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r = rand(1000,1); x = zeros(20,1); for i=1:1000 for j=0:19 if(r(i)>=j*0.05 && r(i)<(j+1)*0.05) x(j+1)=x(j+1)+1; end end end a=0; for k = 1:20 a = a + (x(k)-50)*(x(k)-50)/50; end a

Pokers_test.m r = randi([0,999],1000,1); a=0; b=0; c=0; for i = 1:1000 if r(i)==0 b=b+1; elseif (r(i)>0)&&(r(i)<=10) c=c+1; elseif (r(i)>10)&&(r(i)<100) x=mod(r(i),10); y=(r(i)-x)/10; if (x==y || x==0) c=c+1; else a=a+1; end else x=mod(r(i),10); y=mod(((r(i)-x)/10),10); z=(((r(i)-x)/10)-y)/10; if (x==y)&&(y==z) b=b+1; elseif (x~=y)&&(x~=z) a=a+1; else c=c+1; end end end a b c chisq = (a-720)*(a-720)/720 + (b-10)*(b-10)/10 + (c-270)*(c-270)/270

normal.m

r=rand(200,1); r=sort(r); x=zeros(100,1); a=zeros(100,2); for n=1:100 i=2*n-1 x(n,1)= sqrt(-2*log(r(i,1)))*cos(2*pi*r(i+1,1)); a(n,2)=x(n,1); a(n,1)=exp(-x(n,1)*x(n,1))/(2*pi); end plot(a(:,1),a(:,2))

Exp_dist.m

r=rand(100,1); r=sort(r); x=zeros(100,1); a=zeros(100,2); for n=1:100 x(n,1)=-log(1-r(n,1)); a(n,2)=x(n,1); a(n,1)=exp(-x(n,1)); end plot(a(:,1),a(:,2))

r = rand(1000,1); x = zeros(20,1); for i=1:1000 for j=0:19 if(r(i)>=j*0.05 && r(i)<(j+1)*0.05) x(j+1)=x(j+1)+1; end end end a=0; for k = 1:20 a = a + (x(k)-50)*(x(k)-50)/50; end a

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...questions. Use only blanks left to answer lab questions. Provide all histograms you are asked to print, but DO NOT print data you are asked to generate. 1. Continuous distributions: Generate and store in column c1 10,000 values from the uniform distribution on the interval [3,7] as follows: random 10000 c1; uniform 3 7. [3] a. Use mean command to ﬁnd the sample mean x of these data———————– ¯ [2] b. What is the mean µ of the uniform distribution on the interval [3,7]?————[1] c. Compare µ to the value x you found in part a). ———————– ¯ Generate and store in column c2 1,000 values from exponential distribution with parameter λ = .125 as follows: random 1000 c2; exponential 8. Note: The mean µ and the standard deviation σ of such distribution are both equal to 1/λ = 8 and this is the value you are asked to enter in the command above. [3] d. Use desc command to ﬁnd the sample mean x and sample standard deviation s for ¯ these 1,000 data —————– and —————— Are x and s close to the value 1/λ = 8?———————– Why?——————————¯ [3] e. Print (and include in your assignment) the histogram of the 1,000 values you generated from this exponential distribution. What is the shape of this distribution?———————– 2. Normal distribution: Generate and store in column c3 10,000 values from the standard normal distribution as follows: random 10000 c3; normal. [3] a. Print (and include in your assignment) the histogram for these data. What is the shape of this histogram?———————————– [3] b. What is the value...

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...taken from past examinations. The weight of topics in these sample questions is not representative of the weight of topics on the exam. The syllabus indicates the exam weights by topic. Copyright 2013 by the Society of Actuaries and the Casualty Actuarial Society C-09-08 PRINTED IN U.S.A. 1. You are given: (i) Losses follow a loglogistic distribution with cumulative distribution function: bx / θ g F b xg = 1+ bx / θ g γ γ (ii) The sample of losses is: 10 35 80 86 90 120 158 180 200 210 1500 Calculate the estimate of θ by percentile matching, using the 40th and 80th empirically smoothed percentile estimates. (A) (B) (C) (D) (E) Less than 77 At least 77, but less than 87 At least 87, but less than 97 At least 97, but less than 107 At least 107 2. You are given: (i) The number of claims has a Poisson distribution. (ii) (iii) (iv) Claim sizes have a Pareto distribution with parameters θ = 0.5 and α = 6 . The number of claims and claim sizes are independent. The observed pure premium should be within 2% of the expected pure premium 90% of the time. Determine the expected number of claims needed for full credibility. (A) (B) (C) (D) (E) Less than 7,000 At least 7,000, but...

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...Erlang Distribution If interarrival or service times are not exponential, an Erlang random variable can often be used to model them. If T is an Erlang random variable with rate parameter R and shape parameter k, the density of T is given by R(Rt)kϪ1eϪRt f (t) ϭ ᎏᎏ (k Ϫ 1)! (t Ն 0) and k E(T) ϭ ᎏᎏ R and k var T ϭ ᎏᎏ R2 Birth–Death Processes For a birth-death process, the steady-state probability (pj) or fraction of the time that the process spends in state j can be found from the following ﬂow balance equations: ( j ϭ 0) ( j ϭ 1) ( j ϭ 2) и и и ( jth equation) p0l0 ϭ p1m1 (l1 ϩ m1)p1 ϭ l0p0 ϩ m2p2 (l2 ϩ m2)p2 ϭ l1p1 ϩ m3p3 (lj ϩ mj)pj ϭ ljϪ1pjϪ1 ϩ mjϩ1pjϩ1 The jth ﬂow balance equation states that the expected number of transitions per unit time out of state j ϭ (expected number of transitions per unit time into state j). The solution to the balance equations is found from l0l1 и и и ljϪ1 pj ϭ p0 ᎏᎏ m1m2 и и и mj ( j ϭ 1, 2, . . .) and the fact that p0 ϩ p1 ϩ и и и ϭ 1. Notation for Characteristics of Queuing Systems pj L Lq Ls W Wq Ws l m ϭ ϭ ϭ ϭ ϭ ϭ ϭ ϭ ϭ steady-state probability that j customers are in system expected number of customers in system expected number of customers in line (queue) expected number of customers in service expected time a customer spends in system expected time a customer spends waiting in line expected time a customer spends in service average number of customers...

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...be measured over an extended period of time. • We assume arrival times and service times are random. • • • • Assumptions Independent arrivals Exponential distributions Customers do not leave or change queues. Large queues do not discourage customers. Many assumptions are not always true, but queuing theory gives good results anyway Queuing Model Q W λ Tw Tq S Interesting Values • Arrival rate (λ) — the average rate at which customers arrive. • Service time (s) — the average time required to service one customer. • Number waiting (W) — the average number of customers waiting. • Number in the system (Q) — the average total number of customers in the system. More Interesting Values • Time in the system (Tq) the average time each customer is in the system, both waiting and being serviced. Time waiting (Tw) the average time each customer waits in the queue. Tq = Tw + s Arrival Rate • The arrival rate, λ, is the average rate new customers arrive measured in arrivals per time period. Common units are access/second • The inter-arrival time, a, is the average time between customer arrivals. It is measured in time per customer. A common unit would be seconds/access. • a=1/λ Random Values • We assume that most of the events we are interested in occur randomly. – Time of a request to a device – Time to service a request – Time user makes a request Exponential...

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