Exam 2 practice problems - 1

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End of Packet Problems 1. Assume that you are given the following historical data for Security Y. What is the coefficient of variation for Security Y? CVeE S/ ¢ Year Security Y 1 0.30 o 2 0.08 = o et 3 -0.10 S = O VSonlSiatel 4 0.05 5 0.19 : EV - .14 9 6bouy 2. Assume that you are given the following historical data for Security Y and the market on average What is the covariation between Security Y and the market? COViny coR < m CO\/Wb 2.6 .O~\ (p(o] N X Year Security Y Market 1 0.30 0.22 2 .. 0.08 0.10 3 -0.10 0.03 - 0.05 0.09 g -0 0.15 N 2056764111 CV\; = L.HU (oY COQ«:) 2 s\'\,(’-\ o =WV eTy) COR= 0«99 7300‘17'57 8 St #0:0N13039E CVerz 0.0 5N

e— o 3. Assume that security A has an expected return of 15% and a standard deviation of 9.6%. Given this information, determine the probability of observing a value between 18% and 25%. ! - P i O Fas)-Hed 5 15 16 as 0-8512... - O.6326... [o.228547 57 ] ® Z:= O-i8—0-55 = 0.?)"35 2O P2 0.2 N (to Hre [¢FH) Z- 0-35-045 — = L. oW (pppb) ©-0%( P> 0 8312163706 4. “How’d you do on Banko’s FIN3403 test?” Not bad. Out of the 100 possible points, I scored 73. Banko standardizes each exam score via Z- scores to a mean of 77.50% and a standard deviation of 12.5%. This just means that whatever your z-score was before the curve, your z-score will be the same after the curve as well... but with a different mean and standard deviation. The raw class average was 53.43%. And I think Banko said the standard deviation was 20% (out of 100). So, with the curve, my score is Raw . Curue - 5234 - - 13 5 > z = X <2775 Q.5 & i N [y 0.9785 - *- 1.5 1.5 X2 87.73[;25

1 5. You are given the following information concerning the returns for Security J over the coming year. Asyou can calculate, the standard deviation is 15.82 percent. Based on this inforrTlafion, d-'%, and assuming the actual distribution is normal, use the Z-table to calculate the probability that the o< actual return will be positive. = Qs | 589, State Probability Return J M 0.0\ 1 2000% | 0.30 20.00% | 0.08 7 60.00% 2 -0.10 7 — 0 o l.b% = X e " O -0,0l (D =z -——/: = = 2 "‘O.lougjg o o. 1583 =0 Llschm Cleft op Z Score) RIGEIE D Gods i S5 Loy, :'Ep.sqoaaﬂ 6. You are given the following information concernin el” year. Asyou can calculate, the standard deviation | O'CAL and assuming the actual distribution is normal, us actual return will be positive. g the returns for Security ] over the coming is 16.16 percent. Based on this information, e the Z-table to calculate the probability that the O = O. Il 33Ns State Probability Return] | M ool 9 1 21.30% 0.30 2 wil7:55% 0.08 < 61.15% -0.10 ® Bl o ot USing Voriancy P g:f%i‘l& TO.lo3g1511y Lo ( Vz02) 7 cagicmy k) =200 (- -y > _ o».»a\zgo,g- F)ﬂ ®1- 1S BIua = * o 185{(0.08-7) O.54\1 o5 5 = % C 2 [Camy(Lay , P T ‘”% o (155 ( '08) 5 V= O. oab‘&GQQS'ﬂ (.01555(-,‘3] 4—;,.;0.:0‘ i = Bt PV P r = O-O‘hmﬂﬂw L @2023 SHadvy RAma® Ot man

