This document contains sample exam questions and solutions from past Society of Actuaries and Casualty Actuarial Society exams. It includes 17 multiple choice probability questions with detailed step-by-step solutions. The questions cover a range of probability concepts including independent and dependent events, conditional probability, Venn diagrams, and probability distributions.
Systematic literature reviews - Moving offline communities onlineJenine Beekhuyzen
This Systematic Literature Review presentation at the University of Lapland on 22.5.15 is based on this paper http://www98.griffith.edu.au/dspace/bitstream/handle/10072/54874/88574_1.pdf?sequence=1
هیدراتهای گازی به عنوان یکی از منابع عظیم امیدوارکننده گاز طبیعی در جهان به شمار میروند. سازندهای حاوی هیدرات گازی همچنین میتوانند خطری عمده هنگام حفاری چاههای نفت و گاز متعارف به شمار بروند
Systematic literature reviews - Moving offline communities onlineJenine Beekhuyzen
This Systematic Literature Review presentation at the University of Lapland on 22.5.15 is based on this paper http://www98.griffith.edu.au/dspace/bitstream/handle/10072/54874/88574_1.pdf?sequence=1
هیدراتهای گازی به عنوان یکی از منابع عظیم امیدوارکننده گاز طبیعی در جهان به شمار میروند. سازندهای حاوی هیدرات گازی همچنین میتوانند خطری عمده هنگام حفاری چاههای نفت و گاز متعارف به شمار بروند
Frontiers of data-driven property prediction: molecular machine learningIchigaku Takigawa
Innovation Camp 2018 for Computational Materials Science(ICCMS2018)
January 23rd(Tue.)-25th(Thu.), 2018
The Jozankei View Hotel, Sapporo, Hokkaido, Japan.
http://ccms.issp.u-tokyo.ac.jp/events/eventsfolder/ICCMS2018
In materials science, data-centric science is becoming one of the major approaches along with theoretical, experimental, and computational sciences. The main purpose of this camp is that we learn the basics of the machine learning as data-centric science and use it to solve problems in our researches through group works. We will also have lectures on advanced researches in computational and data-centric sciences and discuss future perspectives. Furthermore, we learn innovation minds by inviting lecturers who are at the forefront beyond the industry-government-academia framework.
計算物質科学イノベーションキャンプ2018
物質科学の課題を解決する際、理論科学、実験科学、計算科学に加え、データ科学の活用が盛んになっている。本キャンプでは、そのデータ科学として機械学習の手法を学び、チームでの実習を通し手法を身に着け、各自の研究やプロジェクトの課題解決に役立てることを主目的とする。また、講師を招いて計算科学やデータ科学の最先端の研究成果に関する講義と今後の発展の可能性などについて議論する。さらに、産官学や学問領域を超えて活躍する方々のレクチャーと意見交換などでイノベーションマインドを学ぶ。
Quantum mechanics 1st edition mc intyre solutions manualSelina333
Quantum Mechanics 1st Edition McIntyre Solutions Manual
Download at: https://goo.gl/SdC7Ef
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quantum mechanics mcintyre solutions pdf
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hidden life of prayer
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
How to Create Map Views in the Odoo 17 ERPCeline George
The map views are useful for providing a geographical representation of data. They allow users to visualize and analyze the data in a more intuitive manner.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
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The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
Ethnobotany and Ethnopharmacology:
Ethnobotany in herbal drug evaluation,
Impact of Ethnobotany in traditional medicine,
New development in herbals,
Bio-prospecting tools for drug discovery,
Role of Ethnopharmacology in drug evaluation,
Reverse Pharmacology.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
2. 1. Solution: D
Let
G event that a viewer watched gymnastics
B event that a viewer watched baseball
S event that a viewer watched soccer
Then we want to find
Pr ª G ‰ B ‰ S
12. 1 0.48 0.52
--------------------------------------------------------------------------------------------------------
2. Solution: A
Let R = event of referral to a specialist
L = event of lab work
We want to find
P[RˆL] = P[R] + P[L] – P[R‰L] = P[R] + P[L] – 1 + P[~(R‰L)]
= P[R] + P[L] – 1 + P[~Rˆ~L] = 0.30 + 0.40 – 1 + 0.35 = 0.05 .
