) = 600 − 20 ⋅
80
2
= 66 .
3
3
Answer C.
Posted April 29, 2007
The second formula to the last in the solution of Problem No. 6 in Practice
Examination No. 3 should be:
8
t 7
⎛ ⎛ 2 + et ⎞ 8
⎞
d 2 M (t )
d t ⎛ 2 + et ⎞
2
t
t⎛2+e ⎞
t
E X =
=
3e
=
3e
+
8e
⋅
e
⎜
⎟
⎜⎝ 3 ⎟⎠
⎜⎝ 3 ⎟⎠
⎜⎝ 3 ⎟⎠
dt 2 t = 0 dt
⎝
⎠
t =0
( )
= 11.
t =0
instead of
( )
E X
2
d 2 M (t )
d t ⎛ 2 + et ⎞
=
= 3e ⎜
dt 2 t = 0 dt
⎝ 3 ⎟⎠
8
t =0
t
⎛ ⎛ 2 + et ⎞ 8
⎞
t
t⎛2+e ⎞
t
= ⎜ 3e ⎜
+
8e
⋅
e
⎟
⎜⎝ 3 ⎟⎠ ⎟
⎝ ⎝ 3 ⎠
⎠
= 11.
t =0
Posted April 27, 2007
In the solution of Problem No. 9 in Practice Examination No. 3, the expression
max ( X,1) should be min ( X,1) .
Posted April 20, 2007
Problem No. 16 in Practice Examination No. 1 should be:
May 2001 Course 1 Examination, Problem No. 26, also Study Note P-09-05, Problem
No. 109
A company offers earthquake insurance. Annual premiums are modeled by an
exponential random variable with mean 2. Annual claims are modeled by an exponential
random variable with mean 1. Premiums and claims are independent. Let X denote the
ratio of claims to premiums. What is the density function of X?
A.
1
2x + 1
B.
2
( 2x + 1)
C. e− x
2
D. 2e−2 x
E. xe− x
Solution.
Let U be the annual claims and let V be the annual premiums (also random), and let
fU ,V ( u, v ) be the joint density of them. Furthermore, let f X be the density of X and let FX
be its cumulative distribution function. We are given that U and V are independent, and
hence
v
−
1 −v 1
fU ,V ( u, v ) = e−u ⋅ e 2 = e−u e 2
2
2
for 0 < u < ∞, 0 < v < ∞. Also, noting the graph below,
v
u = vx or v = u/x
u
we have:
⎛U
⎞
FX ( x ) = Pr ( X ≤ x ) = Pr ⎜ ≤ x ⎟ = Pr (U ≤ Vx ) =
⎝V
⎠
∞ vx
+∞
u = vx
0
u=0
⎛ 1 −u − 2v ⎞
∫ ⎜⎝ − 2 e e ⎟⎠
v
−
1
= ∫ ∫ e−u e 2 dudv =
2
0 0
∞ vx
∫ ∫ f (u, v )dudv =
U ,V
0 0
∞
v
−
⎛ 1
1 −v ⎞
dv = ∫ ⎜ − e− vx e 2 + e 2 ⎟ dv =
2
2
⎠
0⎝
⎛ 1⎞
v
⎛ 1 − v ⎛⎝⎜ x + 12 ⎞⎠⎟ 1 − v ⎞
⎛ 1
− v⎜ x + ⎟
− ⎞
⎝ 2⎠
2
2
= ∫⎜− e
+ e ⎟ dv = ⎜
e
−e ⎟
2
2
⎠
⎝ 2x + 1
⎠
0⎝
∞
+∞
=−
0
1
+ 1.
2x + 1
Finally,
f X ( x ) = FX′ ( x ) =
2
( 2x + 1)2
.