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You are given the following information concerning the returns for Security J over the com-ing year. Asyou can calculate, the standard deviation is 15.82 percent. Based on this information, d-'%, and assuming the actual distribution is normal, use the Z-table to calculate the probability that the o< actual return will be positive. = Qs | 589, State Probability Return J M 2 0.0\ 1 2000% | 0.30 20.00% | 0.08 7 60.00% 2 -0.10 ', — 0 o l.b% = X s fi O -0,0l (D =z —_— il 0NV ST E= ©.1583 =0 Llschm Cleft op Z Score) RIGEIE D Gods i S5 Loy, :'Ep.sqoaaﬂ 6. You are given the following information concernin year. Asyou can calculate, the standard deviation and assuming the actual distribution is normal, us actual return will be positive. g the returns for Security ] over the coming is 16.16 percent. Based on this information, e the Z-table to calculate the probability that the o = O lbl3307S | Bk § 0 State Probability Return'J P ivp) 9} 1 21.30% 0.30 ZEN 2 “17.55% 0.08 e o ) 3 61.15% -0.10 Mz L% ? 2z (o0 VP fodusvtatte. - : 20, i ® B0l o ot USing Voriancy o oM ©. 10381511y e (0 vy D\ OUSBoIug < °SM\ 365197 ] = [Gay(.ay ("755) (.08) » V= (. 6185) (-, V] F = OIO‘hy it

Fall 2015 Exam 1 #20. Assume that you are given the following historical returns for Security A. Given this information, determine the coefficient of variation for Security A. Q - D—— = O, CV = z 321\ 087319 Year Return on Security A 20.00% ) — - 16.00% . X = 0.l 12.00% . G 15.00% | S = O.»O?,?,;waqjc' 12.00% o g o syl dar ks G B Wl N - . A. 0.1534 0.2211 C. 0.2888 D. 0.3565 E. 0.4242 Answer: B.

Summer 2015 Exam 1 #20. Assume that you are given the following probability distribution for the returns on Security A. Given this information, determine the coefficient of variation for Security A. State of Probability Return on Nature Security A 1 20.00% 2 | 15.00% 2 30.00% 3 | 10.00% 3 50.00% 5§ |30.00% A. 0.3280 B. 0.3551 C. 0.3822 D. 0.4093 0.4364 Answer: E CV= == = 0.HAY3EIS ? O = 0.09W515129. Fall 2014 Exam 2 #9. You are given the following possible returns for Security J. Given this information, determine the coefficient of variation for Security J. Yo - ' — = o O State | Probabiity |~ R - CV: c oee b 386351 1 30.00% 10.00% = O.lFaa 2 25.00% 18.00% 3 18.00% 24.00% , 4 27.00% -0.00% 1| (D N @0.2970 €= étg%a‘jo.\\+ (0.35.0.8) + (0@ 0. 24) + 7 0.2997 _ . . - - C, 0.3024 i . e ‘10-3.\ = o1 733 D. 0.3051 Py = Ay I = 5. 03078 @ [0'3[“’" R ]] * [‘?-35[.' (0.35- ?\a]] ¥ Answer: A o [ovelcoan- 2] *parteoasy] - v o Ve 0. 0086 1516c |

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Summer 2013 Exam 1 #24. Assume that you are given the following probability distribution for the returns on Security A. Given this information, determine the standard deviation for Security A. State of .| Probability Return on Nature Security A 50.00% & | 11.00% Ao 2 30.00% % | 15.00% ( e\ e\ oS s 10O 3 20.00% 2. | 20.00% Qe+ A () A. 4.359% . 3.912% 3.464% D. 3.017% E. 2.569% Answer: C Modute 4 (Risk and Rates of Return) #71. You are given the following mformatlon concerning the possible returns for the Market and Security A for the coming year: - 3 . i T (.m = o ° 0-7 State of Nature Probability Market . Security A | T =2 00184919 3 1 30.00% 15.00% 14.00% R 2 40.00% 7.00% 20.00% 2 30.00% 9.00% 18.00% Caz 0.170 Th - 00971395 As you can calculate, the Market has an expected return.of 7.00 percent and a standard deviation of 1.5491933 percent, while Security A has an expected return of 17.60 percent and a standard deviation of 2.497992 percent. Given this data, determine the correlation between the Market and Security A. = A o. OO0 O 3;_\ - O-[ ; :ns - COVQM A. 05237 Cokam = - (0.00497192) (00154 ua33 C. 04228 D. 0.7255 ©°'6202~ - @ 8solve Rse covu 2P (e~ Fe) (Fa - Ca) Answer: E 0. o0 MO [_O 3(0005— o. 07)(o.lq Ocl'?(p)] + O [o.t—\(o.o7~ 0.07) (c.a-o-\‘l(o)] [0:306.09- 0:01)(0.18 - 0.196)] -5 COVan = - O.0o003Y4