--------------------------------------------------------------------------------------------------------
3. Solution: D
First note
P A ‰ B@ P A@ P B @ P A ˆ B @
P A ‰ B '@ P A@ P B '@ P A ˆ B '@
Then add these two equations to get
P A ‰ B @ P A ‰ B '@ 2 P A@ P B @ P B '@
17. 4. Solution: A
For i 1, 2, let
Ri event that a red ball is drawn form urn i
Bi event that a blue ball is drawn from urn i .
Then if x is the number of blue balls in urn 2,
0.44 Pr[ R1 R2
20. 7. Solution: D
Let
A event that a policyholder has an auto policy
H event that a policyholder has a homeowners policy
Then based on the information given,
Pr A ˆ H
41. 9. Solution: B
Let
M event that customer insures more than one car
S event that customer insures a sports car
Then applying DeMorgan’s Law, we may compute the desired
probability as follows:
Pr M c ˆ S c
53. 0.205
--------------------------------------------------------------------------------------------------------
10. Solution: C
Consider the following events about a randomly selected auto insurance customer:
A = customer insures more than one car
B = customer insures a sports car
We want to find the probability of the complement of A intersecting the complement of B
(exactly one car, non-sports). But P ( Ac ˆ Bc) = 1 – P (A ‰ B)
And, by the Additive Law, P ( A ‰ B ) = P ( A) + P ( B ) – P ( A ˆ B ).
By the Multiplicative Law, P ( A ˆ B ) = P ( B _ A ) P (A) = 0.15 * 0.64 = 0.096
It follows that P ( A ‰ B ) = 0.64 + 0.20 – 0.096 = 0.744 and P (Ac ˆ Bc ) = 0.744 =
0.256
--------------------------------------------------------------------------------------------------------
11. Solution: B
Let
C = Event that a policyholder buys collision coverage
D = Event that a policyholder buys disability coverage
Then we are given that P[C] = 2P[D] and P[C ˆ D] = 0.15 .
By the independence of C and D, it therefore follows that
0.15 = P[C ˆ D] = P[C] P[D] = 2P[D] P[D] = 2(P[D])2
(P[D])2 = 0.15/2 = 0.075
P[D] = 0.075 and P[C] = 2P[D] = 2 0.075
Now the independence of C and D also implies the independence of CC and DC . As a
result, we see that P[CC ˆ DC] = P[CC] P[DC] = (1 – P[C]) (1 – P[D])
= (1 – 2 0.075 ) (1 – 0.075 ) = 0.33 .
Page 5 of 67
54. 12. Solution: E
“Boxed” numbers in the table below were computed.
High BP Low BP Norm BP Total
Regular heartbeat 0.09 0.20 0.56 0.85
Irregular heartbeat 0.05 0.02 0.08 0.15
Total 0.14 0.22 0.64 1.00
From the table, we can see that 20% of patients have a regular heartbeat and low blood
pressure.
--------------------------------------------------------------------------------------------------------
13. Solution: C
The Venn diagram below summarizes the unconditional probabilities described in the
problem.
