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Unformatted text preview: Random Variables 7.1 (a) P(less than 3) = P(l or 2) = g = (b)—(c) Answers vary. 7.2 (a) BBB, BBC, BCB, CBB, CCB, CBC, BCC, CCG. Each has probability 1/8. (b) Three
of the eight arrangements have two (and only two) girls, so P(X = 2) = 3/8 = 0.375. (c) See
table. Value ofX 0 1 2 3
Probability. 1/8 3/8 3/8 1/8 7.3 (a) 1%. (b) All probabilities are between 0 and 1; the probabilities add to 1. (c) P(X S 3) = 0.48
+ 0.38 + 0.08 = 1 — 0.01 — 0.05 = 0.94. (d) P(X < 3) = 0.48 + 0.38 = 0.86. (e) Write either X 2
4 or X > 3. The probability is 0.05 + 0.01 = 0.06. (f) Read two random digits from Table B. Here
is the correspondence: 01 to 48 <=> Class 1, 49 to 86 <=> Class 2, 87 to 94 <=> Class 3, 95 to 99 <=>
Class 4, and 00 <=> Class 5. Repeatedly generate 2 digit random numbers. The proportion of num
bers in the range (Mgwﬂl be an estimate of the required probability. 7.4 515’ 0 0.4 Probability
o .O .0
,_. 0.0—=l:3l:l 1 2 3 4 5 6 7 8
Number of rooms in owner—occupied units 124 Random Variables 125 0.4 P
w Probability
O .0
H 0.0 [la 1 2 3 4 5 8 9 10
Number of rooms in renter—occupied units The rooms distribution is skewed to the right for renters and roughly symmetric for owners. This sug—
gests that renter—occupied units tend, on the whole, to have fewer rooms than owneroccupied units. 7.5 (3) {X2 5}.P(X2 5) = P(X = 5)+ P(X: 6) +... + P(X = 10) = 0.868. 78 (a) (b) {X > S} = the event that the unit has more than five rooms. P(X > 5) = P(X = 6) + P(X
=7)+...+P(X=10)=0.658. (c) A discrete random variable has a countable number of values, each of which has a distinct probability (P(X = P(X 2 5) and P(X > 5) are different because the first event contains the
value X = 5 and the second does not. P(O s X s 0.4) = 0.4. (b) P(O.4 s X s 1) = 0.6. ( P(0.3 s X s 0.5) = 0.2. (d) P(0.3 < X < 0.5) = 0.2. (e) P(O.226 s X s 0713) = 0.713 — 0.226 = 0.487. (f) A continuous distribution assigns probability 0 to every individual outcome. In this case,
the probabilities in (c) and (d) are the same because the events differ by 2 individual values, 0.3
and 0.5, each of which has probability 0. 7 7 (a) P(X s 0.49) = 0.49. (b) P(X 2 0.27) = 0.73. (c) P(0.27 < X < 1.27) = P(0.27 < X < 1) = 0.73. (d) P(0.1 s X s 0.2 or 0.8 s X s 0.9) = 0.1+ 0.1: 0.2.
(e) P(not [0.3 s X s 0.8]) = 1 — 0.5 = 0.5. (f) P(X = 0.5) = 0. A 2 0.45) = P(Z 2 94373304) = P(Z 2 2.17) = 00150. P
(b) 1303 < 0.35) = 142 < —2.17)= 0.0150.
(c) P(0.35 sf; 3 0:45) = P(—Z.17s z s 2.17) = 0.9700. 126 Chapter 7 7.9 For a sample simulation of 400 observations from the N(0.4, 0.023) distribution, there were 0
observations less than 0.25, so the relative frequency is 0/400 = 0. The actual probability that
f) < 0.25 is P(Z < —6.52) z 3.5 X 10‘“, essentially O. 7.10 (a) Both sets of probabilities sum to 1. (b) Both distributions are skewed to the right; how
ever, the event {X = 1} has a higher probability in the household distribution. This reflects the
fact that a family must consist of two or more persons. Also, the events {X = 3} and {X = 4}
have slightly higher probabilities in the family distribution, which may reﬂect the fact that
families are more likely than households to have children living in the dwelling unit. 0.5
0.4
>5
g 0.3
.5
C6
,0
8 0.2
9..
