Economics 391 (Spring 2013) Professor Lamarche, University of Kentucky Sample Midterm Exam Questions 1. Review Examples 5, 8, and 10 in Lecture 1. 2. Review Examples 1, 3, 5 and 9 in Lecture 2. 3....

SOLUTIONS
5. Consider an investment portfolio of $50,000 in stock A and $50,000 in stock B. The expected value of
A is 9.5% and B is 6%. The variance of A is 13% and the variance of B is 8%. The covariance between A
and B is 18.6%.
a) Since the amount of investment is same in both the stocks, we assign equal weights to both the
stocks.
Weight to Stock A = 0.50
Weight to Stock B = 0.50
b) Portfolio Expected Return (E(RP) ) = wARA + wBRB = 0.50*9.5% + 0.50*6% = 4.8% + 3.0% = 7.8%
E (A) = 50,000*9.5% = $ 4,750 and E (B) = = 50,000*6% = $ 3,000
Portfolio expected return = 7.8%
c) Portfolio variance σp2 = wA
2σA
2 + wB
2σB
2 + 2Cov(A,B) =
= (0.5^2)(0.13) + (0.5^2)(0.08) + 2(0.5)(0.5)(0.186) = 0.033 + 0.020 + 0.093 = 0.146 = 14.6%
6. Let p be the probability of workers being sure that they will be able to retire at 65 years of age and X
be the number of workers retiring at 65.
Using binomial distribution with n =10, p =0.80 and q =1- p = 1 – 0.80 = 0.20
( ) ∑(


) ( )
a) P (X = 0) = ∑ (

)( ) ( )

= 0.0000001
b) P (X = 2) = ∑ (

)( ) ( )

= 0.0000779
c) Expected Value (X) = np = 10*0.80 = 8

Variance = npq = 10*0.80*0.20 = 1.60
Standard deviation = √Variance = √1.60 = 1.265
7. Average visit per year = 4
a) Average visit per month = 4/12 = 1/3 =0.33
Mean = np = 12*0.33 = 3.96 or 4
Standard deviation = √Variance = √npq = √8/3 = 1.63
b) P (X ≥ 1) = 1 – P (X <1) = 1 – P(X =0)
P (X = 0) = ∑ (

)( ) ( )

= 0.00818
P (X ≥ 1) = 1 – 0.00818 = 0.9918
8. P (Y > 10) = 0.16 and P (5 ≤ Y ≤ 10) = 0.15
a. P (Y < 10) = 1 – P (Y> 10) = 1 – 0.16 = 0.84
b. P (Y< 5) =
Since P (5 ≤ Y ≤ 10) = 0.15, we have
P (Y ≤ 10) – P (Y ≥5) = 0.15 => 0.84 – 0.15 = P (Y ≥5)
P (Y ≥5) = 0.69
Therefore, P (Y< 5) = 1- 0.69 = 0.31
c. P (Y = 0) = 1 – 0
9. Expected value of the antique gift = $ 2,000 * 0.4 + $ 200 *0.5 + $ 20*0.10
= $ 800 + $ 100 + $ 2 = $ 902
Since the expected value of the antique gift is $ 902 which is less than $ 1,000 cash, Jenny should accept
cash instead of the gift.
10. Mean time = 25 minutes
Using exponential distribution,
f(x) = λe-λx, 0 where E(X) = 1/λ
When mean = 25 minutes, then λ = 0.04
a) P (no arrivals for more than one hour or 60 minutes) = P (X ≤ 60)
f(X ≤ 60)) = 0.04e-0.04*60
P (X ≤ 60) = 0.9093
b) P (Arrival within 10 minutes) = P (X ≤ 10)
(X ≤ 10)) = 0.04e-0.04*10
P (X ≤ 10) = 0.3297
11.
a) The sample mean x is what ou compute from our random sample in order to STIMATE the
population mean μ.
The sample mean is summed over the sample size n. The population mean is summed over the
population size N or, if you knew the whole population. Usually you do not, which is why you take a
sample in the first place.
b) Concept of estimation Biases: The bias (or bias function) of an estimator is the difference between
this estimator's expected value and the true value of the parameter being estimated. An estimator or
decision rule with zero bias is called unbiased. Otherwise the estimator is said to be biased.
c) Properties of Sampling...