In addition, we are told that
1 P A ˆ B ˆ C@ x
P A ˆ B ˆ C | A ˆ B@
3 P A ˆ B@ x 0.12
It follows that
1 1
x x 0.12
55. x 0.04
3 3
2
x 0.04
3
x 0.06
Now we want to find
P ª A ‰ B ‰ C
86. 0.40
--------------------------------------------------------------------------------------------------------
18. Solution: D
Let X1 and X2 denote the measurement errors of the less and more accurate instruments,
respectively. If N(P,V) denotes a normal random variable with mean P and standard
deviation V, then we are given X1 is N(0, 0.0056h), X2 is N(0, 0.0044h) and X1, X2 are
X1 X 2 0.00562 h 2 0.00442 h 2
independent. It follows that Y = is N (0, ) = N(0,
2 4
0.00356h) . Therefore, P[0.005h d Y d 0.005h] = P[Y d 0.005h] – P[Y d 0.005h] =
P[Y d 0.005h] – P[Y t 0.005h]
ª 0.005h º
= 2P[Y d 0.005h] – 1 = 2P « Z d 1 = 2P[Z d 1.4] – 1 = 2(0.9192) – 1 = 0.84.
¬ 0.00356h »¼
Page 8 of 67
87. 19. Solution: B
Apply Bayes’ Formula. Let
A Event of an accident
B1 Event the driver’s age is in the range 16-20
B2 Event the driver’s age is in the range 21-30
B3 Event the driver’s age is in the range 30-65
B4 Event the driver’s age is in the range 66-99
Then
Pr A B1
125. 22. Solution: D
Let
H Event of a heavy smoker
L Event of a light smoker
N Event of a non-smoker
D Event of a death within five-year period
1
ª º
Now we are given that Pr ¬ D L ¼ 2 Pr ª D N º and Pr ª D L º
¬ ¼ ¬ ¼ Pr ª D H º
2 ¬ ¼
Therefore, upon applying Bayes’ Formula, we find that
Pr ª D H º Pr H @
¬ ¼
Pr ª H D º
¬ ¼ Pr ª D N º Pr N Pr ª D L º Pr L Pr ª D H º Pr H
¬ ¼ @ ¬ ¼ @ ¬ ¼ @
2 Pr ª D L º 0.2
129. 0.25 0.3 0.4
1
2 ¬ ¼ ¬ ¼ ¬ ¼
--------------------------------------------------------------------------------------------------------
23. Solution: D
Let
C = Event of a collision
T = Event of a teen driver
Y = Event of a young adult driver
M = Event of a midlife driver
S = Event of a senior driver
Then using Bayes’ Theorem, we see that
P[C Y ]P[Y ]
P[Y~C] =
P[C T ]P[T ] P[C Y ]P[Y ] P[C M ]P[ M ] P[C S ]P[ S ]
(0.08)(0.16)
= = 0.22 .
(0.15)(0.08) (0.08)(0.16) (0.04)(0.45) (0.05)(0.31)
--------------------------------------------------------------------------------------------------------
24. Solution: B
Observe
Pr 1 d N d 4@ ª1 1 1 1 º ª1 1 1 1 1º
Pr ¬ N t 1 N d 4 º
ª ¼ « 6 12 20 30 » « 2 6 12 20 30 »
Pr N d 4@ ¬ ¼ ¬ ¼
10 5 3 2 20 2
30 10 5 3 2 50 5
Page 10 of 67
130. 25. Solution: B
Let Y = positive test result
D = disease is present (and ~D = not D)
Using Baye’s theorem:
P[Y | D]P[ D] (0.95)(0.01)
P[D|Y] = = 0.657 .
P[Y | D]P[ D] P[Y |~ D]P[~ D] (0.95)(0.01) (0.005)(0.99)
--------------------------------------------------------------------------------------------------------
26. Solution: C
Let:
S = Event of a smoker
C = Event of a circulation problem
Then we are given that P[C] = 0.25 and P[S~C] = 2 P[S~CC]
P[ S C ]P[C ]
Now applying Bayes’ Theorem, we find that P[C~S] =
P[ S C ]P[C ] P[ S C C ]( P[C C ])
2 P[ S C C ]P[C ] 2(0.25) 2 2
= .
2 P[ S C ]P[C ] P[ S C ](1 P[C ])
C C
2(0.25) 0.75 23 5
--------------------------------------------------------------------------------------------------------
27. Solution: D
Use Baye’s Theorem with A = the event of an accident in one of the years 1997, 1998 or
1999.
