Q1.
You are trying to develop a strategy for investing in t different stocks. The anticipated annual return for a $1,000 investment in each stock under four different economic conditions has the following probability distribution: Probabilities & Outcomes: recession slow growth moderate growth fast growth
a. b. c. d.
P
X
Y
0.1 0.3 0.3 0.3
-100 0 80 15 150
50 50 150 -20 -100
Compute the expected return for X and Y. Compute the standard deviation for X and Y. Compute the covariance of X and Y. Would you invest in X or Y? Explain. N
∑1 X P( X ) = 59
(a) E ( X X ) =
i
i
i=
N
E (Y ) =
∑1 Y P ( Y ) = 14 i
i
i=
(b) σ X
N
∑1 X − E ( X )
=
i
2
P ( Xi )
= 78.6702
i=
σ Y
N
∑1 Y − E ( Y )
=
i
2
P ( Yi )
= 99.62
i=
N
(c)
σ
XY
=
∑1 X i=
i
− E ( X ) Yi − E ( Y ) P ( X i Yi ) = 6306 6306
X gives the investor a lower standard deviation while yielding a higher (d) Stock X expected return so the investor should select stock X X .
Q2. Half the portfolio assets are invested in X and half in Y. E(X)=$105, E(Y)=$35, σ X
= 14, 725, σ = 11, 025 , Y
σ
XY
= −12,675 .Calculate the portfolio expected return
and risk if a. 30% of the portfolio assets are invested in in X and 70% in Y. in X and 30% in Y. b. 70% of the portfolio assets are invested in c. Which of the two investment strategies (30%, 70% in X) would you recommend? (a)
E ( P P ) = 0.3(105) + 0.7(35) = $56 σ P
= (0.3) (14,72 ,725) + (0.7) (11,02 ,025) + 2(0.3)(0.7)(− 12,67 ,675) = $37.47 2
2
37.47 ( 100% ) = 66.91% 56 E ( P ) P ) = 0.7(105) + 0.3(35) = $84 (b) E ( P CV =
σ P
=
σ P
= (0.7) (14,72 ,725) + (0.3) (11,02 ,025) + 2(0.7)(0.3)(− 12,67 ,675) = $53.70 2
2
53.70 ( 100% ) = 63.93% 84 E ( P ) (c) Investing 70% in X will yield the lower risk per unit average return. CV =
σ P
=
Q3. A study showed that in 2004 only 64% of US income earners aged 15 and older had
a bank account. If a random sample of 20 US income earners aged 15 and older is selected, what is the probability that: a. All All 20 20 hav havee a bank bank accou account nt?? b. No more more than than 15 have have a bank bank account account?? c. More More tha thann 10 hav havee a bank bank acc accoun ount? t? The assumptions needed are (i) there are only two mutually exclusive and collective exhaustive outcomes – “have a bank account” and “do not have a bank account”, (ii) the probabilities of “have a bank account” and “do not have a bank account” are constant, and (iii) the outcome of “have a bank account” from one income earner is independent of the outcome of “have a bank account” from any other income earners. Data Sample size Probability of an event of interest Binomial Probabilities Table
20 0.64
X 10 15 20
P(X) 0.07 .07788 0.11 .11605 3 0.00 .00013 3
P(<=X) 0.142 .1423 399 0.89 0.8989 8937 37 1
P(X) 0.06 .064519 0.857 .8576 601 0.78 0.7828 2885 85 0.10 0.1010 1063 63
P(>=X) 0.9 0.935481 0.21 0.2171 7115 15
0.999867
0. 0 .000133
P ( X X = 20) = 0.0001329
(a)
n! p (1− p) p) − X!(n − X)! 20! = (0.64) (1 − 0.64) 20!(20 − 20)! = (1)(0.64) = 0.0001329
P(X = 20) =
X
n X
20
20
20 −20
0
(b)
P ( X X ≤ P ( X X ≤
15) = 0.8989 X > 15)=1-[P(X=16)+ P(X=17)+ P(X=18)+ 15)=1- P ( X P(X=19)+P(X=20)]
(c)
P ( X X > 10) = 0.8576 (similarly calculated)
Q4. When the cookie production process is in control, the mean number of chip parts per
cookie is 6. What is the probability that in any particular cookie being inspected: a. Less Less than than four four chip chip parts parts wil willl be found? found? b. Exactl Exactlyy three three chip chip parts parts will will be be found? found? c. Four or more more chip chip parts parts will will be be found found?? d. Either Either two two or three three chip chip parts parts will will be foun found? d? P ( X ) =
− e λ λ x
X !
