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Mid2B_EC41_key_F09

Mid2B_EC41_key_F09 - EC 41 UCLA Winter 2009 Name(print#61 2...

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Unformatted text preview: EC 41, UCLA Winter 2009 Name (print) #61 2 B . Midterm #2 - Version B — 11/9/09 RE Ch 4, 5, 6.1, 6.2 TA: Section Time - The normal table and useful formulas are on the last page of this exam. — Only pens, pencils and erasers may be used, this is a closed book, closed note, exam. - Students may use a calculator, but nothing that can access the internet. - Write non—integer answers to 3 significant digits, e.g., 303, or 3.03 or .0303, or .00303 - This exam consists of 10 True/False (20 points), 10 short answer (40 points) and 4 longer questions (40 points) - Clearly write answers on this exam. No points are awarded for illegible answers. - Be prepared to show a photo ID during the exam (e.g., UCLA ID) — You may leave when finished. Do not disrupt those still taking the exam. 1. Circle T for True or F for False (2 points each) 1) T oré‘) If the random variable Z = X + Y, then the mean of Z is the weighted average: 112 = (1/2)nx + (1/2)ny 2)@0r F A confidence interval gets narrower (smaller) as the confidence level C is reduced. 3) T [email protected] Flip a coin twice. Two events, A = head on lst toss and B = head on 2nd toss; are disjoint events. 4) T o A hi her —value will rovide better evidence that the null h othesis is false. g P P YP 5 T r F Probability involves making statements about sample statistics, given that population parameters are known, and is an example of deductive reasoning. 6) T or® If two investments had perfect negative correlation so correlation between their returns = —1, then any portfolio comprised of these two assets will have no risk (standard deviation and variance = 0). 7©or F Rejecting H0 when 0L = .01 provides stronger evidence against H0 than rejecting H0 when (X = .05 8) T 0r® It is possible for a particular positive sample mean, J—C to reject a null hypothesis H0: 110 = 0 in favor of a two—sided alternative HA: “A 75 0; but not reject this H0 in favor of the one sided alternative HA: uA>0 9®0r F If two assets have identical mean returns and identical standard deviations, it is still possible to benefit by forming a portfolio with both assets, rather then just investing in one of the assets. 10) T [email protected] When calculating 95% confidence intervals with a given known population standard deviation and different samples of the same size, the size (or width) intervals will differ. 11. Briefly, clearly, and correctly answer the following 10 questions (4 pts. each) 1) Consider two disjoint events, A and B, where P(A) = .2 and P(B) = .2 a) What is the probability that A occurs given that B occurred = P(AIB)? Q b) Are A and B independent events (Yes or c) What is the probability of both A and B occur? ‘0\ d) What is the probability that neither A nor B occur? I _, Z «1 l :@ 32 21; Consider two events, A = a perso as disease X; and B = person tests positive for disease X. (A) 2.002 (= 2 in 1,000); P(BIA) Q P(BIAC ) =.10 Find the following: a) Probability of the complement of A: P(AC ) = t thg b) Probability of A given B: P(AlB): P63 14} PW , l Cl‘l (90 Ll _ 31,4 pet); (MW—9 ,qcrééml + JDJC W93 T mm c - ’ W Jalig ‘ ~-7 @ m“ «w? ”my WW” , c) Probability of A and B: P(A n B. d) Probability of B, P(B) = W9 Will “(L‘ii‘éth/HDIL JD; Mflw‘flf‘v‘ L1 CHM”? [Mfr/mm Warm; 3) The table below shows number of auto insurance customers in each category. Realize total number of {WM Ml r customers is 1000 and drivers may file a claim after an accident that requires repairs. by 3 10401-4 Filed Claim Did Not File Claim Female. .100 600 ’7 V 1‘ Male: 100 200 3w . N a) The conditional probability P(Filed Claim I Female) we“ 6 b) The conditional probability P(Filed Claim I Male) c) Are gender and likelihood of filing a claim independent for these customers? @ d) The conditional probability P(Female I Filed Claim) = i DE) 1.; I 202? H 4) X and Y represent the return (in %) on two individual investments. LLX = l