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Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page
1
Errata and updates for ASM Exam C/Exam 4 Manual (Sixteenth Edition
Second Printing) sorted by page
Note the change in wording in Practice Exam 1:1 (page 1261), and the changed answer choices for Practice
Exam 10:8 (page 1359).
[3/25/2014]
On page 11, in the third displayed line of Section 1.4, change e t x to e t X .
[6/12/2014]
On page 23, in the solution to exercise 1.15, the second displayed line should be
0.16 = exp −
2
z 2 /2
z2
z
2
(ln 5 − ln 2) = exp
ln 2/5 =
2
2
5
[1/31/2014]
On page 53, two lines above “Continuity correction”, change “call” to “called”.
[6/27/2014]
On page 99, on the first line of the answer to Example 6F, the first sentence is inaccurate; E[(X − 500)+ ] is not
calculated previously. However, it is easy to calculate: E[(X − 500)+ ] = 0.2(1000 − 500) + 0.1(2000 − 500) = 250.
[2/22/2014]
On page 99, 2 lines after the answer to Example 6H, change θ − d to θ = d .
[3/27/2014]
On page 103, in exercise 6.4, in the second bullet, change r % to 100r %.
[3/27/2014]
On page 112, in the solution to exercise 6.2, on the second displayed line, place a parenthesis around
[2/5/2014]
On page 165, in exercise 9.13, although the exercise can be solved mechanically, the values in the table are
impossible if losses are nonnegative. If F (100) = 0.3, then E[X ∧ 100] ≥ (1 − 0.3)(100) = 70.
[4/1/2014]
On page 171, in the solution to exercise 9.4, on the first line, add “is” after “it”.
[6/2/2014]
On page 176, in the solution to exercise 9.19, on the last displayed line, there should be a 0.8 on the left hand
side, so that the part preceding the first equals sign is
20,000
.
20,000 + d 2
0.8(E[X ∧ 5000] − E[X ∧ 100])
[9/10/2014]
On page 206, in the solution to exercise 11.26, on the third line, add “1 plus” before the second “mean”: “. . . so
its mean is 1 plus the mean of an unshifted . . . ”.
[4/1/2014]
On page 237, in the warning box on the second line, change “variance distribution” to “variance formula”.
[8/1/2014]
On page 274, in the solution to exercise 15.29, on the third line from the end, change 1.3965 to 1.9501.
[7/3/2014]
On page 332, in the solution to exercise 19.1, on the last line, change 50 to 40.
[6/17/2014]
On page 332, in the solution to exercise 19.3, on the last line, change X > 3000 to S > 3000.
[4/18/2014]
On page 339, in exercise 20.11, delete the first line with “s”.
[6/23/2014]
On page 361, in exercise 21.9, on the sixth line, replace MSEβ (ψ)/ MSEβ (φ) with MSEψ (β )/ MSEφ (β ).
[6/23/2014]
On page 365, in the solution to exercise 21.9, on the first displayed line, replace MSEβ (ψ) with MSEψ (β ). On
the second displayed line, replace
M S Eβ (φ) with MSEφ (β ).
[7/29/2014]
On page 382, in exercise 23.11, on the last line, delete the hat on top of v (θ ).
[6/18/2014]
On page 386, in the solution to exercise 23.11, on the second line, change “θn is binomial” to “θn is a binomial
proportion”.
Updated 10/12/2014
2
Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page
1
7
to q = 67 .
[7/29/2014]
On page 386, in the solution to exercise 23.13, on the fifth line, change q =
[6/19/2014]
On page 403, in exercise 24.29, on the fourth line, change “states” to “times”.
[4/23/2014]
On page 475, in the solution to exercise 27.9, on the third displayed line, change Fn (17) = Fn (16) to Fn (17) −
Fn (16).
[6/11/2014]
On page 499, in the solution to exercise 28.5, on the second line, replace “toal” with “total”.
