Course 1 Solutions November 2001 Exams

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1 Course Solutions November Exams

2 . A For i =,, let R = event that a red ball is drawn form urn i i B = event that a blue ball is drawn from urn i. i Then if x is the number of blue balls in urn, ( R R) ( B B) R R [ B B] [ R ] [ R ] + [ B ] [ B ].44 = Pr[ ] = Pr[ ] + Pr = Pr Pr Pr Pr x = + x+ 6 x+ 6 Therefore, 3 3x 3x+ 3. = + = x+ 6 x+ 6 x+ 6.x+ 35. = 3x+ 3.8x = 3. x = 4. C We are given that f x, y da= 6 and da= R R It follows that 4 (, ) 4 (, ) R f xy da= f xyda da R R = 4 6 = Course Solutions November

3 3. A Since S = 75L A, we have S 4 = L A > for L L >, A > S 4 5 = 3.5 L A > for L L>, A> S 3 = 4 L A 5 > for A L >, A > S = 8 L A < for A L>, A> It follows that S increases at an increasing rate as L increases, while S increases at a decreasing rate as A increases. 4. B Apply Baye s Formula: Pr Seri. Surv. = Pr Surv. Seri. Pr[ Seri. ] Pr Surv. Crit. Pr[ Crit. ] + Pr Surv. Seri. Pr[ Seri. ] + Pr Surv. Stab. Pr[ Stab. ] (.9)(.3).9 (.6)(.) + (.9)(.3) + (.99)(.6) = = Course Solutions November

4 5. A Let us first determine K. Observe that = K = K = K K = 37 It then follows that Pr N = n = Pr N = n Insured Suffers a Loss Pr Insured Suffers a Loss 6 3 = (.5 ) =, N =,...,5 37N 37N P = E X where [ ] [ ] Now because of the deductible of, the net annual premium [ ], if N X = N, if N > Then, P E [ X 5 ] ( N = = ) = () 3.34 N = = 37N 37 37( 4) 37( 5) 6. E The line 5y 4x= 3 has slope dy dy dx t = = = 5 dx dt dt t + 8t+ 4= t t = 4 t =. It follows that we need to find t such that Course Solutions 3 November

5 7. A Cov C, C = Cov X + Y, X +.Y ( ) = Cov ( X, X) + Cov ( Y, X) + Cov ( X,.Y) + Cov( Y,.Y) = Var X + Cov ( X, Y) +.Cov ( X, Y) +.VarY = Var X +. Cov ( X, Y) +.VarY ( ) ( E Y ) Var X = E X E X = =.4 VarY = E Y = =.4 ( X + Y) = X + Y + ( X Y) Var Var Var Cov, Cov ( XY, ) = Var ( X+ Y) Var X VarY = ( ) =.6 Cov, = = 8.8 ( C C ) Course Solutions 4 November

6 7. Alternate solution: We are given the following information: C = X + Y C = X +.Y E X = 5 [ ] = E X EY = 7 [ ] = 7.4 E Y 5.4 Var[ X + Y] = 8 Now we want to calculate Cov C, C = Cov X + Y, X +.Y ( ) = E ( X + Y)( X +.Y) E[ X + Y] ie[ X +.Y] = + + [ ] + [ ] + = E X +.E[ XY] +.E Y ( 5 + 7) 5 + (.) 7 = E[ XY] +.( 5.4) ( 3.4) =.E[ XY] 7.7 Therefore, we need to calculate E [ XY ] first. To this end, observe 8= Var[ X + Y] = E ( X + Y) ( E[ X + Y] ) [ ] E XY ( [ ] [ ]) E X.XY.Y E X E Y E X.E Y = E X + XY + Y ( E[ X] + E[ Y] ) = E X + E[ XY] + E Y ( 5 + 7) = E[ XY] = E[ XY] 65. ( 8 65.) 36.6 = + = Finally, Cov CC = = 8.8 (, ) Course Solutions 5 November

7 8. D The function F() t, t 7, achieves its minimum value at one of the endpoints of the interval t 7 or at t such that t t t = F () t = e te = e ( t) Since F () t < for t >, we see that F () t is a decreasing function on the interval < t 7. We conclude that F 7 = 7e =.64 7 is the minimum value of F. 9. D Let C = event that patient visits a chiropractor T = event that patient visits a physical therapist We are given that Pr[ C] = Pr[ T] +.4 Pr C T =. c c ( C T ) Pr =. Therefore, c c.88 = Pr C T = Pr C T = Pr C + Pr T Pr C T = Pr[ T] Pr[ T]. = Pr T.8 or [ ] Pr[ T ] = ( ) =.48 [ ] [ ] [ ] [ ] Course Solutions 6 November