P[ A 1997]P[1997]
P[1997|A] =
P[ A 1997][ P[1997] P[ A 1998]P[1998] P[ A 1999]P[1999]
(0.05)(0.16)
= = 0.45 .
(0.05)(0.16) (0.02)(0.18) (0.03)(0.20)
--------------------------------------------------------------------------------------------------------
Page 11 of 67
131. 28. Solution: A
Let
C = Event that shipment came from Company X
I1 = Event that one of the vaccine vials tested is ineffective
P I1 | C @ P C @
Then by Bayes’ Formula, P C | I1 @
P I1 | C @ P C @ P ª I1 | C c º P ªC c º
¬ ¼ ¬ ¼
Now
1
P C @
5
1 4
P ªC c º 1 P C @ 1
¬ ¼ 5 5
P I1 | C @
143. --------------------------------------------------------------------------------------------------------
29. Solution: C
Let T denote the number of days that elapse before a high-risk driver is involved in an
accident. Then T is exponentially distributed with unknown parameter O . Now we are
given that
50
³ Oe
Ot
0.3 = P[T d 50] = dt e O t 50
= 1 – e–50O
0
0
Therefore, e–50O = 0.7 or O = (1/50) ln(0.7)
80
³ Oe
Ot
It follows that P[T d 80] = dt e O t 80
0
= 1 – e–80O
0
(80/50) ln(0.7)
=1–e = 1 – (0.7)80/50 = 0.435 .
--------------------------------------------------------------------------------------------------------
30. Solution: D
e O O 2 e O O 4
Let N be the number of claims filed. We are given P[N = 2] = 3 = 3 ˜ P[N
2! 4!
= 4]24 O = 6 O
2 4
O2 = 4 Ÿ O = 2
Therefore, Var[N] = O = 2 .
Page 12 of 67
144. 31. Solution: 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 @ d 0.01
or, equivalently,
0.99 d Pr X d x @ ¦
152. 2 18
2 0.053 0.993
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
--------------------------------------------------------------------------------------------------------
32. Solution: D
Let
X = number of low-risk drivers insured
Y = number of moderate-risk drivers insured
Z = number of high-risk drivers insured
f(x, y, z) = probability function of X, Y, and Z
Then f is a trinomial probability function, so
Pr z t x 2@ f 0, 0, 4
169. 2 72 9
2
--------------------------------------------------------------------------------------------------------
34. Solution: C
We know the density has the form C 10 x
170. for 0 x 40 (equals zero otherwise).
2
40
First, determine the proportionality constant C from the condition ³ 0
f ( x)dx 1 :
40
C C 2
C 10 x
171. dx C (10 x) 1
40 2
1 ³ 0 0
10 50 25
C
so C 25 2 , or 12.5 . Then, calculate the probability over the interval (0, 6):
1 6 §1 1·
12.5³ 10 x
181. 4 5 1
= P[100,000 Y 40,000] = P[Y 0.4] = 0.4
= (0.6)5 = 0.078 .
0.4
It now follows that P[V 40,000~V 10,000]
P[V ! 40, 000 ˆ V ! 10, 000] P[V ! 40, 000] 0.078
= = 0.132 .
P[V ! 10, 000] P[V ! 10, 000] 0.590
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37. Solution: D
Let T denote printer lifetime. Then f(t) = ½ e–t/2, 0 d t d ¥
Note that
1
1
P[T d 1] = ³ e t / 2 dt e t / 2 1 = 1 – e–1/2 = 0.393
0
2 0
2
1
³ 2e
t / 2
e t / 2
2
P[1 d T d 2] = dt 1 = e –1/2 e –1 = 0.239
1
Next, denote refunds for the 100 printers sold by independent and identically
distributed random variables Y1, . . . , Y100 where
200 with probability 0.393
°
Yi ®100 with probability 0.239 i = 1, . . . , 100
°0 with probability 0.368
¯
Now E[Yi] = 200(0.393) + 100(0.239) = 102.56
100
Therefore, Expected Refunds = ¦ E Y @ = 100(102.56) = 10,256 .