Mod\{le 4 (Risk and Rates of Return) #77. Assume that you are given the following probability distribution for the returns on Security A. Given this information, determine the coefficient of variation for Security A. | State of Nature [ Probability Return on Security A &z CV: 11 | 20.00% 10.00% |2 | 50.00% 14.00% 13 | 30.00% 20.00% A. 0.6784 B. 0.5689 ~ C. 04594 D,_0.3499 (B )0.2404 Answer: E 0= 0.,030555 1.8 l\ r - 0.5 Module 4 (Risk and Rates of Return) #82. Assume that you are given the following historical retlrns for Securities J and K, as well as their average return and their variance of returns over this time period. Given this information, determine the correlation between Securities J and K. Year Security J Security K 1 0.05000 0.10000 2 0.10000 0.08000 3 0.28000 0.32000 4 0.22000 0.08000 5 0.12000 0.18000 Average 0.15400 0.15200 Variance 0.00878 0.01052 g' Y0.644 . 0.568 C. 0.496 COV¢ oz Z (N D. 0.424 E. 0.352 Answer: A @ COld i = : - CoVwe COQ.)\L - SQSK © | Avey = o. 154 03 = 0,093 7016542 Pvew = 0.18& S = 0102856705 @ oWt coV to ‘pluj inte COR Ryrrnulon f')(r,_—r‘)/('n~\) «—-} [(o-os- Os184) (0-1- 0.153)] +[(0.1-0.a54) (0. oQ-o-lsa)] + [{o.88 - 0154 (033 -015)] + [ (0.82-0-154) (.08 0.152)] + [(013-0-154)(0u18 ~o.tsn)]] (5-1Y = 0.00619 o. oc:(p\ﬂ (0.093 £ 0.1085...) = 9644073636

Z-scores and Probabilities Summer 2015 Exam 1 #21, Fall 2014 Exam 2 #16, Spring 2014 Exam 2 #8, Fall 2012 Exam 2 #26, Spring 2006 Exam 2 #20 Module 4 PDF (Risk and Rates of Return) - 9, 58 — Summer 2015 Exam 1 #21. Assume that Security A hés an e>'( Given this information, determine the probability of observi A. 51.74% A B. 57.15% v X D o. - ész_ss% e @ om ... A~ lon pected return of 24 percent and a standard deviation of 8.42 percent, ng a value between 14 percent and 31 percent. 67.97¢ - : 73.38"2 O.48... - ©-C1963716° Answer: D @ Z - \\;-i: = - I‘\87@'48"\5(o Z - 3\~3"| T 8qp T 2832125999 PT- 12129 Fall 2014 Exam 2 #10. Assume that Security K has a mean of 8.32% and using a Z-table or a t-table in the Exam handout (or in this packet...), and 6.8%. . \_1/'\ i g AL ‘ 28.25% Z// S 87 30.42%° asedy, C. 32.59% &.%2 D. 34.76% E. 36.93% a standard deviation of 3.26%. Given this information and determine the probability of observing a return between 2.5% @ 0.025 - O; 0323 Answer: A : - , - | -78 S ANOY Z: Oo 039.(0 P2 0-037108971 7 - &0(08 - 0.0 OO0 = - 0, Uk 25069 P 0.2205 15538 @ oc.5305 - 00327,