8 . E Note that the terms of a n= n + n exceed the terms of the harmonic series for a = n, a = n, or an n = n and will thus fail to converge. Moreover, the terms of the series n ( ) + = n= n= + = n= n n n n n are seen to be identical to the terms of the divergent harmonic series. By contrast, the series n a n= n + n= n= n = + = n n n is seen to converge by the integral test, since x dx = = < x. D If a month with one or more accidents is regarded as success and k = the number of failures before the fourth success, then k follows a negative binomial distribution and the requested probability is 3 4 k 3+ k 3 Pr[ k 4] = Pr[ k 3] = ( k ) k = = ( 3) = =.898 Alternatively the solution is ( 3) = which can be derived directly or by regarding the problem as a negative binomial distribution with i) success taken as a month with no accidents ii) k = the number of failures before the fourth success, and Pr k 3 iii) calculating [ ] Course Solutions 7 November

9 . E Observe that x f x = f t dt For [ ] = (area bounded by f above the x-axis from to x) (area bounded by f below the x-axis from to x) x,5, f is drawn such that the area bounded by f above the x-axis from to x is strictly greater than the area bounded by f below the x-axis from to x. Therefore, f ( x) > for < x 5. We conclude that f is an increasing function on the interval [,5 ], and its maximum value occurs at x = E F y Y y X y X e [ ].8 8 Y = Pr = Pr = Pr Y = Therefore, 4 Y f y = F y = e 8 ( Y ) A L Hôspital s Rule may be applied since f and g are given differentiable and from the diagram lim f( x) = lim g( x) =. x x f ( x) f ( x) Therefore lim = lim x g ( x) x g ( x) Also from the diagram f ( x) f This leads to lim = > x and g ( x) x g f ( x) f ( x) f g( x) x g ( x) g lim = < lim = lim = < x Course Solutions 8 November

10 5. E Let X,..., X denote the number of pensions that will be provided to each new recruit. Now under the assumptions given, with probability.4 =.6 X i = with probability (.4)(.5) =. with probability (.4)(.75) =.3 for i =,...,. Therefore, E X = =.7, [ i ] i [ Xi] E X i { E[ Xi] } E X = =.3, and Var = =.3.7 =.8 Since X,..., X are assumed by the consulting actuary to be independent, the Central Limit Theorem then implies that S = X X is approximately normally distributed with mean E S = E X E X =.7 = 7 [ ] [ ] [ ] and variance Var S = Var X Var X =.8 = 8 [ ] [ ] [ ] Consequently, S Pr[ S 9.5] = Pr 9 9 = Pr[ Z.8] = C Observe Pr 4 < S < 8 = Pr 4 < S < 8 N = Pr N = + Pr 4 < S < 8 N > Pr N > = e e + e e * 3 6 =. *Uses that if X has an exponential distribution with mean µ [ ] [ ] [ ] b t µ t µ µ Pr ( a X b) = Pr ( X a) Pr ( X b) = e dt e dt = e e µ µ µ a b a Course Solutions 9 November

11 7. B Note that [ > ] = ( ) Pr X x.5 t dt x =.5 t t x =.5 4 x + x =.5 x+ x where < x <. Therefore, [ X > ] [ X ] Pr Pr X > 6 X > 8 = = = = Pr > D Note that x for x< and x> f ( x) = for < x < x Furthermore, lim lim = x f ( x) x + f x =+ and lim lim = f ( x ) f ( x + ) = x x so f is not differentiable at x = or x =, although f is differentiable everywhere else. By contrast, f is continuous everywhere since lim f ( x lim ) f ( x + = ) = x x and lim f x lim = f x = + x x Course Solutions November

12 9. B The amount of money the insurance company will have to pay is defined by the random variable x if x< Y = if x where x is a Poisson random variable with mean.6. The probability function for X is.6 k e (.6) p( x) = k =,,,3 $ and k! k EY [ ] = + (.6) e + e k = k! k = (.6) e + e e (.6) e k = k! k.6 (.6) = e e (.6) e = e 6e k = k! = 573 k E Y = (.6) e + e k = k! k = e e (.6) e k = k!.6 = e.6 (.6) e = 86,893 Var [ Y] = E Y { E[ Y] } = 86,893 ( 573) = 488, 564 [ Y ] Var = 699. B The graph of E tells us that employment increases from April through August (because E t E t then). () > then) and does not increase from August through April (since () It follows that employment is a minimum in April. Course Solutions November