[2/2/2014]
On page 499, in the solution to exercise 28.8, replace the bottom 3 lines of the page with
Total exposure in months is 24(36) + 41(34) + 24 + 12(29) + 8 + 27 + 22(25) = 3215. The estimate of 3q 25 , by
formula (28.2), is
nd j
3(1)
=
= 3/(3215/12) = 0.011198
3q 25 =
ej
3215/12
The standard deviation, by formula (28.4), is
q
v
u q (1 − q ) v
t (0.011198)(1 − 0.011198)
t j
j
d
Var(3 qˆ25 ) =
=
= 0.011135
e j /n
(3215/12)/3
[6/11/2014]
On page 500, in the solution to exercise 28.9, on the fifth line, replace (33-5,38-5,x ) with (33.5,38-5,s ).
[4/17/2014]
On pages 501, in the solution to exercise 28.13, on the first line of the page, change e −19/12 to e −12/19 . On the
third line of the page, change the final answer to 0.01175 .
[8/24/2014]
On page 501, in the solution to exercise 28.15, on the third line from the end, change e −17/12 to e −12/17 . On the
second line from the end, change 12/17 to 12/22. Change 0.5456 to 0.5455. On the last line, change 0.0391 to
0.0392.
[4/19/2014]
On page 503, in the solution to exercise 28.26, on the fourth line, change 0.5(350) to 0.5(700).
[8/24/2014]
On page 501, in the solution to exercise 28.14, on the third line from the end, change e −13/12 to e −12/13 .
[3/14/2014]
On page 541, in the solution to exercise 30.32, the answer key should be (B) instead of (C).
[6/24/2014]
On page 579, in exercise 32.15, in answer choice (B), the exponent 3 in the denominator should be outside the
parenthesis, so that the denominator is f (100; θ )3 .
[5/8/2014]
On page 626, in exercise 33.36, the correct solution, starting with the second line, is
ˆ=P
α
n
ln(1 + xi )
P
If the sample mean
Pn xi /n = x¯ , then at least one of the xi s is greater than or equal to x¯ . The other xi s are
ˆ → 0. (A)
nonnegative. So i =1 ln(1 + xi ) ≥ ln(1 + x¯ ). As x¯ → ∞, ln(1 + x¯ ) → ∞. Therefore α
[8/4/2014]
On page 642, 2 lines from the end of the page, change e −8250/10,000 to e −10,000/8250 .
[5/5/2014]
On page 643, on the fifth displayed line of the answer to Example 34E, change the minus sign after
sign.
[1/7/2014]
On page 653, in exercise 34.2II, change Crarner to Cràmer.
[4/23/2014]
On page 661, in exercise 34.32, on the last line, change Var to VaR.
[6/26/2014]
On page 664, in the solution to exercise 34.7, delete “negative”.
m
to a plus
a
Updated 10/12/2014
Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page
[4/23/2014]
3
On page 669, in the solution to exercise 34.32, change the last four lines to
∂g
= (− ln 0.05)1/τ = (− ln 0.05)1/2 = 1.73082
∂θ
θ (− ln 0.05)1/τ
5000(− ln 0.05)1/2
∂g
= ln(− ln 0.05)
= ln(− ln .05)
= −2373.79
∂τ
−τ2
−4
The variance of the estimate is
(1.730822 )(62,500) + (2,373.792 )(0.0008) = 191,741
[6/27/2014]
On page 671, the solution to exercise 34.37 is incorrect. The correct solution is
The length of the 2-sided confidence interval for the proportion is 2 times 1.96 (the 97.5th percentile of a standard normal distribution) times the standard deviation of the estimate for the proportion. We will use the delta
method. In the previous exercise, we estimated θˆ = 125. Let g (θ ) be Pr(X > 250 | θ ). Then
g (θ ) = S (250 | θ ) = e −250/θ
By the delta method
2
Var g (θˆ ) ≈ Var(θˆ ) g 0 (θˆ )
Since θˆ is the sample mean, its variance is the variance of the distribution divided by the size of the sample.
The variance of an exponential with mean θ is θ 2 , so the variance of θˆ is
2
θ2
125
ˆ
Var(θ ) =
≈
100
10
where as usual we approximate θ with its estimate.
250 −250/θ
e
θ2
250 −250/125
2 −2
g 0 (θˆ ) =
e
=
e
1252
125
2 
‹
125
2 −2 2
Var g (θˆ ) ≈
e
10
125
2
= 0.2e −2
g 0 (θ ) =
The standard deviation of g (θˆ ) is 0.2e −2 , and the length of the confidence interval is 2(1.96)(0.2e −2 ) = 0.1061 .
