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Unformatted text preview: EC 41, UCLA \Vinter 2009 Name (print) Midterm #2  Version A — 2/23/09 RE Ch 4,5, 6.1, 6.2 TA: Name & Section Time  The normal table and useful formulas are on the last page ot'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 noninteger answers to 3 signiﬁcant digits, e.g., 333, or 3.33 or .0333  This exam consists of 10 Trues’l‘alse (20 points). l0 short answer (30 points) and 5 longer questions (50 points)
— Clearly “I rite answers on this exam. No points are awarded for illegible ans“ ers.  Be prepared to Show a photo lD during the exam (6.9... UCLA lD) — You may leave when ﬁnished. Do not disrupt those still taking the exam. 1. Circle T fer True or F for False (2 points each) lﬂrF
2)To@ The primary beneﬁt ofa welldiversiﬁed portfolio is that the standard de\ iation of its return will typically be
less than that ofa lessdiversified portfolio. lithe random variable Z = XtY, then standard deviation of Z is the sum ofthe standard deviations of X
and Y (o; ox + 5y) lfthe random variable Z = XlY, then the variance of Z is the sum ofthe variances of X and Y (032 = 6;:2 6Y2) lftw'o events are disjoint, probability ofA or B probability ofA plus B — PM) " P(B) Benford’s Law (used to cheek accuracy ofﬁnancial documents) implies that the probability ofthe lirst digit
of numbers in ﬁnancial documents is equally likely to be any integer from 1 to 9. lftvm assets hm e correlation “  l. any portt'olio with of these tuo assets \\ ill hm e st';tn(lard <ch iation  0 A conﬁdence inten'al gets wider (larger) as the number ofobservations in a sample is increased. For a particular calculated positive sample mean )7, the probability of rejecting a null hypothesis Iii41:0,
is smaller against a twosided alternate Ha: u :t 0; than against a onesided alternate H.._: it I‘ll. 10) T or@ In one calculates a 95% contidenee interval, 95% of the means ofa given population. it. are in this interval. 11. Brieﬂy, clearly, and correctly answer the following 10 questions (3 pts. each) 1) Consider two independent events, A and B, where P(A) = .3 and P(B) = .2.
a) What is the probability that A occurs given that B occurred = P(A l B)? ‘ = b) What is the probability of both A and B occur? Photo: NM—PtBJ: (av2) c) What is the progahility that neither A nor B occur? : PM att‘mm firiﬁr‘iili— pm)? — [,[grz r.0{;i=t.£/£[/ MM 8‘ 2) Consider the random variable X with the density curve below. a) What is the distrihution ofX is called? b) Suppose random variable Y has the same distribution as X.
Dram the distribution ot‘Z — X + Y to the right: o) What is the probability that Z is greater than 1? P(Z> l)'= 3) in upopulation ot' \xorkers, 20% earn $10 and 80% earn $20 per hour.
a) What is the merage “age of \\0l'i\'Cl'.\' in this population? .ZODH r800) ’ 2 H6 [3) What is the population _\_'2n'izinee of the wage for these \x'orket's? GZ '201949J27l9 tum“ = .2tglzttroth attorney 12mm. 0) What is the population standard deviation ofwage for these \x'm'kers? OWNED 4) Suppose the effect ofnn $800 Billion increase in government spending on GDP czm be written as n gemnetric series:
800.00 I 720.00  648.00 : 583.20 '. 524.88 . (the sum of different "rounds" of autonomous spending)
a) What prOpor 'on of each “round ofspending" is spent in the following round"? 9 b) Wt‘i[gimbomsatiies.;13.th€—}3¥Q€l11C~I—Of neonstztntt' ‘ sum of a geometric s ' " ‘*‘».«*"‘"'_ “mac;
x"? 0 l 2 3 a; l Warming .3 2mm r 900 low i o) What is the total increase in GDP caused by the above increase in government spendi _ 5) Suppose a simple random odl]][)lC. si7e 36 is taken; the true population standnrd deviation. ._ is know to equal 4; but the true mean it is unknown: and the sample mean. X: 40.