13 . D Let N and N denote the number of claims during weeks one and two, respectively. Then since N and N are independent, 7 [ N+ N = ] = [ N = n] [ N = n] n= Pr 7 Pr Pr 7 7 = n= n+ 8 n 7 = n= 9 8 = = = D The amount of the drug present peaks at the instants an injection is administered. If A(n) is the amount of the drug remaining in the patient s body at the time of the nth injection, then A = 5 () 6 = 5( + e ) A % n k 6 An = 5 e k = and the least upper bound is k 6 5 lim An = 5 e = = 68 (noting the n k = 6 e 6 series is geometric with ratio e ) Course Solutions November

14 3. C Observe that sin ( θ + π) + 3 cos ( θ + π) = sinθ + 3 cosθ r, θ and r, θ + π define the same point, and we may restrict attention to Therefore, π θ such that θ < π. Now for θ, r > and the graph of r is in the first π π quadrant (i.e., to the right of θ = ). On the other hand, for < θ < π, r > when sinθ + 3 cosθ > sinθ > 3 cosθ sinθ > 3 cosθ tanθ > 3 π This is true in the interval, π when π π < θ <. Consequently, the area bounded 3 π π 3 by the graph of r to the left of θ = is given by rd θ. π [It is not needed to do the problem, but the graph of the curve is a circle of radius 3 centered at x=, y = ]. Course Solutions 3 November

15 4. D 4 x y must be maximized subject to x+ y = 5, Since y = 5, x, this reduces to maximizing 4 S x = x 5, x, x 5, S ( x) x ( x) x ( x) ( 5, x) x= = 5 5, 5 5, = 3x = 5, x = 5, (This value of x is a maximum since S ( x) 5, < x < 5, ). > for x 5, 4 Maximum sales are then Alternate solution using Lagrange Multipliers Solve x+ y 5, = 4 x y = λ x+ y 5, x x 4 x y = λ x+ y 5, y y From the last two equations 5x 5x Eliminating λ y y < for < < and S ( x) 5, 5, 5, = 47,87 = λ = λ x y = 5x y 5y = 5x y = x Using the first equation x+ x 5, = x = 5, y =, 4 The extreme value (which must be a maximum) is 5,, = 47,87 Course Solutions 4 November

16 5. E 3 f Pr < Y < 3 X = = dy f ( ) (, y) f (, y) = y = y fx = y dy = y = 4 4 x 3 3 y dy Finally, Pr < Y < 3 X = = = y = = 4 6. C The increase in sales from year two to year four that is attributable to the advertising campaign is given by t + t + dt = t t dt 3 t t 4 = t = = = 3 3 Course Solutions 5 November

17 7. D Let X denote the number of employees that achieve the high performance level. Then X follows a binomial distribution with parameters n= and p =.. Now we want to determine x such that Pr X > x. [ ] or, equivalently,.99 Pr[ ] x (.) k (.98) X x = k k = k The following table summarizes the selection process for x: x Pr X = x Pr X x [ ] [ ] = 9 = 8 = Consequently, there is less than a % chance that more than two employees will achieve the high performance level. We conclude that we should choose the payment amount C such that C =, or C = 6, Course Solutions 6 November

18 8. B Let X and Y denote the two bids. Then the graph below illustrates the region over which X and Y differ by less than : Based on the graph and the uniform distribution: Shaded Region Area Pr X Y < = = ( ) 8 8 = = (.9) =.9 More formally (still using symmetry) Pr X Y < = Pr X Y = Pr[ X Y ] x x = dydx y dx = = x dx= x 8 = =.9 Course Solutions 7 November

19 9. B Let X and Y denote repair cost and insurance payment, respectively, in the event the auto is damaged. Then if x 5 Y = x 5 if x > 5 and 5 [ ] 5 5 EY = x 5 dx= x 5 5 = = E Y = ( x 5) dx= ( x 5) 5 = = 434, Var[ Y] = E Y { E[ Y] } = 434, 8 ( 5) Var Y = 43 [ ] Course Solutions 8 November