i 1
i
Page 15 of 67
182. 38. Solution: A
Let F denote the distribution function of f. Then
F x
183. Pr X d x @
x x
³1
3t 4 dt t 3
1
1 x 3
Using this result, we see
Pr ª X 2
194. 2 2
2 x dx x
0 0
Therefore, solving for C, we find C r0.8 0.5
Finally, since 0 C 1 , we conclude that C 0.3
Page 16 of 67
195. 41. Solution: E
Let
X = number of group 1 participants that complete the study.
Y = number of group 2 participants that complete the study.
Now we are given that X and Y are independent.
Therefore,
^
P ª X t 9
216. º
9 10 9 10
¬ 9 10
¼¬ 9 10
¼
2 0.376@1 0.376@ 0.469
--------------------------------------------------------------------------------------------------------
42. Solution: D
Let
IA = Event that Company A makes a claim
IB = Event that Company B makes a claim
XA = Expense paid to Company A if claims are made
XB = Expense paid to Company B if claims are made
Then we want to find
¬^
Pr ª I A ˆ I B º ‰ ª I A ˆ I B
224. Pr X B X A t 0@
0.18 0.12 Pr X B X A t 0@
Now X B X A is a linear combination of independent normal random variables.
Therefore, X B X A is also a normal random variable with mean
M E X B X A @ E X B @ E X A @ 9, 000 10, 000 1, 000
and standard deviation V Var X B
229. ª 1000 º
Pr X B X A t 0@ Pr « Z t (Z is standard normal)
¬ 2000 2 »
¼
ª 1 º
Pr « Z t
¬ 2 2»¼
ª 1 º
1 Pr « Z
¬ 2 2»¼
1 Pr Z 0.354@
1 0.638 0.362
Finally,
^
Pr ª I A ˆ I B º ‰ ª I A ˆ I B
242. 44. Solution: C
If k is the number of days of hospitalization, then the insurance payment g(k) is
g(k) =
100k
^ for k 1, 2, 3
300 50 (k 3) for k 4, 5.
5
Thus, the expected payment is ¦ g (k ) p
k 1
k 100 p1 200 p2 300 p3 350 p4 400 p5 =
1
100 u 5 200 u 4 300 u 3 350 u 2 400 u 1
243. =220
15
--------------------------------------------------------------------------------------------------------
45. Solution: D
0 4
x2 4 x
2
x3 x3 8 64 56 28
Note that E X
244. 0
³ 2 10
dx ³
0 10
dx
30 2 30 0
30 30 30 15
--------------------------------------------------------------------------------------------------------
46. Solution: D
The density function of T is
1 t / 3
f t
246. º
¬ ¼
2 2 t / 3 f t
³0 3
e dt ³ e t / 3 dt
2 3
f
2e t / 3 _ 0 te t / 3 _ f ³ e t / 3 dt
2
2
2
2 / 3 2 / 3
2e 2 2e 3e t / 3 _ f
2
2 3e 2 / 3
Page 19 of 67
247. 47. Solution: D
Let T be the time from purchase until failure of the equipment. We are given that T is
exponentially distributed with parameter O = 10 since 10 = E[T] = O . Next define the
payment
x for 0 d T d 1
°x
°
P under the insurance contract by P ® for 1 T d 3
° 2
°0
¯ for T ! 3
We want to find x such that
1 3 1
x –t/10 x 1 –t/10 x
1000 = E[P] = ³ ³ 2 10 e dt = xe
t /10
e dt + e t /10 3
1
0
10 1 0 2
= x e –1/10
+ x – (x/2) e –3/10
+ (x/2) e –1/10
= x(1 – ½ e –1/10
– ½ e–3/10) = 0.1772x .