0.1 l:l l:l D :1 =
1 z 3 4 s 6 7 Number of persons in household 1; Probability l 2 3 4 5 6 7
Number of persons in family 7.11 (a){Y> 1}.P(Y>1)=P(Y=2)+P(Y=3)+...+P(Y=7)=0.75.0r,P(Y>1)= 1 —
P(Y= 1): 1 — .25 = .75.(b)P(Z<Ys4)=P(Y= 3)+P(Y=4)=0.32.(c)P(Y¢2)=1
HY=D=0%. Random Variables _ I 27 7.12 (a) The probabilities sum to 1. 0.4 .0
w Probability
_o
N 0 1 0.1 U
0.0 B 1:1 El
2 3 4 5 Number of cars in householii’l (b) {X 2 1} = the event that the household owns at least one car. P(X Z 1} = P(X = 1) +
P(X=2)+...+P(X=5)=0.91. (c) P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5) = 0.20. 20% of households own more cars
than a two—car garage can hold. 7.13 '(a) The 36 possible pairs of “up faces” are (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (27 1) (Z, 2) (2, 3) (2, 4) (2, 5) (Z, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) (b) Each pair must have probability 1/36.
(0) Let x = sum of up faces. Then Sum Outcomes Probability
x=2 (1, 1) p=1/36
x=3  (1,2)(2, 1) p=2/36
x = 4 (1, 3) (2, 2) (3, l) p = 3/36
x =5 (1, 4) (2, 3) (3, 2) (4, 1) p = 4/36
x = 6 (1, 5) (Z, 4) (3, 3) (4, 2) (5, 1) p = 5/36
x = 7 (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) p = 6/36
x = 8 (2, 6) (3, 5) (4, 4) (5, 3) (6, 2) p = 5/36
x = 9 (3, 6) (4, 5) (5, 4) (6, 3) p = 4/36
x = 10 (4, 6) (5, 5) (6, 4) p = 3/36
x=11 _(5, 6)(6, 5) p=2/36
x =12 (6, 6) p =1/36 128 Chapter 7 23456789101112 (d) P(7 or 11) = 6/36 + 2/36 = 8/36 or 2/9. (e) P (any sum other than 7) = l — P (7) = l —
6/36 = 30/36 = 5/6 by the complement rule. 7.14 Here is a table of the possible observations of Y that can occur when we roll one standard die
and one “weird” die. As in Problem 7.13, there are 36 possible pairs of faces; however, a number
of the pairs are identical to each other. 1 2 3 4 5 6
0 1 ' 2 3 4 5 6
0 1 2 3 4 5 6
0 1 2 3 4 5 6
6 7 8 9 10 11 12
6 7 8 9 10 11 12
6 7 8 9 10 11 12 The possible values of Y are l, 2, 3, 4, ..., 12. Each value of Y has probability 3—6 = i 12 7.15 (a) 75.2%. (b) All probabilities are between 0 and l; the probabilities add to 1. (c) P(X Z 6) =
1 — 0.010 — 0.007 = 0.983. (d) P(X > 6) = 1 ~— 0.010 — 0.007  0.007 = 0.976. (e) Either X a 9 or
X > 8. The probability is 0.068 + 0.070 + 0.041 + 0.752 = 0.931. 7.16 (a) (0.6)(0.6)(0.4) = 0.144. (b) The possible combinations are SSS, SSO, SOS, OSS, SOO,
080, 008, 000 (S = support, 0 = oppose). P(SSS) = 0.63 = 0.216, P(SSO) = P(SOS) = P(OSS)
= (0.62)(0.4) = 0.144, P(SOO) = P(OSO) = P(OOS) = (O.6)(0.42) = 0.096, and P(OOO) = 0.43 =
0.064. (c) The distribution is given in the table: The probabilities are found by adding the proba
bilities from (b), noting that (e.g.) P(X = 1) = P(SSO or SOS or 053). (d) Write either X 2 2 or X
> 1. The probability is 0.288 + 0.064 = 0.352. Value of X 0 1 2 3
Probability 0.216 0.432 0.288 0.064 7.17 (a) The height should be %, since the area under the curve must bell. The density curve is
below. (b) P(y S 1) = a (c) P(0.5 < y < 1.3) = 0.4. (d) P(y _>. 0.8) = 0.6. l—ﬂ—ﬁ l
I
I