Answer B. This problem can also be done with the use of bivariate transformation. We
will now give an alternative solution using that approach. Consider the following
U
transformation X = , Y = V. Then the inverse transformation is U = XY , V = Y . We
V
know that
v
−
1
fU ,V ( u, v ) = e−u e 2
2
for u > 0, v > 0. It follows that
∂ ( u, v ) 1 − xy − 2y
⎡ y x ⎤ 1 − xy − 2y
= e e ⋅ det ⎢
⎥ = ye e
∂ ( x, y ) 2
⎣0 1 ⎦ 2
for xy > 0 and y > 0, or just x > 0 and y > 0. Therefore,
f X ,Y ( x, y ) = fU ,V ( u ( x, y ) , v ( x, y )) ⋅
fX ( x ) =
+∞
1
∫ 2 ye
0
y
−
− xy
2
+∞
e dy =
1
∫ 2 ye
⎛ 1⎞
−⎜ x+ ⎟ y
⎝ 2⎠
1
w= y
2
z=−
dy =
0
1
e
1⎞
⎛
⎜⎝ x + ⎟⎠
2
⎛
⎛ 1⎞
−⎜ x+ ⎟ y
⎝ 2⎠
=
1⎞
−⎜ x+ ⎟ y
1
dw = dy
dz = e ⎝ 2 ⎠ dy
2


INTEGRATION BY PARTS
⎛
⎞
⎛ 1⎞
−
x
+
y
⎜ 1
⎜
⎟ ⎟
1
= ⎜− y⋅
e ⎝ 2⎠ ⎟
⎜ 2 ⎛⎜ x + 1 ⎞⎟
⎟
⎝
⎝
⎠
2⎠
y→+∞
+∞
−
∫
0
y= 0
⎛
⎞
⎛ 1⎞
⎜
1
1 ⎟ − ⎝⎜ x + 2 ⎠⎟ y
dy =
⎜−
⎟e
1⎞ ⎟
2⎜ ⎛
x+ ⎟
⎝ ⎜⎝
2⎠ ⎠
⎛
⎞
⎛
⎞
+∞
⎛ 1⎞
⎛ 1⎞ ⎟
⎜
−
x
+
y
−
x
+
y
⎜
⎟
⎜
⎟
1
1
1
⎝⎜ 2 ⎠⎟
=0+ ∫
e ⎝ 2 ⎠ dy = ⎜
⎟
⎜−
⎟ ⋅e
2x + 1
⎜ 2x + 1 ⎜ ⎛ x + 1 ⎞ ⎟
⎟
0
⎟
⎜⎝
⎟⎠
⎝ ⎜⎝
2⎠ ⎠
y→+∞
=
2
( 2x + 1)2
.
y= 0
Answer B, again. The second approach is probably more complicated, but it is a good
exercise in the use of multivariate transformations.
Posted January 20, 2007
Exercise 2.8 should read as follows:
November 2001 Course 1 Examination, Problem No. 11
A company takes out an insurance policy to cover accidents that occur at its
manufacturing plant. The probability that one or more accidents will occur during any
3
given month is . The number of accidents that occur in any given month is independent
5
of the number of accidents that occur in all other months. Calculate the probability that
there will be at least four months in which no accidents occur before the fourth month in
which at least one accident occurs.
A. 0.01
B. 0.12
C. 0.23
D. 0.29
E. 0.41
Solution.
Consider a Bernoulli Trial with success defined as a month with an accident, and a month
3
with no accident being a failure. Then the probability of success is . Now consider a
5
negative binomial random variable, which counts the number of failures (months with no
accident) until 4 successes (months with accidents), call it X. The problem asks us to find
Pr ( X ≥ 4 ) . But
Pr ( X ≥ 4 ) = 1 − Pr ( X = 0 ) − Pr ( X = 1) − Pr ( X = 2 ) − Pr ( X = 3) =
4
4
4
2
4
3
⎛ 3⎞ ⎛ 3 ⎞
⎛ 4 ⎞ ⎛ 3 ⎞ 2 ⎛ 5⎞ ⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 6⎞ ⎛ 3 ⎞ ⎛ 2 ⎞
= 1− ⎜ ⎟ ⋅⎜ ⎟ − ⎜ ⎟ ⋅⎜ ⎟ ⋅ − ⎜ ⎟ ⋅⎜ ⎟ ⋅⎜ ⎟ − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ≈
⎝ 0⎠ ⎝ 5 ⎠
⎝ 1 ⎠ ⎝ 5 ⎠ 5 ⎝ 2 ⎠ ⎝ 5 ⎠ ⎝ 5 ⎠ ⎝ 3⎠ ⎝ 5 ⎠ ⎝ 5 ⎠
≈ 0.289792.
Answer D.
Posted December 31, 2006
In Section 2, the general definition of a percentile should be
the 100-p-th percentile of the distribution of X is the number x p which satisfies both of
(
)
(
)
the following inequalities: Pr X ≤ x p ≥ p and Pr X ≥ x p ≥ 1 − p.