o x = 3 W = 1 0y = 3 Z = a portfolio, half invested in X and Y each, so Z = .SX + .SY . a) What 1s the variance of the return on the portfolio 1 f/true correlation between the returns, ny— — l? 01 ("a-:12 2,1522? +26 Nils?) L717} +3: its H€@ b) What is the standard ' tion of the return on the portfolio if true correlation between the returns, pxy = l? 0' = r’ ~ c) What is the variance of the return on the portfolio if true correlation between the returns, pXY- — 0? r. d) What is the standard deviation of the return on the portfolio if true correlation between the returns, pxy = 0? gal/ML“ V‘OVWJWj minim/{Dag a M? g [(4qu plifiégwf 0M5 «LL/9V5 5) The variance of the return to any insurance policy i $1,000. Assume the returns on policies for the insurance company are independent. a) What is 'ance of the total return on 2 policies? (T2: 2 000 b) What 1s the standard deviation of the total return on 2 polic. (1‘: 2000 : £14.7Zl 3m c) What is t ' ce of the total return on 3 policies? d) What is the standard deviation of the total return on 3 policies? G: Raw 1 439.772 3: 6) Suppose the probability of receiving an order from any one particular sales call during a given period is exactly equal t .30» or any sales call. What is the prob ‘ ' at the first three sales calls result in: a) no orders? (3&9! 7) ’5 . i 2’ b) one order? [flHCVU 0) two orders? (2;) C§Z[7 J i 7 ‘1 ya 3 '1 a d) three orders? {3) (3J( 7) b > H \l ‘7) Suppose a simple random sample, sizek25 is taken; the true population standard deviation, (5, is know to equal 3'2 but the true mean [1 is unknown; and the sample mean, Y = 1. a) A 95% confidence interval for the sample mean ' ‘ (I 1% (E1; ii int b) A 99% confidence interval for the sample mean is: H, 2.975 6—3: if 1,5451: 8) (Continue with the previous simple random sample) a) If HO: ”X = 0 and Ha: ux 76 0, thei ulied -value 15: f -L) .5 . H A __ £1575”): Zmlpi‘fl’ 3a»: \ $16 I) 2.0 3/13? 3 ’ l) L 7 b) If Ho: ”X = O and Ha: lix > 0, the implied p-value is: 8’ Li a“? 1;“9) Suppose people are equally likely to be born on any one of 365 days in a year (ignore leap years). a) What is the probability that three randomly selected people all have different birthdays? 65 312:! .2. ,_ @1355“ '. b9” ) ‘ ,qq [751(5) 3’@ 2 b) What is the probability that for three rando , . -. : ed people, at least two have the same birthday? - t t "L e @ Z a W 10) A deck of 52 cards has 13 hearts . Clearly answer the following questions. Show your work (10 points each, 40 total) 1) X and Y represent the return (in %) on two individual investments. lix = 1 6 x = 4 M = 3 OY = 6 Z: a portfolio, 1%: invested in X and 3A in Y, so 22 .25X + .75Y ‘ What are variance ~and standard deviation of the return on the portfolio if true correlation 1s: .__’——-—— d) pxy=—.5? 21.25” «i141? e>p =-1° 7"15'01 ”"25? ’2'?) [333/ ,an m‘ £01" WWW 2) Consider two independent random variables. 9/...— X Probability Y Probability .J 0 .5 0 .7 2 .5 E .3 a) Find population mea nd population standard deviation of X (find variance first) U}: 0+2“) b ind population ’- 5{3 5 My 0 ®Z=X+Y. Write a table to the right showing possible values of Z and their probabilities. d) Find population 1-: nd po lation standard deviation of Z find Ml: I+ng e) W = X - Y . Write a table to the right showing possible values of W and their probabilities. f) Find population mean _, o ulation MUM: i " LS L “F 5' “ B e 3) Each time a fair four sided die is tossed, there is .25 = 1/: probability that red shows facing upward. a) Suppose this die is tossed two times. Let X = number of reds showing in 2 tosses. Use the binomial distribution to find the probability that red shows exactly 0, 1, or 2 times in the two tosses. b) The four sided die is tossed 40 times. Use the Normal approximation to the Binomial with continuity correction to find the probability that red shows i) 8 or fewer times; ii) between 9 and 11 times (including 9 and 11) and iii) exactly 10 times. 