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### Fall_2005_Exam_C_solutions

Course: MAT 2070, Spring 2011
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2005 EXAM FALL C SOLUTIONS Question #1 Key: D S (300) = 3 /10 (there are three observations greater than 300) H (300) = ln[ S (300)] = ln(0.3) = 1.204 . Question #2 Key: A E ( X | ) = Var ( X | ) = = v = E ( ) = ; a = Var ( ) = 2 ; k = v / a = 1/ n n Z= = n + 1/ n + 1 + 1 0.15 = (1) + = +1 +1 +1 2 1 4 + 0.20 = (2) + = 2 + 1 2 + 1 2 + 1 From the first equation, 0.15 + 0.15 = + and so = 0.15...

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2005 EXAM FALL C SOLUTIONS Question #1 Key: D S (300) = 3 /10 (there are three observations greater than 300) H (300) = ln[ S (300)] = ln(0.3) = 1.204 . Question #2 Key: A E ( X | ) = Var ( X | ) = = v = E ( ) = ; a = Var ( ) = 2 ; k = v / a = 1/ n n Z= = n + 1/ n + 1 + 1 0.15 = (1) + = +1 +1 +1 2 1 4 + 0.20 = (2) + = 2 + 1 2 + 1 2 + 1 From the first equation, 0.15 + 0.15 = + and so = 0.15 0.85 Then the second equation becomes 0.4 + 0.2 = 4 + 0.15 0.85 0.05 = 2.75 ; = 0.01818 Question #3 Key: E 0.75 = 1 1 ; 0.25 = 1 + (100 / ) 1 + (500 / ) (100 / ) = 1/ 3; (500 / ) = 3 Taking the ratio of these two equalities produces 5 = 9 . From the second equality, 9 = [(500 / )2 ] = 5 ; (500 / ) 2 = 5; = 223.61 Question #4 Key: B f ( x) = a + bx + cx 2 + dx 3 0 = f (0) = a 2 = f (2) = a + 2b + 4c + 8d 1 = f '(0) = b 24 = f ''(2) = 2c + 12d ; c = 12 6d Insert the values for a, b, and c into the second equation to obtain 2 = 2 + 4(12 6d ) + 8d ; 48 = 16d ; d = 3 Then c = 6 and f ( x) = x + 6 x 2 3x3 ; f (1) = 4 Question #5 Key: E Begin with y s r 350 2 10 500 2 8 1000 1200 1500 1 1 1 5 2 1 8 641 Then S1 (1250) = = 0.24 10 8 5 2 The likelihood function is 2 2 2 L( ) = 1e 350 / 1e500 / e500 / 1e 1000 / e 1000 / 1e 1200 / 1e1500 / 7 7900 / = e l ( ) = 7 ln 7900 ; l '( ) = S 2 (1250) = e 1250(7) / 7900 = 0.33 The absolute difference is 0.09. 7 + 7900 2 = 0; = 7900 / 7 Question #6 Key: E f ( x) = S '( x) = 4 x 4 ( 2 + x 2 )3 L( ) = f (2) f (4) S (4) = 4 4(2) 4 4(4) 4 128 12 =2 ( 2 + 22 )3 ( 2 + 42 )3 ( 2 + 42 ) 2 ( + 4)3 ( 2 + 16)5 l ( ) = ln128 + 12 ln 3ln( 2 + 4) 5ln( 2 + 16) 12 6 10 2 = 0;12( 4 + 20 2 + 64) 6( 4 + 16 2 ) 10( 4 + 4 2 ) = 0 l '( ) = 2 + 4 + 16 0 = 4 4 + 104 2 + 768 = 4 26 2 192 26 262 + 4(192) = = 32; = 5.657 2 2 Question #7 Key: A 2x 0 2 E( X | ) = x dx = 2 2x 4 2 2 4 2 2 ; Var ( X | ) = x 2 2 dx = = 0 3 9 2 9 18 1 = (2 / 3) E ( ) = (2 / 3) 4 4 d = 8 /15 0 1 EVPV = v = (1/18) E ( 2 ) = (1/18) 4 5 d = 1/ 27 0 VHM = a = (2 / 3) Var ( ) = (4 / 9) 4 / 6 (4 / 5) 2 = 8 / 675 2 1/ 27 1 = 25 / 8; Z = = 8 / 33 8 / 675 1 + 25 / 8 Estimate is (8 / 33)(0.1) + (25 / 33)(8 /15) = 0.428. k= Question #8 Key: D From the Poisson(4) distribution the probabilities at 0, 1, and 2 are 0.0183, 0.0733, and 0.1463. The cumulative probabilities are 0.0183, 0.0916, and 0.2381. Because 0.0916 < 0.13 < 0.2381 the simulated number of claims is 2. Claim amounts are simulated from solving u = 1 e x /1000 for x = 1000 ln(1 u ) . The two simulated amounts are 51.29 and 2995.73 for a total of 3047.02 Question #9 Key: B It may be easiest to show this by graphing the density functions. For the first function the three components are each constant. One is of height 1/20 from 0 to 2 (representing the empirical probability of 1/10 at 1, one is height 1/20 from 1 to 3 and one is height 8/20 from 2 to 4. The following figure shows each of them and their sum, the kernel density estimator. 