It was mistyped as
the 100-p-th percentile of the distribution of X is the number x p which satisfies both of
(
)
(
)
the following inequalities: Pr X ≤ x p ≥ p and Pr X ≥ x p ≤ 1 − p.
Posted December 28, 2006
On page 3 in the bottom paragraph, the words:
is also defined as the sent that consists of all elementary events that belong to any one of
them.
should be
is also defined as the set that consists of all elementary events that belong to any one of
them.
The word “set” was mistyped as “sent.”
Posted December 1, 2006
In Practice Examination 6, Problem 22, had inconsistencies in its assumptions, and
it should be replaced by the following:
A random variable X has the log-normal distribution with the density:
1
2
− ( ln x− µ )
1
fX ( x ) =
⋅e 2
x 2π
for x > 0, and 0 otherwise, where µ is a constant. You are given that Pr ( X ≤ 2 ) = 0.4.
Find E ( X ) .
A. 4.25
Solution.
B. 4.66
C. 4.75
D. 5.00
E. Cannot be determined
(
)
X is log-normal with parameters µ and σ 2 if W = ln X  N µ, σ 2 . The PDF of the lognormal distribution with parameters µ and σ is
2
fX ( x ) =
1
−
1 ( ln x− µ )
2 σ2
2
e
xσ 2π
for x > 0, and 0 otherwise. From the form of the density function in this problem we see
that σ = 1. Therefore
⎛ ln X − µ ln 2 − µ ⎞
Pr ( ln X ≤ ln 2 ) = Pr ⎜
≤
⎟ = 0.4.
⎝
1
1 ⎠
Let z0.6 be the 60-th percentile of the standard normal distribution. Let Z be a standard
ln X − µ
normal random variable. Then Pr ( Z ≤ −z0.6 ) = 0.40. But Z =
is standard normal,
1
thus ln 2 − µ = −z0.6 , and µ = ln 2 + z0.6 . From the table, Φ ( 0.25 ) = 0.5987 and
Φ ( 0.26 ) = 0.6026. By linear interpolation,
0.6 − 0.5987
1
z0.6 ≈ 0.25 +
⋅ ( 0.26 − 0.25 ) = 0.25 + ⋅ 0.01 ≈ 0.2533.
0.6026 − 0.5987
3
This gives
µ = ln 2 + z0.6 ≈ 0.9464805.
The mean of the log-normal distribution is E ( X ) = e
E(X) = e
Answer A.
1
0.9464805+
2
1
µ+ σ 2
2
, so that in this case
≈ 4.2481369.
Posted November 25, 2006
The third equation from the bottom in the solution of Problem No. 10 in Practice
Examination No. 2 should be
Pr ( A ∩ E99 ) = Pr ( A E99 ) ⋅ Pr ( E99 ) = 0.03⋅ 0.20 = 0.006.
Previously it had a typo:
Pr ( A ∩ E99 ) = Pr ( A E99 ) ⋅ Pr ( E99 ) = 0.03⋅ 0.20 = 0.0036.
Posted November 22, 2006
The third sentence of Problem No. 12 in Practice Examination No. 7 should read:
Let X and Y be the times at which the first and second circuits fail, respectively, counting
from the moment when a circuit is activated.
Posted August 3, 2006
In the solution of Problem No. 4 in Practice Examination No. 1, the formula
fY ( y ) = k2 ⋅ y,
should be
fY ( y ) = k2 ⋅1.
Posted July 3, 2006
The relationship between the survival function and the hazard rate, stated just after
the definition of the hazard rate, should be
x
sX ( x ) = e
instead of
−
∫ λ X (u ) du
−∞
x
∫
− λ X ( u ) du
sX ( x ) = e 0
.
The second equation applies in the case when the random variable X is nonnegative
almost surely.
Posted July 3, 2006
The condition concerning a continuous probability density function in its first
definition, in Section 1, should be
f X ( x ) ≥ 0 for every x ∈
instead of
0 ≤ f X ( x ) ≤ 1 for every x ∈.
Posted May 16, 2006
The solution of Problem No. 13 in Practice Examination No. 3 should be (some of
the exponents contained typos)
Solution.