—‘—V—1_l——V—l—V_r_—l—V——l_l—V—:’ 0 0.5 l 1.5 2 Random Variables 129 7.18 (a) The area ofa triangle is §bh = %(2)(1) = 1. (b) P(Y < 1) = 0.5. (c) P(Y < 0.5) = 0.125. 7.19 The resulting histogram should approximately resemble the triangular density curve of
Figure 7.8, with any deviations or irregularities depending upon the specific random numbers gen—
erated. 7.20 (a) Pg; 2 0.16) = P(Z 2 9481,39—315) = P(Z 2 1.09) = 0.1379. (b) P(0.14 sf; 3 0.16) =
P(—l.09 s z s 1.09) = 0.7242. 7.21 In this case, we will simulate 500 observations from the'3.N(.15, .0092) distribution. The
required TI—83 commands are as follows: ClrList L1
randNorm (.15, .0092, 500) —> L1
sortA(L1) Scrolling through the 500 simulated observations, we can determine the relative frequency of
observations that are at least .16 by using the complement rule. For a sample simulation, there are
435 observations less than .16, thus the desired relative frequency is 1 — 435/500 = 65/500 = .13. The actual probability that p 2 .16 is .1385. 500 observations yield a reasonably close approxima—
tion. 7.22 p. = (0)(0.10) + (1)(0.15) + (2)(0.30) + (3)(0.30)+ (4)(0.15) = 2.25. 7.23 Owneroccupied units: ,u. = (1)(.003) + (2)(.002) + (3)(.023) + (4)(.104) + (5)(.210) + (6)(.224) +
(7)(.197) + (8)(.149) + (9)(.053) + (10)(.035) = 6.284. Renteroccupied units: [.L = (1)(.008) + (2)(.027) + (3)(.287) + (4)(.363) + (5)(.164) + (6)(.093) +
(7)(.039) + (8)(.013) + (9)(.003) + (10)(.003) = 4.187. The larger value of p. for owneroccupied units reflects the fact that the owner distribution was
symmetric, rather than skewed to the right, as was the case with the renter distribution. The “cen ter” of the owner distribution is roughly at the central peak class, 6, whereas the “center” of the
renter distribution is roughly at the class 4. 7.24 (a) If your number is abc, then of the 1000 threedigit numbers, there are six—abc, acb, bac, bca, cab, cba—for which you will win the box. Therefore, we win nothing with probability
3330 = 0.994 and win $83.33 with probability ﬂ : 0.006. b) The expected payoff on a $1 bet is p. = ($0)(0.994) + ($83.33)(0.006) = $0.50. (
(c) The casino keeps 50 cents from each dollar bet in the long run, since the expected payoff
= 50 cents. 7.25 (a) The payoff is either $0 or $3; see table on next page. (b) For each $1 bet, ,ux = ($0)(0.75) +
($3)(0.25) = $0.75. (0) The casino makes 25 cents for every dollar bet (in the long run). 150 Chapter 7 Value ofX 0 3
Probability 0.75 0.25 7.26 In 7.22, we had 11 = 2.25, so of = (0 — 2.25)2(0.10) + (1 —~ 2.25)2(0.15) + (2 — 2.25)2(0.30) + (3
— 2.25)2(0.30) + (4 — 2.25)2(0.15) = 1.3875, and a, = v—138‘7—5 = 1.178. 7.27 Mean:
Household size: p. = (1)(.25)+(2)(.32)+(3)(.17)+(4 (.15)+(5)(.07)+(6)(.03)+(7)(.01) = 2.6.
Family size: ,1]. = (l) 0)+(2)(.42)+(3)(.23)+(4)(. 1)+(5)(.09)+(6)(.03)+(7)(.02) = 3.14.
Standard deviation: ‘
Household size: 02 = (1 — 2.6)2(.25) + (2  2.6)2(.32) + (3 — 2.6)2(.17) + (4 — 2.6)2(.15) +
(5 — 2.6)2(.07) + (6 — 2.6)2(.03) + (7 — 2.6)2(.01) = 2.02, and 0' = m = 1.421.