a. If λ = 6.0, P ( X X < 4) = P ( X X = 0) + P ( X X = 1) + P ( X X = 2) + P ( X X = 3) = 0.0025 + 0.0149 + 0.0446 + 0.0892 = 0.1512 b. P ( X X = 3)= 0.0892 c. P ( X <4)=1-0.1512 X ≥ 4)= 1- P ( X X <4)=1-0.1512 X = 2) + P ( X X = 3)= 0.0446 + 0.0892 d. P ( X Q5. Assume that the number of flaws per foot in rolls of grade 2 paper follows a Poisson
distribution with a mean of 1 flaw per 5 feet of paper (0.2 flaw per foot). What is the probability that in a : a. 1-foot 1-foot roll roll,, there there will will be at least least 2 flaws? flaws? b. 12-foot 12-foot roll, roll, there there will will be be at least least 1 flaw flaws? s? c. 50-foot roll, there will will be greater greater than than or equal equal to 5 and and less than than or equal to 15 flaws? Poisson Probabilities Data Aver Averag age/ e/Ex Expe pect cted ed nu numb mber er of succ succes esse ses: s: Poisson Probabilities Table X P(X) 2 0.01637 5 X ≥ a. If λ = 0.2, P ( X
P(<=X) 0.99885 2
0.2 0.2
P(
P(>X) 0.00114 8
P(>=X) 0.01752 3
2) = 1 – [ P P ( X X = 0) + P ( X X = 1)] = 1 – [0.8187 + 0.1637] = 0.0176
b. If there are 0.2 flaws per foot on the average, then there are 0.2•(12) or 2.4 flaws
on the average in a 12-foot roll,
λ = 2.4.
Poisson Probabilities Data Aver Averag age/ e/Ex Expe pect cted ed nu numb mber er of succ succes esse ses: s:
2.4 2.4
Poisson Probabilities Table X P(X) P(<=X) P(X) 0 0.09071 0.09071 0.00 0.0000 0000 00 0.90 0.9092 928 8 8 8 2 1 0.21772 0.30844 0.09 0.0907 0718 18 0.69 0.6915 155 5 3 1 9 If λ = 2.4, P ( X X ≥ 1) = 1 – P ( X X = 0) = 1 – 0.0907 = 0.9093
P(>=X) 1.00000 0 0.90928 2
c. If there are 0.2 flaws per foot on the average, then there are 0.2•(50) or 10 flaws on
the average in a 50-foot roll,
λ = 10.
Poisson Probabilities Data Aver Averag age/ e/Ex Expe pect cted ed nu numb mber er of succ succes esse ses: s: Poisson Probabilities Table X P(X) 5 0.03783 3 6 0.06305 5 7 0.09007 9 8 0.11259 9 9 0.12511 0 10 0.12511 0 11 0.11373 6 12 0.09478 0 13 0.07290 8
P(<=X) 0.06708 6 0.13014 1 0.22022 1 0.33282 0 0.45793 0 0.58304 0 0.69677 6 0.79155 6 0.86446 4
10
P(
P(>X) 0.93291 4 0.86985 9 0.77977 9 0.66718 0 0.54207 0 0.41696 0 0.30322 4 0.20844 4 0.13553 6
P(>=X) 0.97074 7 0.93291 4 0.86985 9 0.77977 9 0.66718 0 0.54207 0 0.41696 0 0.30322 4 0.20844 4
14
0.05207 7 0.03471 8
15
0.91654 2 0.95126 0
0.86446 4 0.91654 2
0.08345 8 0.04874 0
0.13553 6 0.08345 8
Sum = 0.9220 If λ = 10, P (5 (5 ≤ X ≤ 15) = 0.9220 Q6. The increase or decrease in the price of a stock between the beginning and the end of
a trading day is assumed to be an equally likely random event. What is the probability that a stock will show an increase in its closing price on five consecutive days? Given p = 0.5 and n = 5, P ( X X = 5) = 0.0312. n! p (1− p) p) − X!(n − X)! 5! = (0.5) (1 − 0.5) − 5!(5 − 5)! = (1)( (1)(0. 0.5) 5) (1) (1) = 0.0312
P(X = 5) =
X
n X
5
5
5 5