(E)
[6/30/2014]
On page 753, in exercise 39.1, at the end of the last bullet, change “good to fit” to “goodness of fit”.
[6/30/2014]
On page 763, in exercise 39.23, after the table, add “A Pareto distribution is fit to the data.”
[9/27/2014]
On page 770, in the solution to exercise 39.6, on the sixth line, change 4041 to 4047. On the last line, change
4046 to 4047 in two places, and change the final answer to 258.
[6/19/2014]
On page 778, in the paragraph beginning “Thus if an exponential model”, on the fourth line, delete one “that”.
[4/23/2014]
On page 780, on the second-to-last line of the answer to Example 40C, replace 2(−17.8631 + 17.8625) with
2(17.8631 − 17.8625).
[5/19/2014]
On page 799, in the solution to exercise 41.5, on the first displayed line, change F ( x ) to F ∗ (x ).
Updated 10/12/2014
4
[6/9/2014]
Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page
On page 801, in the solution to exercise 41.13, in the table, change “Transformed gamma” to “Generalized
Pareto”.
[10/12/2014] On
page 852, in the solution to exercise 44.18, replace the line that is three lines from the bottom with
yp2 =
2831.25(0.12 )
= 3.539
8
[8/21/2014]
On page 890, in the paragraph beginning with 1., on the first line, change x to θ .
[5/28/2014]
On page 971, exercise 51.26 is a duplicate of exercise 51.7.
[8/30/2014]
On page 1045, in the solution to exercise 53.17, replace the 9 lines starting with “We want to calculate” with
We want to calculate
Pr(1, 0 and p < 0.1)
Pr(1, 0)
Pr(1, 0 and p < 0.1)
=
Pr(1, 0 and p < 0.1) + Pr(1, 0 and p ≥ 0.1)
Pr(p < 0.1 | 1, 0) =
We will obtain the numerator by integrating, from 0 to 0.1, Pr(1, 0) = p (1−p ) over the density function of p . We
will obtain the second term of the denominator by integrating the same integrand from 0.1 to 0.2.
Replace the left side on the next line with Pr(1, 0 and p < 0.1. Replace the left side three lines after that line
with Pr(1, 0 and p ≥ 0.1). Replace the final line of the solution with
Pr(p < 0.1 | 1, 0) =
37/1200
37
37
=
=
37/1200 + 0.0575 37 + 69
106
(B)
[5/6/2014]
On page 1079, in the solution to exercise 55.16, on the second line, change 0.2(0.5) + 0.8(3) to 0.8(0.5) + 0.2(3).
[6/26/2014]
On page 1154, in the solution to exercise 59.16, on the second-to-last line, change 17,0460,223 to 17,046,223.
[6/17/2014]
On page 1160, on the first displayed line, change the integrand to
5
du
u 1.5
[6/13/2014]
On page 1169, exercise 60.19 is a duplicate of exercise 60.2.
[7/9/2014]
On page 1177, in the solution to exercise 60.17, on the last line, change
[1/14/2014]
On page 1186, two lines from the bottom, change 1 − 0.8 − 0.2 to 1 − 0.08 − 0.02.
[5/26/2014]
On page 1192, in exercise 61.2, on the displayed line, e x /3 should be e −x /3 .
[6/23/2014]
On page 1196, in exercise 61.21, on the first line, change X (l ) to X (1).
[12/3/2013]
[6/16/2014]
On page 1198, in the solution to exercise 61.13, on the last displayed line, change 2.842434 to 2.482433.
p
On page 1200, in the solution to exercise 61.22, on the last line, change 2e to (2e ).
[6/17/2014]
On page 1225, in exercise 63.16, two lines after the enumerated list, change “longer times” to “shorter times”.
[6/2/2014]
ˆ 2 to s 2 .
On page 1240, 2 lines above Example 64F, change σ
[12/4/2013]
On page 1261, in question 1, on the last line, change “counts” to “costs”.
p
p
3
2
0.125 to 0.125.