Family size: 01 = (1 — 3.14)Z(0) + (2 — 3.14)2(.42) + (3 — 3.14)2(.23) + (4  3.14)Z(.21) + (5 —
3.14)2(.09) + (6 — 3.14)2(.03) + (7 — 3.14)2(.02) = 1.5604, and a = m = 1.249.,
The family distribution has a slightly larger mean than the household distribution,
reﬂecting the fact that the family distribution assigns more “weight” (probability) to the
values 3 and 4. The two standard deviations are roughly equivalent; the household stan— dard deviation may be larger because of the fact that it assigns a nonzero probability to
the value 1. 7.28 We would expect the owner distribution to have a wider spread than the renter distribution. The central “peak” of the owner distribution is more spread out than the lefthand “peak” of the renter distribution, and as a result the average distance between a value and the mean is slightly larger in the owner case. Owneroccupied units: 02 = (1 — 6.284)2(.003) + (2 — 6.284)2(.002) + (3 — 6.284)2(.023) + (4 — 6.284)2(.104) + (5 — 6.284)Z(.210) + (6 — 6.284)2(.224) + (7 —' 6.284)Z(.197)+ (8 — 6.284)2(.149)+ (9 — 6.284)2(.053)+ (10 V— 6.284)2(.035) = 2.689344, arid o : V2.689344 = 1.64. Renteroccupied units: ,1. = (1 — 4.187)2(.008) + (2 — 4.187)Z(.027) + (3 — 4.187)2(.287) + (4 — 4.187)2(.363) + (5 — 4.187)2(.164) + (6 — 4.187)2(.093) + (7 — 4.187)2(.039)+ (8 — 4.187)2(.013)+ (9 — 4.187)2(.003)+ (10 — 4.187)2(.003) = 1.710031, and a = V1.710031 = 1.308. 7.29 (a) [1.x = (0)(0.03) + (1)(0.16) + (2)(0.30) + (3)(0.23) + (4)(0.17) + (5)(0.11) = 2.68. of
2.68)2(0.03) + (1 — 2.68)2(0.16) + (2 — 2.68)Z(0.30) + (3 — 2.68)2(0.23) + (4 — 2.68)Z(0.17)
(5 — 2.68)2(0.11) = 1.7176, and 0,, = V1.7176 = 1.3106. (b) To simulate (say) 500 observations of x, using the T183, we will first simulate 500 random
integers between 1 and 100 by using the command: (0—
+ randInt(1,100,500) —> L1 The command sortA(L1) sorts these random observations in increasing order. We now iden
tify 500 observations of x as follows: '0 Integers 1 to 3 correspond to x = 0
4 to 19 x = 1
20 to 49 x = 2
50 to 72 x = 3
73 to 89 x = 4
90 to 100 x = 5 Random Variables . 131 For a sample run of the simulation, we obtained 12 observations of x = 0
86 x =1
155 x = 2
118 x =3
75 x = 4
54 x =5 These data yield a sample mean and standard deviation of E = 2.64, s = 1.292, very close to
p. and 0'. 7.30 The graph for xmax = 10 displays visible variation for the first ten values of x, whereas the graph
for xmax = 100 gets closer and closer to ,u. = 64.5 as x increases. This illustrates that the larger the
sample size (represented by the integers 1, 2, 3, . . . . in L1), the closer the sample means x get to the
population mean p. = 64.5. (In other words, this exercise illustrates the law of large numbers in a
graphical manner.) 7.31 Below is the probability distribution for L, the length of the longest run of heads or tails. P(You
win) = P(run of 1 or 2) = % = 0.1738, so the expected outcome is p. = ($2)(0.1738) + (—$1)(0.8262)
= —$O.4785. On the average, you will lose about 48 cents each .time you play. (Simulated results
should be close to this exact result; how close depends on how many trials are used.) Value of L 1 2 3 4 5 6 7 8 9 10
6_  1 88 185 m 3 A. i i A ._1_
PTObablhtY m T m 51 512 512 51 512 512 512 7.32 (a) The wheel is not affected by its past outcomes—it has no memory; outcomes are inde—
pendent. So on any one spin, black and red remain equally likely. (b) Removing a card changes the composition of the remaining deck, so successive draws are
not independent. If you hold 5 red cards, the deck now contains 5 fewer red cards, so your
chance of another red decreases. 7.33 No: Assuming all “atbat"s are independent of each other, the 35% figure only applies to the