Let X be the random number of passengers that show for a flight. We want to find
Pr(X = 31) + Pr(X = 32).
We can treat each passenger arrival as a Bernoulli Trial, and then it is clear that X has
binomial distribution with n = 32, p = 0.90. Therefore,
⎛ 32 ⎞
Pr ( X = x ) = ⎜
⋅ 0.90 x ⋅ 0.10 32−x.
⎟
x
⎝
⎠
The probability desired is:
⎛ 32 ⎞
⎛ 32 ⎞
Pr ( X = 31) + Pr ( X = 32 ) = ⎜
⋅ 0.90 31 ⋅ 0.101 + ⎜
⋅ 0.90 32 ⋅ 0.10 0 ≈
⎟
⎟
31
32
⎝
⎠
⎝
⎠
≈ 0.12208654 + 0.03433684 = 0.15642337.
Answer E.
Posted March 1, 2006
The formula for the variance of a linear combination of random variables should
be:
Var ( a1 X1 + a2 X2 +…+ an Xn ) =
(
(
)
)
T
= ⎡ a1 a2 … an ⎤ ⋅ ⎡⎣Cov Xi , X j ⎤⎦
⋅ ⎡ a1 a2 … an ⎤ =
⎣
⎦
⎣
⎦
i, j=1,2,…,n
(
(
)
= ⎡ a1 a2 … an ⎤ ⋅ ⎡⎣Cov Xi , X j ⎤⎦
⎣
⎦
i, j=1,2,…,n
n
n
n
(
)
⎡
⎢
⋅ ⎢⎢
⎢
⎢⎣
a1 ⎤
⎥
a2 ⎥
=
 ⎥
⎥
an ⎥
⎦
)
= ∑ a Var ( Xi ) + ∑ ∑ 2ai a j ⋅Cov Xi , X j .
2
i
i=1
i=1 j=i+1
The only incorrect piece was in the last double sum, where a1a2 appeared instead of
ai a j .
Posted March 1, 2006
The moment generating function of the chi-square distribution should be:
n
⎛ 1 ⎞2
M X (t ) = ⎜
.
⎝ 1− 2t ⎟⎠
(its argument was mistakenly written as x instead of t).
Posted February 18, 2006
Problem 18 of Practice Examination Number 7 and its solution should be:
The number of warranty claims in a month follows a Poisson distribution with mean 4.
Each warranty claim requires a payment of $1200 by the manufacturer to a repair facility.
Find the probability that the total payment by the manufacturer in a month is less than one
standard deviation away from the expected value of that payment.
A. 0.1225
B. 0.3907
C. 05470
D. 0.6936
E. 0.7977
Solution.
Let X be the random number of the warranty claims. The expected value of monthly
payments is 4 ⋅$1200 = $4800, and the standard deviation is 2 ⋅$1200 = $2400. The
probability that the total payment is between $4800 – $2400 = $2400 and $4800 + $2400
= $7200, not including the endpoints (because we are supposed to be less than one
standard deviation away from the mean) is the sum of probabilities of making 3 payments
(cost of $3600), 4 payments (cost of $4800), or 5 payments (cost of $6000), i.e.,
k
3
4
5
5
5
−4 4
−4 4
−4 4
−4 4
∑ Pr ( X = k ) = ∑ e ⋅ k! = e ⋅ 3! + e ⋅ 4! + e ⋅ 5! ≈ 0.54702708.
k=3
k=3
Answer C.
February 11, 2006
The solution of Problem No. 18 in Practice Examination No. 4 should be:
Let X be the claim for employee who incurred a loss in excess of 2000, and Y be the
claim for the other employee. The joint distribution of X and Y , given that each loss
occurs, is uniform on [1000, 5000 ] . This means that we can calculate all probabilities by
comparing areas. The probability X + Y > 8000, i.e., total losses exceed reimbursement,
given that X > 2000 is the ratio of the area of the triangle in the right upper corner of the
2
square [1000, 5000 ] in the figure below to the area of the rectangle
1
⋅ 2000 ⋅ 2000 1
= .
[ 2000, 5000 ] × [1000, 5000 ], and that ratio is 2
4000 ⋅ 3000
6
2
Y
5000
X + Y = 8000
3000
1000
X
1000
2000
5000
8000
However, this is conditional on the occurrence of the loss of the employee whose loss is
Y (and not conditional on the occurrence of the loss for the employee whose loss is X,
because the event considered is already conditional on X > 2000, so the loss has
occurred). The probability of the loss occurring is 40%. Therefore the probability sought
is
1 2 1 1
40% ⋅ = ⋅ = .