20 3. C Let (, ) below: and f t t denote the joint density function of T T. The domain of f is pictured Now the area of this domain is given by A = = 36 = 34 Consequently,, < t < 6, < t < 6, t+ t < f ( t, t) = 34 elsewhere and [ + ] = [ ] + [ ] = [ ] ET T ET ET ET (due to symmetry) t = t dt dt + 34 t 4 dt dt 34 4 t 6 6 t t = t 34 dt + t 4 34 dt 43t 6 = dt + ( t t ) dt t = + 5t t = = 5.7 Course Solutions 9 November

21 3. E The general solution of the differential equation may be determined as follows: dy = ( k k) dt y ln y = k k t+ C (C is a constant) ( ) c ( k k) t () = e e y t When t =, c y = e, so () ( k ) k t y t = y e Now we are given that if k =, y( 8) = y. Therefore, 8k y = y( 8) = y e 8k = e 8k = ln k = ln ( ) 8 4( k k) [Note the problem also gives y( 4) = y = y e, but that information is not needed to determine k ]. 3. B Observe [ N ] [ N ] Pr 4 Pr N N 4 = = Pr = = = Course Solutions November

22 33. 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 x f ( x) = (.6) (.4 ), x=,,3... and ( 5 x) if x=,,3, 4 It follows that y = if x > 4 3 [ ] 4(.4) 3(.6)(.4) (.6) (.4) (.6) (.4) EY = = D The diagram below illustrates the domain of the joint density f xy, of Xand Y. We are told that the marginal density function of X is f ( x) =,< x< while f ( yx) =, x< y< x+ yx It follows that f ( x, y) f ( x) f ( y x) Therefore, if < x <, x< y< x+ = x yx = otherwise [ ] [ ] x Pr Y >.5 = Pr Y.5 = dydx = y x dx= x dx 7 = x x = + = [Note since the density is constant over the shaded parallelogram in the figure the solution is also obtained as the ratio of the area of the portion of the parallelogram above y =.5 to the entire shaded area.] x Course Solutions November

23 35. E If X is the random variable representing claim amounts, the probability that X exceeds the deductible is [ ] x x x Pr X > = xe dx= xe + e dx (integration by parts) x = e e = e + e = e =.736 It follows that the company expects (.736) = 74 claims. 36. C Let Then x = price in excess of 6 that the company charges, p(x) = price per policy that the company charges, n(x) = number of policies the company sells per month, and R(x) = revenue per month the company collects = 6 + x = 8 x = = ( 8 )( 6 + ) = ( 6 + ) + ( 8 ) = ( x) = < p x n x R x n x p x x x R x x x x R It follows that R(x) is a maximum when R ( x) x = x = R = = 49 and when R(x) is a maximum. =. We conclude that Course Solutions November

24 37. A The joint density of T and T is given by t t f t, t = e e, t >, t > Therefore, Pr X x = Pr T + T x [ ] [ ] = = x ( x t ) x ( x t ) t t t t e e dtdt e e dt = = x x+ t x x t t t e e dt e e e dt = e + e e = e + e e + e x t x x x t x x x x x x x = e + e e = e + e, x> It follows that the density of X is given by d x x g ( x) = e + e dx x x = e e, x> 38. E X, X, and X denote annual loss due to storm, fire, and theft, respectively. In Let 3 addition, let Y Max( X, X, X ) Then =. 3 [ Y > ] = [ Y ] = [ X ] [ X ] [ X ] Pr 3 Pr 3 Pr 3 Pr 3 Pr 3 ( e )( 3 e )( 3 e ) ( e 3 )( e )( 5 e ) = = 3 =.44 * Uses that if X has an exponential distribution with mean µ Pr ( X x) = Pr ( X x) = e dt = ( e ) x = e µ x * t µ t µ x µ Course Solutions 3 November

25 39. D We are given that P x = x + 5x 5 before the translation occurs. The revised profit function P ( x) follows: P* x = P x 3 ( x ) ( x ) = = x + x + x = x + x * may be determined as 4. B Observe that E X + Y = E X + E Y = 5 + = 7 [ ] [ ] [ ] [ ] [ ] [ ] [ ] Var X + Y = Var X + Var Y + Cov X, Y = = for a randomly selected person. It then follows from the Central Limit Theorem that T is approximately normal with mean ET = 7 = 7 [ ] and variance Var T = = [ ] Therefore, T Pr[ T < 7] = Pr < = Pr[ Z < ] =.843 where Z is a standard normal random variable. Course Solutions 4 November

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