“long run” of the season, not to “short runs.” 7.34 (3) Independent: Weather conditions a year apart should be independent. (b) Not independ
ent: Weather patterns tend to persist for several days; today’s weather tells us something about
tomorrow’s. (0) Not independent: The two locations are very close together, and would likely have
similar weather conditions. 7.35 (a) Dependent: since the cards are being drawn from the deck without replacement, the
nature of the third card (and thus the value of Y) will depend upon the nature of the first two
cards that were drawn (which determine the value of (b) Independent: X relates to the outcome of the first roll, Y to the outcome of the second
roll, and individual dice rolls are independent (the dice have no memory). 7.36 The total mean is 40 + 5 + 25 = 70 minutes. 7.37 (a) The total mean is 11 + 20 = 31 seconds. (b) No: Changing the standard deviations does not affect the means. (0) No: The total mean does not depend on dependence or independence of
the two variables. 132 Chapter 7 7.38 Assuming that the two times are independent, the total variance is 010,312 = ape} + cat} = 22
+ 42 = 20, so atom] = V20 = 4.472 seconds. Assuming that the two times are dependent with cor—
relation 0.3 the total variance is atotalz = 01,052 + am2 + 2paposcratt = 22 + 42 + 2(0.3)(2)(4) = 24.8, so
atom = V 24.8 = 4.98 seconds. The positive correlation of 0.3 indicates that the two times have
some tendency to either increase together or decrease together, which increases the variability of
their sum. 4/ 7.39 Since the two times are independent, the total variance is am} = aﬁrst2 + o 2 = 22 + 12 = second
5, so atom] = \/5 = 2.236 minutes. 7.40 (a) 6,2: (300 — 445)2(0.4) + (500 — 445)2(0.5) + (750 — 455)2(0.1) = 19.225 and a, =
138.65 units. (b) a“; = 0X2 + 6,2 = 7,800,000 + 19,225 = 7,819,225, so am = 2796.29 units.
(C) 0'22 : O'ZOOOXZ ‘1‘ (73500172 = + (3500)20'Y2, SO 02 = m 7.41 If F and L are their respective scores, then F — L has a N(0, V 22 + 22% =JN(0, 2\/2) dis—
: <
a? t'b t' , P F—L >5 =P z > 1.7678 =0.0771 t bl~ 1 :0.07 8.. “W”*§V5")’
r1u10n SO 1 ) 1 9657.42 (a) Let X = the value of the stock after two days? e possible combinations of gains and loss
es on two days are presented in the table below, together with the calculation of the corre— '\/ sponding values of X.
W l M W 5!,» 1st day 2nd day Value of X Gain 30% Gain 30% 1000 + (.3)(1000) = 1300
f” 1300 + (.3)(1300) = 1690
Gain 30% Lose 25% 1000 + (.3)(1000) = 1300
1300 — (.25)(1300) = 975
Lose 25% Gain 30% 1000 — (.25)(1000) = 750
750 + (.3)(750) = 975
Lose 25% Lose 25% 1000 — (.25)(1000) = 750
750 T (.25)(750) = 562.50 ,1.» Since the returns on the two days are independent and P(gain 30%) = P(lose 25%) = 0.5, the
probability of each of these combinations is (.5)(.5) = .25. The probability distribution of X is
therefore x 1690 975 562.5
P(X=x) 0.25 0.5 0.25 The probability that the stock is worth more than $1000 = P(X = 1690) = 0.25.
(b) p. = (1690)(.25) + (975)(.5) + (562.5)(.25) = 1050.625, or approximately $1051. 7.43 The probability distribution of digits under Benford’s Law reﬂects longterm behavior. For
each digit v, P(V = v) z the long—term relative frequency of V, with the agcuracy of the approxi—
mation improving as the number of observations increases. We would expect a large number of
items to reﬂect Benford’s Law more accurately than a small number of items. 744 (a) First die: M = (1X6) + 5%) + (41(6) + (51(6) + (6X6) + (8X6) = 45
Second diet M = (1X6) + (ZXl/a) + (390/3) + (4X6) = 25
(b) The table on the facing page gives the distribution of X = sum of spots for the two dice.
