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:

P X Y

recession 0.1 -100 50slow growth 0.3 0 150moderate growth 0.3 80 -20fast growth 0.3 150 -100

a. Compute the expected return for X and Y.b. Compute the standard deviation for X and Y.c. Compute the covariance of X and Y.d. Would you invest in X or Y? Explain.

(a) E(X) = = 59

E(Y) = = 14

(b)

= = 78.6702

= = 99.62

(c) XY = = 6306

(d) Stock X gives the investor a lower standard deviation while yielding a higher expected return so the investor should select stock X.

Q2. Half the portfolio assets are invested in X and half in Y. E(X)=$105, E(Y)=$35, , XY = .Calculate the portfolio expected return and

risk ifa. 30% of the portfolio assets are invested in X and 70% in Y.b. 70% of the portfolio assets are invested in X and 30% in Y.c. Which of the two investment strategies (30%, 70% in X) would you recommend?

(a) E(P) = 0.3(105) + 0.7(35) = $56

(b) E(P) = 0.7(105) + 0.3(35) = $84

(c) Investing 70% in X will yield the lower risk per unit average return.

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 20 have a bank account?b. No more than 15 have a bank account?c. More than 10 have a bank account?

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.

DataSample size 20Probability of an event of interest

0.64

Binomial Probabilities Table

X P(X) P(<=X) P(<X) P(>X) P(>=X)10 0.07788 0.142399 0.064519 0.857601 0.93548115 0.11605

30.898937 0.782885 0.101063 0.217115

20 0.000133

1 0.999867 0 0.000133

(a) P(X = 20) = 0.0001329

(b) P(X 15) = 0.8989P(X 15)=1- P(X > 15)=1-[P(X=16)+ P(X=17)+ P(X=18)+ P(X=19)+P(X=20)]

(c) P(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 than four chip parts will be found?b. Exactly three chip parts will be found?c. Four or more chip parts will be found?d. Either two or three chip parts will be found?

a. If = 6.0, P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0025 + 0.0149 + 0.0446 + 0.0892 = 0.1512

b. P(X = 3)= 0.0892c. P(X 4)= 1- P(X<4)=1-0.1512d. P(X = 2) + P(X = 3)= 0.0446 + 0.0892

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 roll, there will be at least 2 flaws?b. 12-foot roll, there will be at least 1 flaws?c. 50-foot roll, there will be greater than or equal to 5 and less than or equal to 15

flaws?

Poisson Probabilities

DataAverage/Expected number of successes: 0.2

Poisson Probabilities TableX P(X) P(<=X) P(<X) P(>X) P(>=X)2 0.01637

50.99885

20.98247

70.00114

80.01752

3

a. If = 0.2, P(X 2) = 1 – [P(X = 0) + P(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, = 2.4.

Poisson Probabilities

DataAverage/Expected number of successes: 2.4

Poisson Probabilities TableX P(X) P(<=X) P(<X) P(>X) P(>=X)0 0.09071

80.09071

80.000000 0.90928

21.00000

01 0.21772

30.30844

10.090718 0.69155

90.90928

2If = 2.4, P(X 1) = 1 – P(X = 0) = 1 – 0.0907 = 0.9093

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, = 10.

Poisson Probabilities

DataAverage/Expected number of successes: 10

Poisson Probabilities TableX P(X) P(<=X) P(<X) P(>X) P(>=X)5 0.03783

30.06708

60.02925

30.93291

40.97074

76 0.06305

50.13014

10.06708

60.86985

90.93291

47 0.09007

90.22022

10.13014

10.77977

90.86985

98 0.11259

90.33282

00.22022

10.66718

00.77977

99 0.12511

00.45793

00.33282

00.54207

00.66718

010 0.12511

00.58304

00.45793

00.41696

00.54207

011 0.11373

60.69677

60.58304

00.30322

40.41696

012 0.09478

00.79155

60.69677

60.20844

40.30322

413 0.07290

80.86446

40.79155

60.13553

60.20844

4

14 0.052077

0.916542

0.864464

0.083458

0.135536

15 0.034718

0.951260

0.916542

0.048740

0.083458

Sum = 0.9220If = 10, P(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 = 5) = 0.0312.