a) .\ 9591. COl'lllthllCC lI‘ltCl'VLtl for the sample mean is: ,—.._.. __ " i! (P +
xiZ 1,} 2‘) Wilda—9%: Hula/36 b) lflln: ll); = 39 and Hﬁ.’ In}; 3‘). the implied p—vnlnc is: I —~ [V : LS’ c) [fllg: tlx' = 39 and IL: ll); 1?: 30, the implied pvnlue is: Puutw [l5 6) A random Vnt'inhle. Kean take (tn three. different values with the following probabilities: l_Vaue__of_i: \ O 06154 '3 : ~36 2 0 IPrOPalftitity 0.6. 0—2.; 02'5 a) What is the mean of X? __ _ _ .
. [bitoti‘16:2.4l b) What is the variance. ol~ _ c) What is the standard deviation ol‘ : t'fﬁ‘n
7') At Loon] State [TY lO“"' ofthc students come. from each of ll) counties (all students come from one of these 10 counties} and assume the student pn'xpulzttion is mueh larger than size ot‘nny‘ sample size). Ltlé E [339W MV Pkwy—in
n‘) lf‘twn ' 1 its me randnmly ohusen. What is the probability they are from the. same county? ()ka MW” 7
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u iﬁﬁrl'he followit'lg table is for a sample 0' anncs w 10 completed boot camp together
White (W) NonWhite (N W) Total:
Sent to Iraq (5): 10 20 30
Not sent to Iraq (NS): 40 10 50
Total: 50 30 80 a) Find the following Conditional probabilities: 13(5an ; White) ’0 t ‘ Z_ P (Sent ' Not White) = 9) X and Y represent the return on two individual investments. 1.1.x = 2 ox = 3 My " 8 G ,  4
Z '— a portfolio lialt‘invested in X and Y each, so Z = 5x + ,SY . Truc correlation between the returnsT pigt .2 a What i' th" m ‘an return of the ortfolio, ? s _, a
> M ° P “2 .€[2J+.s[ﬂt~ W!—
b) What are the variance and the standard deviation ofthe return 0n the portfolio? “iii”: @1332 M5)2(HJ?+__2 (ZlCﬁ‘lé’JLsM/J FT is
: 219+ H +i.z Gail/Ms : 272% 10) A deck of 52 cards has 13 hearts. _
a) What is the probability that a randomly drawn card will be a heat? '3 l 52 L t/ b) What is the probability that two eal'cls randomly drawn with replacement are both hearts?
_____‘\ ..__ c) What is the probability that two cards randome drawn without ' u laconient are both hearts? III. Clearly answer the following questions. Show your work (10 points each, 50 total) 1) Suppose a simple random sample of size 64 is taken; the true population standard deviati o, is know to equal 9 but the true mean u is unknown; and the sample mean, if = 52. Si i 2_ 21) Find a 95% conﬁdence interval for the mean value of X. 52 ‘l‘ 3/ 1,63?
m Lani it 220‘) 1 150;“ “(iii ‘ _ . What is the zcalculated titan b) Consider the Null Hypothesis, II;.: ll“: 50
observed sample mean X — 52? ; ism a go ' WT— “ ‘l/g‘lt7?) c) 1151' the previous problem, what is the *—eritical ’lien HU: .ui. = 5.0 ; H“: ‘11; .'> 50 and significance level, a = .05?