0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 f1 f2 f3 f 0 1 2 3 4 The triangular one is similar. For the triangle from 0 to 2, the area must be 1/10. With a base of 2, the height is 1/10. the same holds for the second triangle. The third has height 8/10. When added they look as follows; 0.9 0.8 0.7 0.6 f1 0.5 f2 0.4 f3 0.3 f 0.2 0.1 0 0 1 2 3 4 The question asks about cumulative probabilities. From 0 to 1 the first is linear and the second is quadratic, but by x = 1 both have accumulated 0.05 of probability. Because the cumulative distribution functions are the same at 1 and the density functions are identical from 1 to 2, the distribution functions must be identical from 1 to 2. Question #10 Key: D and E For the Poisson distribution, the mean, , is estimated as 230/1000 = 0.23. # of Days Poisson Probability 0.794533 0.182743 0.021015 0.001709 0 1 2 3 or more Total Expected # of Workers 794.53 182.74 21.02 1.71 Observed # of Workers 818 153 25 4 1000 2 0.69 4.84 0.75 3.07 9.35 The 2 distribution has 2 degrees of freedom because there are four categories and the Poisson parameter is estimated (d.f. = 4 1 1 = 2). The critical values for a chi-square test with two degrees of freedom are shown in the following table. Significance Level 10% 5% 2.5% 1% Critical Value 4.61 5.99 7.38 9.21 9.35 is greater than 9.21 so the null hypothesis is rejected at the 1% significance level. Question #11 Key: D 25(480 472.73) 2 + 30(466.67 472.73) 2 = 2423.03 where 480 = 12,000/25, 2 1 466.67 = 14,000/30, and 472.73 = 26,000/55. 55 k = 2423.03 / 254 = 9.54; Z = = 0.852 55 + 9.54 EVPV = v = Question #12 Key: C 1 2 0 < x <1 1.6 x, f ''( x) = S = (1.6 x) 2 dx + (2.4 0.8 x) 2 dx = 2.56 0 1 1.6 0.8( x 1) = 2.4 0.8 x, 1 < x < 3 Question #13 Key: C Relative risk = e 1 2 which has partial derivatives e0.2 at 1 = 0.05 and 2 = 0.15 Using the delta method, the variance of the relative risk is 0.2 7e 0.4 1 0.2 0.2 2 1 e e e = 0.000469 = 10, 000 1 3 e 0.2 10, 000 Std dev = 0.0217 upper limit = e 0.2 + 1.96 ( 0.0217 ) ( ) = 0.8613 Alternatively, consider the quantity 1 + 2 . The variance is 2 1 1 1 7 = 0.0007 . The lower limit for this quantity is (1 1) = 10, 000 1 3 1 10, 000 0.2 1.96 0.0007 = 0.1481 and the upper limit for the relative risk is e 0.1481 = 0.8623 . Question #14 Key: C ln 5000 The quantity of interest is P = Pr( X 5000) = . The point estimate is 6.84 ln 5000 = (1.125) = 0.87 . 1.49 For the delta method: P (1.125) P 1.125 (1.125) 1 z2 / 2 = = 0.1422; = = 0.1600 where ( z ) = e . 