4) Al is about to turn 91. At the time A1 will buy insurance, on his 9lst birthday, the probability he will die while 91 is 10%, and while 92 is 20%. An insurance company sells a two year insurance policy to Al which will pay $1000S if he diesn dwhile insured. It charges Al $300 on his birthday (for the subsequent year— on his 91S birthday for his 91S yearn,d 92n dbirthday for his 92n d.year) If he dies while 91 the $300 need not be paid on what would have been his 92n dbirthday. Dies when: Lives until 300 r-mw we "(em a) List the three possible outcomes Net Return .. 70 0 1 L130 and their net return to the insurance company. Probability .1 .2 .7 b) Let X be the net return to the insurance company from the policy sold to A1. What is the mean of random variable X? ”0003+ ~40£*(2l+ 600673: ~79 30 + L120 - 270 c) What are the variance and standard deviation of random variable X? (70"‘27337U1LF‘W 27032[Z}7‘6 @DD 27mzf7) “090510 4 84720 + 7122720 - >0} 2mm 1 @ Now the company insures two people identical to Al. Probability of death is independent for each. d) What is the mea _ a1 expected net return from the two policies? e) What are the variance and standard deviation of t xpected return from til/\eg’e two policies? (:2: 250101? at Noam : 5707700 . > n U... Wi,‘ Suppose instead that Al’s insurance was for twice as much with the company paying $2000 in the event of his death and receiving $600 at the start of each year he is alive. 0 What is the expect etum from this one large policy? g) What are the variance and standard deviation of the total net expected return from this one large policy? C” l’il’£’~5ilo3 .. 110%? CW) 2 + (W éqdi “576550 + 35an 13045720 ‘5 187 Population mean and variance of discrete random variable formulas : ,uX = 2 xi P1 0'; = 2 (x1 — fl) 2 p I. Dispersion of linear combinations of random variables: 2 _ 2 2 2 _ 2 2 2 __ 2 2 0u+bX ‘b O-X O-X+Y ‘0} +017 +2pO-X0-Y 0;(4 —0X +0} ‘2/30'X0'y For a “weighted average” where a+b = 1: O-ZXHJY = aZO'i + 1920'; + ZpaO'XbO'Y If X1 & X2 are uncorrelated (so p = 0), then variance of a portfolio of n investments denoted X1, X2, X3, ...Xn; each with . 2 2 2 _ 2 2 2 2 2 2 2 a1/nshare,1s: 001m. +0'(1,n)X2+...+0'(1,n)X” —(1/n) O'Xl+(1/n) O'XZ+...+(1/n) 0X” =(%1)0'X P(A n B) = P(B | A) - P(A) = P(B | A) . P(A) a es’ru : P A B = —— By le ( l ) 13(3) 15(3) p(B|A)-P(A)+P(B|AC)-P(AC) n For X with Binomial distribution: 11 = np ; 62 = np(l-p) and the probability that X=k is k pk (1— p) "_k J; 0 * * margin of error: m = z * — z- calculated = Z = 7—0- Critical Zs: Z1025 = 1-96; Z005 = 2.576 42 Table entry for z is the area under the standard normal curve to the left of 2‘. 515115515 narmal 19113335757 0.0 5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 5398 .5438 .5478 .5517 .5557 .5595 .5636 .5575 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .5020 .5004 .5103 .6141 0.3 .5179 .5217 .0255 .5293 .5331 .6368 .6406 .5413 .6480 .5517 0.4 .5554 .6591 .6528 .5554 .0700 .6736 .5772 .6808 .6844 .5879 0.5 1 .5915 .6950 .5985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.5 1 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7985 .7517 .7549 0.7 .7580 .75 11 .7542 .7573 .7704 .7754 .7754 .7794 .7823 .7852 0.8 l .7881 .7910 .7939 .7957 .7995 .8023 .8051 .8078 .8106 .5133 0.9 1 .8159 .8181, .8212 .8238 .8254 .8289 .8315 .8340 .8355 .8389 1.0 a .8413 .8138 .8451 .8435 .8508 .8531 .8554 .8577 .8599 .8521 1.1 .8513 .8065 .8585 .8708 .8729 .8749 .877 .8790 .8810 .8830 1.2 3 .3849 .8869 .8888 .8907 .8925 .8944 .8952 .8980 .8997 .901 5 1.3 .9032 .9049 .9050 .9082 .9099 .91 15 .9131 .91 .17 .9152 .9177 1.4 ’ .9192 .9207 .9222 .9235 .9251 .9255 .9279 .9292 .9306 .9319 1.5 .9332 .9345 .9357 .9370 .9352 .9394 .9405 .9418 .9429 .9411 1.1» 193152 , 7 .9495 .9525 .9535 .9545 1.7 "1” .9551 6 .9625 .9533 1.8 1 .9641 .9699 .9705 1.9 1 .9713 .9761 .9767 2.0 .9772 .9812 .9817 2.1 9821 .9854 .9857 2.2 .9861 .9887 .9890 2.3 9893 .9913 .9915 2.4 .9918 .9934 .9935 2.5 1 .9938 .9951 .9952 2.5 E .9953 .9953 .9954 2.7 I .9955 .9973 .9971 2.8 1 .9974 .9980 .9981 2.9 5 .9981 .9986 .9986 3.0 § .9987 .9990 .9990 3. 1 .9990 .9993 .9993 3.2 .9993 .9995 .9995 3.3 .9995 .9998 .9997 3.4 .9997 .9997 .9998 ...
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