' (an: gulp 2} ( 'onsider 2 assets X; & X2 with the same expected return (1.1.1 = 112 = 10%) and standard deviation (0 h or. = 2%). YOU
invest 50% in asset X; and 50% in asset X; a) What is the standard deviation oftlris portfolio when p = 0? p = .5? and p l? “i
’L 2 r 1 4— 22 D _ G. ‘ Z n it Hi
(9‘0 0}: ’5. 2 ' 'l ‘ 2 p ' ll (31.? {it}; 1 + 1+ 269)§{2i.s’{2) = NH] = 3 (fiszlG) = 1» . —  HH : ~ _
p  Q, iiii'zm were 2 Li {Wm {29 h) Suppose one invests 25% in asset X1; 25% in asset X3; 259/0 in asset X3; and 25% in asset X4; true Incan ol‘all
indii‘idual investments X is p;._ r: 10% true standard deviation 01" all individual investments is ox  2% and returns are uncorrelated, p = 0. What is the standard deviation ofthis portfolio?
'L . ’1» 2. .1 ..  7— ._
GP sCzSS 224,629)? 71(25)22 ir[zsj? 22 —LIC?C‘J 22; Lips/2] c) Suppose one invests 10% in IO assetsjust like X] with true mean .ux— 10% true standard deviation ox ' 2% and all
returns uncorrelated (p — 0). What is the standard deviation of this portfolio? G7: D(i_)72_ *LLai;2 445:: l6'32
P‘l’PZ’iol'I—s‘ 4' 3) Suppose the probability ofreceiving an order from any One particular sales call during a given period is exactly equal
to .25 for (my sales call. What is the probability that: a) the ﬁrst two sales calls result in x" .s a . .9. ‘ " 2 e
no orders 67 ‘2)(3 ‘3‘) ‘ x 7 q ‘ /K _
one order JQ‘IS‘) l(2 {ll 1 Z (7 96.757 tnoorders? Z 0 ' . '2“ Z L
[25 ,7s 61s} _ .23 b) the tirst three sales calls result in two orders? (2) [Zfizékf “— 3®le33rs> "' .Ile6’2s‘ 3: e) the ﬁrst order is from the 41h sales call (so the ﬁrst three calls do not result in orders)? {7§§7§)[7 s [29/ 0 4) For an auto insurance company, three outcomes are possible for any policy (measured in thousands of dollars) are: big loss %8 '0 l i0“ Ia {1'
small less 8 .04 _ t q S?  :33 2. + I ,q 3 ill'vl'y‘ﬂ
no loss *2 .95 __~ a) What is the expected net return from a policy on one driver, ELK)? it); What is the standard deviation ofthe return from a poliq on one dri\ er"? (7;; IQ:
(ﬁnd variance first) 1
([1; .Bl (—qsaéil sputteatjz +9; (2‘16) _ ,_ . _ A I. “ _ L
£17.22 4— 2.5158! +1.»?b2 1020'! llwo
b) It‘they have 10 customers what is the to r ' greetIto) expected net return from all 10 customers? rm : t O 6 w 0) Assume the outcomes ofdilTerent cuslmners are. independent. l’ind the standard {lcx'iution ofthe total aggregate. L as.
[D t) {DLD.L{ _/ 70. ID m 5) a) At a given l‘uctory 10% ofthe thumb drives are defective. 10 thumb drives are. randomly chosen. Let X = number
defective in this sample of 10. Use the binomial distributio to the probability that exactly one is defective in this sample. Ci ; _ Q ;
(’Pltti'tn WW“ 0) Consider the factor} le)t)Ve where p = 10%. Suppose a sample of 100 drives is taken, Use the Normal approximation
to the Binomial with continuitv correction to ﬁnd the probability that i) 0 or fewer in the sample are defective; ii) 10 or fewer in the sample are defective; iii) exactly 10 are d Afective.  q S
W ' Q’ 3 Z 5‘ =  ' "a 7
tiPVLP/b (I) wig—“ll ~ 169.» * l5) *‘lz = iﬁ = l.’“b7‘5 .\
: :3); 0‘ (it) {mat} ; 1369... <105j:i3é<wig“’”  péyt H' ...
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 Spring '07
 Guggenberger

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