We conclude that x = 5644 .
--------------------------------------------------------------------------------------------------------
48. Solution: 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
f x
258. 2 3
2694
--------------------------------------------------------------------------------------------------------
49. Solution: D
Define f ( X ) to be hospitalization payments made by the insurance policy. Then
100 X
° if X 1, 2,3
f (X ) ®
¯300 25 X 3
264. , N 1,...,5
137 N 137 N
Now because of the deductible of 2, the net annual premium P E X @ where
0 , if N d 2
X ®
¯N 2 , if N ! 2
Then,
3 § 1 · 3 ª 3 º ª º
EX @ ¦ N 2
287. 0.328 (in thousands)
It follows that the expected claim payment is 328 .
--------------------------------------------------------------------------------------------------------
55. Solution: C
k
The pdf of x is given by f(x) = , 0 x f . To find k, note
(1 x) 4
f
k k 1 f k
1=³ (1 x)4 dx
0
3 (1 x)3
0
3
k=3
f
3x
It then follows that E[x] = ³ (1 x)
0
4
dx and substituting u = 1 + x, du = dx, we see
f f f
3(u 1) ª u 2 u 3 º ª1 1º
E[x] = ³ du = 3 ³ (u 3 u 4 )du 3« » 3 « » = 3/2 – 1 = ½ .
1
u4 1 ¬ 2 3 ¼1 ¬ 2 3¼
Page 23 of 67
288. 56. Solution: C
Let Y represent the payment made to the policyholder for a loss subject to a deductible D.
0 for 0 d X d D
That is Y ®
¯ x D for D X d 1
Then since E[X] = 500, we want to choose D so that
1 ( x D) 2 1000 (1000 D) 2
1000
1 1
500 ³ ( x D)dx
4 D
1000 1000 2 D 2000
(1000 – D)2 = 2000/4 ˜ 500 = 5002
1000 – D = r 500
D = 500 (or D = 1500 which is extraneous).
--------------------------------------------------------------------------------------------------------
57. Solution: B
1
We are given that Mx(t) = for the claim size X in a certain class of accidents.
(1 2500t ) 4
(4)(2500) 10, 000
First, compute Mxc(t) =
(1 2500t ) 5
(1 2500t )5
(10, 000)(5)(2500) 125, 000, 000
Mxs(t) =
(1 2500t )6 (1 2500t )6
Then E[X] = Mxc (0) = 10,000
E[X2] = Mxs (0) = 125,000,000
Var[X] = E[X2] – {E[X]}2 = 125,000,000 – (10,000)2 = 25,000,000
Var[ X ] = 5,000 .
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58. Solution: E
Let XJ, XK, and XL represent annual losses for cities J, K, and L, respectively. Then
X = XJ + XK + XL and due to independence
M(t) = E ªe xt º E ªe J K L
289. º E ªe xJ t º E ª e xK t º E ªe xLt º
x x x t
¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼
= MJ(t) MK(t) ML(t) = (1 – 2t)–3 (1 – 2t)–2.5 (1 – 2t)–4.5 = (1 – 2t)–10
Therefore,
Mc(t) = 20(1 – 2t)–11
Ms(t) = 440(1 – 2t)–12
Msc(t) = 10,560(1 – 2t)–13
E[X3] = Msc(0) = 10,560
Page 24 of 67
290. 59. Solution: B
The distribution function of X is given by
2.5 x
2.5 200
301. 25 25
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60. Solution: E
Let X and Y denote the annual cost of maintaining and repairing a car before and after
the 20% tax, respectively. Then Y = 1.2X and Var[Y] = Var[1.2X] = (1.2)2 Var[X] =
(1.2)2(260) = 374 .