Each of the 36 observations in the table has probability %. Random Variables 133 l 3 4 5 6 8 1 2 4 5 7 9 2 3 5 6 7 8 10 2 3 5 6 7 8 10 3 4 6 7 8 9 11 3 4 6 7 8 9 11 4 5 7 8 9 10 12 The probability distribution of X is: x 2 3 4 5 6 7 8 9 10 11 12
P<x=x)as—8:—zéaiaeaas—sa (C) It = (2)616) + (3)618) + me) + (5)(%) + (6)055) + (7)(%) + (8)(%) + (9)(%) +
(10)(r12) +(11)(fs) + (12)(3—16) = 7, Using addition rule for means: p. = mean from first die + mean from second die = 4.5 + 2.5
=~7. 7.45 (a) Randomly selected students would presumably be unrelated. (b) nf_ m = M — hm = 120 —
105 = 15. af_mz = 0% + 02m: 282 + 352 = 2009, so a'f_m = 44.82f‘(,c) Knowing only the mean and
standard deviation, we cannot find that probability (unless we assume that the distribution is nor—
mal). Many different distributions can have the same mean and standard deviation. 7.46 (a) Mx = 550° Celsius; of, = 32.5, so ox = 5.701°C. (b) Mean: 0°C; standard deviation: 5.701°C.
(c) ,1, = 2,1, + 32 = 1022°F, and a, = ga, = 10.26°F. 7.47 Read twodigit random numbers. Establish the correspondence 01 to 10 <:> 540°, 11 to 35 <=>
545°, 36 to 65 <=> 550°, 66 to 90 <=> 555°, and 91 to 99, 00 <=> 560°. Repeat many times, and record
the corresponding temperatures. Average the temperatures to approximate ,u.; find the standard
deviations of the temperatures to approximate 0'. 7.48 (a) The machine that makes the caps and the machine that applies the torque are
not the same. (b) T (torque) is N(7, 0.9) and .5 (cap strength) is N(10, 1.2), so T — S is N(—3, V0.92 + 1.22) = N(—3, 1.5). Then P(T > S) = P(T — s > 0) = P(Z > 2) = 0.0228. 7.49 (a) Yes: This is always true; it does not depend on independence. (b) No: It is not reasonable
to believe that X and Y are independent. 7.50 (a) R1 + R2 is normal with mean 100+250 = 3500. and s.d. \/2.52 + 2.8? = 3.75370.
(b) P(345 5 R1 + R2 5 355) = P(—1.3320 5 Z 3 1.3320) = 0.8172 (table value: 0.8164). 7.51 The monthly return on a portfolio of 80% Magellan and 20% Japan can be written as 0.8W +
0.2Y. The mean return is ,LLOBWWJY = (1.8,le + 0.2;Ly = (0.8)(1.14) + (0.2)(1.59) = 1.23%, which
is higher than MW. The variance of return is oﬁgwwzy = (0.8)201W + (0.2)2021; +
pry(0.8O'W)(0.Za'y) = (0.64)(4.64)2 + (0.04)(6.75)2 + 2(0.54)(08)(4.64)(0.2)(6.75)= 21.01354. The standard deviation of return is (row/+0.2}; = V 21.01354 = 4.584%, which is lower than ch. 7.52 Assuming that pWY = 0, the variance of return is 0%));me =(0.8)201W+ (0.2)201Y =
(0.64)(4.64)2+(O.04)(6.75)2= 15.601444, and the standard deviation is 0'0.SW+0.2Y:. \/ 15.601444 = 3.95%. This is smaller than the result 4.584% from Problem 7.51. The mean return is not affected by the zero correlation, since it only depends upon the means of the indi
vidual returns. 134 Chapter 7 7.53 The monthly return on a portfolio of 60% Magellan, 20% Real Estate, and 20% Japan can
be expressed as 0.6W + 0.2X + 0.2Y. The mean return is uoﬁwwtzxwly = 0.6p.W +
0.2p.X + 0.2,uy = (0.6)(1.14) + (0.2)(0.16) + (0.2)(1.59) = 1.034%. The variance of return is
00.6W+0.2X+0.2Y = (06)20'ZW + (0'2)2...
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