6 5 6 15
Answer B.
Posted January 31, 2006
In the solution of Problem No. 21 (May 2000 Course 1 Examination, Problem No.
19) in Practice Examination No. 2, the fourth sentence should begin
1 48
The mean of the 48 rounded ages, E =
∑ Ei
48 i =1
Posted January 26, 2006
In the solution of Problem No. 15 (May 2000 Course 1 Examination, Problem No. 7)
in Practice Examination No. 2, the survival function should be written as:
−3
s X ( x ) = (1 + x ) .
Posted January 18, 2006
In the description of the geometric distribution:
pet
M X (t ) =
. Some texts use a different version of this distribution, which only
1 − qet
counts failures until the first success. That’s just too pessimistic for this author. The two
distributions differ by one. For this other form of random variable (similar in its design to
q
q
the negative binomial distribution, just below) Y = X – 1, and E (Y ) = , Var ( X ) = 2 ,
p
p
p
M X (t ) =
.
1 − qet
should be
pet
M X (t ) =
. Some texts use a different version of this distribution, which only
1 − qet
counts failures until the first success. That’s just too pessimistic for this author. The two
distributions differ by one. For this other form of random variable (similar in its design to
q
q
the negative binomial distribution, just below) Y = X – 1, and E (Y ) = , Var (Y ) = 2 ,
p
p
p
M Y (t ) =
.
1 − qet
Posted January 18, 2006
Problem No. 6 in Practice Examination No. 7 and its solution should be:
Let X be a random variable representing the number of times you need to roll (including
the last roll) a fair six-sided dice until you get 4 consecutive 6's. Find E ( X ) .
A. 125
B. 1024
C. 1554
C. 2048
D. 15447
Solution.
Note that if X = n, then the last four rolls are 6’s, the one just before was not a 6, and there
were no four consecutive 6’s on the n − 5 rolls before that. Therefore,
4
4
⎛ 1⎞ 5
⎛ 1⎞ 5
Pr ( X = n ) = ⎜ ⎟ ⋅ ⋅ Pr ( X > n − 5 ) = ⎜ ⎟ ⋅ ⋅ Pr ( X ≥ n − 4 ) .
⎝ 6⎠ 6
⎝ 6⎠ 6
We have also X ≥ 4 with probability 1 and
4
⎛ 1⎞
Pr ( X = 4 ) = ⎜ ⎟ ,
⎝ 6⎠
4
5 ⎛ 1⎞
Pr ( X = 5 ) = ⋅ ⎜ ⎟ ,
6 ⎝ 6⎠
4
5 ⎛ 1⎞
Pr ( X = 6 ) = 1⋅ ⋅ ⎜ ⎟ ,
6 ⎝ 6⎠
4
5 ⎛ 1⎞
Pr ( X = 7 ) = 1⋅1⋅ ⋅ ⎜ ⎟ .
6 ⎝ 6⎠
Therefore, recalling that for a discrete random variable, whose only possible values are
+∞
possible integers, E ( X ) = ∑ Pr ( X ≥ n ), we get
n=1
+∞
+∞
3
E ( X ) = ∑ Pr ( X ≥ n ) = ∑ Pr ( X ≥ n ) + ∑ Pr ( X ≥ n ) =
n=1
3
n=1
+∞
n=4
+∞
6
= ∑1 + ∑ Pr ( X ≥ n − 4 ) = 3 + ∑ ⋅ 6 4 ⋅ Pr ( X = n ) =
n=1
n=8
n=8 5
Answer C.
= 3+
⎞
⎞
6 5 ⎛ +∞
6 5 ⎛ n=7
⋅ ⎜ ∑ Pr ( X = n )⎟ = 3 + ⋅ ⎜ 1− ∑ Pr ( X = n )⎟ =
⎠
⎠
5 ⎝ n=8
5 ⎝ n=4
= 3+
65 ⎛
6 5 5 5⎞
6 5 − 21
⋅ ⎜ 1− 5 − 5 − 5 − 5 ⎟ = 3 +
= 1554.
5 ⎝ 6 6 6 6 ⎠
5