Econ_221-2_F2007_Assignment_2_Answers
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Concordia University
Department of Economics
ECON 221/2 – Sections A, B, BB, C
Instructors: N. Islam, I. Jamali, G. Tsoublekas
Fall 2007 - ASSIGNMENT 2 - ANSWERS
Due at noon on Friday, October 12, 2007
1.
In a recent survey, 90 top executives were asked how many hours they spend each year in community service. The data are presented below. [The calculations and the graph for this exercise must be done on Excel. As soon as you get the data on class intervals and frequencies that are shown below, you will create a table
on Excel with the necessary detailed column calculations that will be used as a basis to answer sub-questions (a), (d) and (e).
You will paste your table here.]
(20 points)
a)
Construct a cumulative frequency distribution, and a cumulative relative frequency distribution. (3 points)
See table above.
b)
Using Excel, draw a histogram of the frequencies and paste it in the space provided below. (4 points)
c)
On the basis of the shape of the histogram, make a rough assessment about symmetry in the data and the implication for the mean time spent in community services. (3 points)
The histogram indicates a fairly symmetric distribution. Therefore, the mean is not distorted by skews (or outliers) in the data.
d)
Using the appropriate formula for grouped data, calculate the mean time spent in community services. Use the table provided above for your computations. (3 points)
Using the data calculated above,
Mean =
m
i
* f
i
/ n = 6300 / 90 = 70
e)
Using the appropriate formulas for grouped data, calculate the variance and the standard deviation. Use the table provided above for your computations. (4 points)
Using the data calculated above,
Variance = s
2
=
(m
i
–mean)
2
f
i
/ ( n-1) = 76000 / 89 = 853.93
Standard Deviation =
853.93 = 29.22
f)
Calculate the Coefficient of Variation. (3 points)
The coefficient of variation expresses in a percentage form the standard deviation in terms of the mean.
CV = (s / mean) * 100 = ( 29.22 / 70)*100 = 41.7%
2.
A general contractor has submitted two bids for two projects; A
and B
. The probability of getting project A
is 0.60. The probability of getting project B
is 0.75. The probability of getting at least one of the projects is 0.85. (15 points)
a.
Construct the table or tree of joint probabilities. (3 points)
We know that P (YA
YB) = P (YA) + P (YB) – P (YA
YB) = 0.60 + 0.75 – P (YA
YB) = 0.85
Therefore, P (YA
YB) = 0.50
Project B
Project A
Total
Yes (YA)
No (NA)
Yes (YB)
No (NB)
0.50
0.10
0.25
0.15
0.75
0.25
Total
0.60
0.40
1.00
2
b.
What is the probability that this contractor will not get any project? (3 points)
P (NA
NB) = P (NA) * P (NB
NA) = (0.40)*(0.15 / 0.40) = .15 = 15%
c.
If this contractor gets project B, what is the probability that he will not get project A? (3 points)
P (NA
YB) = P (NA
YB) / P (YB) = 0.25 / 0.75 = 0.333 = 33.3%
d.
If there is a limit on the number of projects awarded to each contractor, what is the probability that this contractor will get only one (anyone) project? (3 points)
P [(YA
NB)
(YB
NA)] = P (YA
NB) + P (YB
NA) – P [(YA
NB)
(YB
NA)] =
= 0.10 + 0.25 – 0.0 = 0.35
e.
If the probabilities of getting these projects are the same for all contractors who bid for them, what is the probability that two randomly chosen contractors who are bidding for them will get one contract each?
(3 points)
The probability that a contractor will get one project is found to be 0.35. Therefore, the probability that two randomly chosen contractors who are bidding for them will get one contract each
is (0.35)
2
= 0.1225 or 12.25%
3.
Your uncle has asked you to analyze his stock portfolio, which contains 10 shares of stock A and 10 shares of stock B. The joint probability distribution of the stock prices is shown below: (30 points)
a.
Find the probability that the price of stock A will be $40. (2 points)
P
A
(40) = 0.00+ 0.05 + 0.10 + 0.20 = 0.35
3
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