1.49 1.49 2 Then the variance of P is estimated as (0.1422) 2 0.0444 + (0.16) 2 0.0222 = 0.001466 and the lower limit is PL = 0.87 1.96 0.001466 = 0.79496 . Question #15 Key: A Pr( = 0.1| X 1 = 1) = Pr( X 1 = 1| = 0.1) Pr( = 0.1) Pr( X 1 = 1| = 0.1) Pr( = 0.1) + Pr( X 1 = 1| = 0.3) Pr( = 0.3) 0.1(0.8) 4 = 0.1(0.8) + 0.3(0.2) 7 Then, E ( X 2 | = 0.1) = 0(0.2) + 1(0.1) + 2(0.7) 1.5 = E = ( X 2 | = 0.3) = 0(0.6) + 1(0.3) + 2(0.1) = 0.5 4 3 E ( X 2 | X 1 = 1) = (1.5) + (0.5) = 1.071 7 7 Question #16 Key: D The requirement is that F (1500) S (1500) 0.01F (1500) 1.96 N 2 P P( N P) 0.0001 2 3.8416 N N3 NP 38, 416. N P For the five answer choices, the left hand side is 34,364, 15,000, 27,125, 39,243, and 37,688. Only answer D meets the condition. Question #17 Key: D ^ ^ s4 = H ( y 4 ) H ( y 3 ) = 0.5691 0.4128 = 0.1563 . r4 s4 ^^ ^^ = V [ H ( y 4 )] V [ H ( y 3 )] = 0.014448 0.009565 = 0.004883 . 2 r4 ( s4 / r4 ) 2 0.15632 = = 5. Therefore, s4 = s4 / r42 0.004833 Question #18 Key: A ln f ( x) = ln 2 ln( + x) ln f ( x) 1 2 = +x 2 ln f ( x) 1 2 = 2 + 2 ( + x) 2 2 ln f ( x) 1 2 1 2 1 2 1 E dx = 2 + = 2 + 2 = 2 = 2 + 0 2 4 3 3( + x) 0 3 ( + x) 3 I ( ) = n 3 2 ; Var = 3 2 n Question #19 Key: B = E[ E ( X | )] = E ( ) = 1(0.9) + 10(0.09) + 20(0.01) = 2 EVPV = v = E[Var ( X | )] = E ( ) = 2 VHM = a = Var[ E ( X | )] = Var ( ) = 1(0.9) + 100(0.09) + 400(0.01) 22 = 9.9 Z= 1 = 0.83193; 11.983 = 0.83193 x + 0.16807(2); x = 14 1 + 2 / 9.9 Question #20 Key: A The given interval for H can be written as 0.775 1.96 0.063 and therefore the estimated variance of H is 0.063. To apply the delta method, dS = e H ; Var ( S ) (e H ) 2 Var ( H ) = (e0.775 ) 2 (0.063) = 0.134 . S = e H ; dH The point estimate of S is e 0.775 = 0.4607 and the confidence interval is 0.4607 1.96 0.0134 = 0.2269 or (0.23, 0.69). Question #21 Key: B The first step is to trend the year 1 data by 1.21 and the year 2 data by 1.1. The observations are now 24.2, 48.4, 60.5, 33, 44, 99, and 132. The first two sample moments are 63.014 and 5262.64. The equations to solve are 2 63.014 = e + 0.5 ; 4.14336 = + 0.5 2 5262.64 = e 2 + 2 ; 8.56839 = 2 + 2 2 . Taking four times the first equation and subtracting the second gives 2 and therefore 4(4.14336) 8.56839 = = 4.00 . 2 2 Question #22 Key: A = x = 12 / 60 = 0.2, EVPV = v = x = 0.2 10(0.4 0.2) 2 + 20(0.25 0.2) 2 + 30(0.1 0.2) 2 (3 1)(0.2) = 0.009545 102 + 202 + 302 60 60 10 = 0.323 k = 20.9524; Z = 10 + 20.9524 VHM = a = Question #23 Key: E By elimination, (A) is incorrect because f ''(3) = 1.833 0 , (B) is incorrect because f ''(0) = 2 0 , (C) is incorrect because f ''(0) = 1 0 , and (D) is incorrect because f ''(0) = 1 0 . Therefore (E) must be correct. Also, this function does meet all the requirements: f ''(0) = 0; f 0 (1) = f1 (1) = 2; f 0' (1) = f1' (1) = 0; f 0'' (1) = f1'' (1) = 3; f (3) = 6; f ''(3) = 0 Question #24 Key: B and C For males, c j = 1 and for females, c j = e0.27 = 1.31 . Then, H (20) = 1.31 1 1 1 + + = 0.6965 and S female (20) = ( e 0.6965 ) = 0.402 . 