Updated 10/12/2014
Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page
[1/27/2014]
5
On page 1359, in question 8, replace the answer choices with
(A) 43.8
(B) 44.1
(C) 44.4
(D) 44.7
(E) 45.0
[6/5/2014]
On page 1395, in the solution to question 16, on the second line, in the exponent on (1 − e −1000/θ ), change 94
to 84.
[9/18/2014]
On page 1398, in the solution to question 26, on the first line, change 1000 to 1600 in two places.
[10/7/2014]
On page 1399, in the solution to question 29, on the seventh line, change
[3/22/2014]
On page 1416, in the solution to question 5, change the two “Age 45”s in the heading of the table to “Age 35”.
Change the three subscripts “45” in the two lines after the table, and the subscript “40” in the displayed formula, to “35”.
[5/19/2014]
On page 1476, the solution to question 8 has typos and is unclear. Here is a clearer solution:
R∞
d
x f (x ) dx to
R∞
q
(x −q )f (x ) dx
0.05
.
The function transforming X to Y is y = g (x ) = x −1 = 1/x . Notice that x = g −1 (y ) = 1/y . The derivative of
g −1 (y ) is
1
dg −1
=−
dy
y2
By the formula for transforming densities of random variables,
1
(1/(100y ))4 e −1/100y 1
=
f Y (y ) = f X (1/y )
y2
(1/y )Γ (4)
y2
−1/100y
1 e
=
1004 Γ (4)y 5
Not surprisingly, inverting X results in an inverse gamma. If you check the tables, you will see that the new
parameters are α = 4 and θ = 1/100. The mode of an inverse gamma according to the tables is θ /(α + 1) =
1/(100 · 5) = 1/500 . (B)
If you did not recognize the density as an inverse gamma, you could still find the mode by differentiating it
and set the derivative equal to 0. It is easier to differentiate the log:
1
+ ln y − ln Γ (4) − 2 ln y
100y
400
1
1
5
1
=−
+
− =− +
=0
100y 100y 2 y
y 100y 2
ln f Y = −4 ln 100y −
d ln f Y
dy
1
−5y +
=0
100
y=
1
500
[1/29/2014]
On page 1485, in the solution to question 30, on the last line, interchange 0.213061 and 0.5.
[1/27/2014]
On page 1489, in the solution to question 5, on the second displayed line, change (1−0.117503) to (1−0.117503)2 .
[1/27/2014]
On page 1501, the solution to question 8 is incorrect. The correct solution is
From the tabular values with d = 20.
E[X ] = E[X ∧ 20] + Pr(X > 20)e (20) = 19 + 0.9(200) = 199
Updated 10/12/2014
6
Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page
We also have
E[X ] = E[X ∧ 50] + Pr(X > 50)e (50)
199 = E[X ∧ 50] + 180 Pr(X > 50)
(*)
Let’s get a lower bound for Pr(X > 50) by using the values of E[X ∧ 50].
E[X ∧ 20] = 20 Pr(X > 20) +
20
Z
x f (x ) dx
0
19 = 20(0.9) +
20
Z
x f (x ) dx
0
20
Z
x f (x ) dx = 1
0
E[X ∧ 50] = 50 Pr(X > 50) +
50
Z
x f (x ) dx
0
= 50 Pr(X > 50) +
20
Z
x f (x ) dx +
0
Z
50
x f (x ) dx
20
≥ 50 Pr(X > 50) + 1 + 20 Pr(20 < X ≤ 50)
= 50 Pr(X > 50) + 1 + 20 1 − 0.1 − Pr(X > 50)
= 30 Pr(X > 50) + 19
(**)
Plugging this into (*), we get
199 ≤ 30 Pr(X > 50) + 19 + 180 Pr(X > 50)
= 210 Pr(X > 50) + 19
210 Pr(X > 50) ≥ 180
Pr(X > 50) ≥
6
7
Then plugging into (**),
6
E[X ∧ 50] ≥ 30
+ 19 = 44 57
7
To prove that this lower bound is attained, here is a discrete random variable satisfying the question’s assumptions with E[X ∧ 50] = 44 75 :
x
Pr(X = x )
10
20
230
[6/17/2014]
0.1
6
7
− 0.1
6
7
On page 1522, in the solution to question 30, on the last line of the page, change 0.109329 to 0.019329.
Updated 10/12/2014