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% a Spring 2014 Exam 2 #8. Assume that Security A has an expected return of 18 percent and a standard deviation of 7.40 p Given this information, determine the probability of observing a value between 11 percent and 29 percent. A Bazrt LD ercent. | To0on noag M- E. 83.35% Answer: D ‘.“‘ '8 _ . L ZLS 2 -0 qUSTHS MO ©-o1 P= 0. 1N20EFONT 0.29-0.18 Z: e o ° \.t—\&pqg@.qg(o PR P:‘O..qg)%QL{j.sb 3 Exam 2 #8. Assume that you are given the following information concerning the returns for Security J over the coming dard deviation is 3.7416574 percent. Based on this information and using the Z-table (or t-table) that the actual return will be between 10 percent and 20 percent. Fall 201 year. As you can calculate, the stan in the exam handout (or in this packet...), calculate the probability E* . State Probability | Return J O(-\\o- 1 20.00% T | 10.00% 2 40.00% Y 15.00% 3 40.00% 4 | 20.00% A. 66.17% O Perd X AT B. 69.70% R X cale = xzo. C. 73.23% ' ‘ . D. 76.76% ' (B 80.29% @ Zio = °:1-0.1b _ | 4 ) 0.5 M5 ‘%35@’1»{5[ Answer: E . . P’-'- o. qu"‘()b\"l \ S- 2. - o. a“Oslb ©.03741(,. .. .00 oMY U | Df O B5THI 629, ® % -9, - 10 2 ©-802670 53

-t X o FI?“ ?012 Exgm 2 #26. As_sume that a security has an expected return of 8.0 percent and a standard deviation of 3.20 percent. Given this information, determine the probability of observing an actual return between 8.80 percent and 11.40 percent. 0.0 [{ A. 19.39% 88 ' B. 22.53% \ D1y .)25.67% 3 (/4 . 28.81% 4 E. 31.95% 0.09 Answer: C © Z 0.08% - 0.08 0% — 5. 0% z 085 P20.5137003206 ) Z . o.l\L{'-o;o | = Yy - ow@j 2 L0635 i o0.8557956a | Spring 2006 Exant 2 #20. Assume that you took an exam in FIN 3403 and earned a raw score of 50, where the average was 68 and the standard deviation was 15 (you should now be able to calculate the Z-value for your score). Assume that grades are adjusted so that the new distribution has an BV'érage score of 75 and a standard deviation of 12.50, but that your Z-score remains the same. Given this information, determine what your adjusted score will be. o - ay X250 %- 7\ sa2 Doals M0 s 2 so i egna DND By x=7 X:7S gota, then use same E. 70.00 ocoa. s - ot solve R X Answer: C @ 2 ?-.SO ‘.‘98 - : ; “..7 o . ® -1.5: 2= S 0w

Q. 0.00195 Al X:o.1085 Collpg = B. Xz 0.105 . 0. C. 0,007 Pz o-.0l5 ¢o ©.75 p:o.o4dol ¢ o D. 0.00285 E. 0.00245 Z = 0.5059 8371\ 2-20-515992299 Answer: C Module 4 (Risk and Rates of Return) #58. Assume that you are given the following historical return information for Security I: %*fi Year Security J Y :z O. \7“‘ 12.00% Vi 19.00% : 34.00% 10.00% 12.00% $:0.09829138838 A (WIN| = o.\Y Given this data, and assuming a normal distribution, determine the probability of observing a return less than 17.40 percent (2- tables are provided in the Exam Handout). 50.00% : 45.00% 60%3 NnCe Wwe o, afled C. 40.00% . 35.00% . g. gg.OO% foc infs afiter Pne. ean . Answer: A So/50 chonw of Reing belaus e aweage

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