3 + 2(1.31) 2 + 2(1.31) 2 + 1.31 . Question #25 Key: C 5 5 j =1 j =1 l ( , ) = ln f ( x j ) = ln + ( 1) ln x j ln ( x j / ) . Under the null hypothesis it is 5 l (2, ) = ln 2 + ln x j 2 ln ( x j / ) 2 . Inserting the maximizing value of 816.7 for gives j =1 35.28 . The likelihood ratio test statistic is 2( 33.05 + 35.28) = 4.46 . There is one degree of freedom. At a 5% significance level the critical value is 3.84 and at a 2.5% significance level it is 5.02. Question #26 Key: C It is given that n = 4, v = 8, and Z = 0.4. Then, 0.4 = 4 8 4+ a which solves for a = 4/3. For the covariance, Cov( X i , X j ) = E ( X i X j ) E ( X i ) E ( X j ) = E[ E ( X i X j | )] E[ E ( X i | )]E[ E ( X j | )] = E[ ( ) 2 ] E[ ( )]2 = Var[ ( )] = a = 4 / 3. Question #27 Key: A u 0.6217 0.9941 0.8686 0.0485 z 0.31 2.52 1.12 1.66 x 5.8325 7.49 6.44 4.355 lognormal 341.21 1790.05 626.41 77.87 Average with deductible 241.21 1690.05 526.41 0 614.42 The value of z is obtained by inversion from the standard normal table. That is, u = Pr( Z z ) . The value of x is obtained from x = 0.75 z + 5.6 . The lognormal value is obtained by exponentiating x and the final column applies the deductible. Question #28 Key: B MSE = E[( X 2 2 ) 2 ] = E ( X 4 2 X 2 2 + 4 ) = 24 4 2(2 2 ) 2 + 4 = 21 4 Question #29 Key: C 157(0) + 66(1) + 19(2) + 4(3) + 2(4) = 0.5 is the maximum likelihood 248 estimate of the geometric parameter as well as the method of moments estimate of the Poisson parameter . Then, P = (1 + 0.5) 1 = 0.6667 and Q = e 0.5 = 0.6065 . The absolute difference is 0.0602. The sample mean of Question #30 Key: D 5000(0) + 2100(1) + 750(2) + 100(3) + 50(4) = 0.5125 and 8000 5000(0.5125) 2 + 2100(0.4875) 2 + 750(1.4875) 2 + 100(2.4875) 2 + 50(3.4875)2 s2 = = 0.5874 . 7999 Then, = v = x = 0.5125 and a = s 2 x = 0.0749 . The credibility factor is 1 Z= = 0.1275 and the estimate is 0.1275(1) + 0.8725(0.5125) = 0.5747. 1 + 0.5125 / 0.0749 x= Question #31 Key: B s = Fn (3000) = 4 / 8 = 0.5 because for the p-p plot the denominator is n+1. t = F (3000) = 1 e 3000 / 3300 = 0.59711 . For the difference plot, D uses a denominator of n and so D = 4 / 7 0.59711 = 0.02568 and the answer is 0.5 0.59711 + 0.02568 = 0.071. Question #32 Key: B (q | 2, 2) f (2 | q ) f (2 | q) (q) = q(q)(q 2 / 0.039) q 4 . Because 0.5 0.2 q 4 dq = 0.006186 , (q | 2, 2) = q 4 / 0.006186 . Given q, the expected number of claims is E ( N | q) = 0(0.1) + 1(0.9 q ) + 2q = 0.9 + q . The Bayesian estimate is 0.5 E ( N | 2, 2) = (0.9 + q) 0.2 q4 dq = 1.319. 0.006186 Question #33 Key: E 200 + 110 310 + x 0.689 = F500 (1500) = 0.5 F500 (1000) + 0.5 F500 (2000) = 0.5 + x = 69 500 500 310 + 69 379 + y 0.839 = F500 (3500) = 0.5F500 (2000) + 0.5 F500 (5000) = 0.5 + y = 81 500 500 Question #34 Key: A A is false because the test works best when the expected number of observations is about the same from interval to interval. B is true (Loss Models, 427-8), C is true (Loss Models, 428), and D is true (Loss Models, 430). Question #35 Key: E 2 2 n 0 1 + Y ; Y = = 10, 000 ; Y = 2 = 108 Y 1.96 n 0.1 2 108 1 1 + 108 2 = 384.16(1 + ) Because is needed, but not given, the answer cannot be determined from the information given.