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z
61. Solution: A
f
The first quartile, Q1, is found by ¾ = f(x) dx . That is, ¾ = (200/Q1)2.5 or
Q1
Q1 = 200 (4/3)0.4 = 224.4 . Similarly, the third quartile, Q3, is given by Q3 = 200 (4)0.4
= 348.2 . The interquartile range is the difference Q3 – Q1 .
Page 25 of 67
302. 62. Solution: C
First note that the density function of X is given by
1
°2 if x 1
°
°
f x
303. ® x 1 if 1 x 2
°
°
°0
¯ otherwise
Then
2
1 §1 3 1 2·
EX
314. 64. Solution: A
Let X denote claim size. Then E[X] = [20(0.15) + 30(0.10) + 40(0.05) + 50(0.20) +
60(0.10) + 70(0.10) + 80(0.30)] = (3 + 3 + 2 + 10 + 6 + 7 + 24) = 55
E[X2] = 400(0.15) + 900(0.10) + 1600(0.05) + 2500(0.20) + 3600(0.10) + 4900(0.10)
+ 6400(0.30) = 60 + 90 + 80 + 500 + 360 + 490 + 1920 = 3500
Var[X] = E[X2] – (E[X])2 = 3500 – 3025 = 475 and Var[ X ] = 21.79 .
Now the range of claims within one standard deviation of the mean is given by
[55.00 – 21.79, 55.00 + 21.79] = [33.21, 76.79]
Therefore, the proportion of claims within one standard deviation is
0.05 + 0.20 + 0.10 + 0.10 = 0.45 .
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65. Solution: B
Let X and Y denote repair cost and insurance payment, respectively, in the event the auto
is damaged. Then
0 if x d 250
Y ®
¯ x 250 if x ! 250
and
1 1 12502
E Y @ ³ x 250
318. 1500
1500
E ¬Y 2 º ³
2 3
ª ¼ 250 434, 028
250 1500 4500 4500
Var Y @ E ªY 2 º ^ E Y @` 434, 028 521
319. 2 2
¬ ¼
Var Y @ 403
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66. Solution: E
Let X1, X2, X3, and X4 denote the four independent bids with common distribution
function F. Then if we define Y = max (X1, X2, X3, X4), the distribution function G of Y is
given by
G y
334. dy
5/ 2
³
3
3/ 2 4
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67. Solution: 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 t 2
where x is a Poisson random variable with mean 0.6 . The probability function for X is
e 0.6 0.6
356. 68. Solution: C
Note that X has an exponential distribution. Therefore, c = 0.004 . Now let Y denote the
x for x 250
claim benefits paid. Then Y ® and we want to find m such that 0.50
¯250 for x t 250
m
m
= ³ 0.004e 0.004 x dx e 0.004 x 0 = 1 – e–0.004m
0
This condition implies e–0.004m = 0.5 Ÿ m = 250 ln 2 = 173.29 .
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69. Solution: D
The distribution function of an exponential random variable
T with parameter T is given by F t
357. 1 e t T , t ! 0
Since we are told that T has a median of four hours, we may determine T as follows:
1
F 4
363. e 5 T e 4
25 4 0.42
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70. Solution: E
Let X denote actual losses incurred. We are given that X follows an exponential
distribution with mean 300, and we are asked to find the 95th percentile of all claims that
exceed 100 . Consequently, we want to find p95 such that
Pr[100 x p95 ] F ( p95 ) F (100)
0.95 = where F(x) is the distribution function of X .
P[ X ! 100] 1 F (100)
Now F(x) = 1 – e–x/300 .
1 e p95 / 300 (1 e 100 / 300 ) e 1/ 3 e p95 / 300
Therefore, 0.95 = 1 e1/ 3e p95 / 300
1 (1 e 100 / 300 ) e 1/ 3
e p95 / 300 = 0.05 e –1/3
p95 = 300 ln(0.05 e–1/3) = 999
Page 29 of 67
364. 71. Solution: A
The distribution function of Y is given by
G y