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00 introductionmy expectationsyour expectationsgradingon-line quizzeswhat I expect from you: active learningbefore classread chapter before we beginwarm-up quiz (on-line) on reading*during classthink/pair/share quizzes*be prepared with question
Kettering - MECH - 115
01 Motiondiagramsuse movie to studymotionfixed time betweenframesdon't pan the camera!usually 1/30th of a secstrobe pictures arefasterfaster objectshave greater positionchange between framesWhich car is going faster, A or B? Assume there are
Kettering - MECH - 115
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03 Motion in1 dimensioncoordinate systemoriginpositive direction: conventionx: right y: upposition-time graphsposition on vertical axis(dependent variable)even if position ishorizontaltime on horizontal axis(independent variable)can pick off
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05 problem solving: constant amodel: particle modelvisualize: shift between these as neededmotion diagrampictorial representation: labelv fs = v is a s tsolve12s f = si v is t a s t 2assess22v fs = v is 2 a s s units? sensible? believable?
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06 Instantaneousaccelerationa s=d vsdt v s = a s dtfollow up:workbook 2.28page 2-16Rank in order, from largest to smallest,the accelerations aA aC at points A C.A) aA &gt; aB &gt; aCB) aC &gt; aA &gt; aBC) aC &gt; aB &gt; aAD) aB &gt; aA &gt; aCWhich velocity-vers
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10 Applying Newton's LawsThe basic equations areNewton's 1st law0F = F i =netior Newton's 2nd lawF = F i = m anetiEquation hunting won't solve these problemsNeeded: Problem solving strategyapply the same patternvary the approach to match th
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11 Mass and weightmass:a F i=m iweight: force of gravityw=m gsame mass m in both formulasdirection: downwardscoincidence? General Relativity?apparent weight: normal force of scale on feettry bouncing on a scale!Inclined plane#1#2#3#1: (A)
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12 frictioncontact forces at an interface between surfacesnormal force acts perpendicular to interfacefriction force acts parallel to interfacefriction equationsMODEL observed behaviorfriction caused bysurface bondingtypes of frictionstatic: no m
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13motion in 2drva , , are vectors, with componentstrajectory (path) graph: y(x)rv avg =trecall: vector subtraction infollow up: workbook 6.2 (page 6-1)instantaneousvelocity &amp;accelerationdrv=dtdva=dtv is tangent to trajectorycompo
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14 projectile motionno horizontal acceleration: constant horizontal velocitydemo: knock 2 balls off a table (different v)mga==g=mmacceleration is vertically downwardsF netlisten for the impacts. does one hit first?follow ups:workbook probl
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15 UniformCircular Motionangular position qpositive =counterclockwise (ccw)from +x axisalways measured in radians2p rad = 1 rev = 360arc length s = r which looks bigger?a dime at arm's length s 2 cm, r 0.8 mthe moon s = 3.5 x 106 m, r = 3.8 x 1
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n16 Circular dynamicsTdemo: object moves on string in horizontal circle wnet force is towards center, constantthis produces constant acceleration towards centerthis results in uniform circular motiontension can also create linear acceleration!circ
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17 Non-uniform circular motiondemo: loop-the-loopat the bottom: FBDnormal force is upweight is downnet force is towards centernwat the top: FBDnormal force is downnet force is towards centernweight is downwvcritical issues2mat the bottom