Download Midterm - University of New Brunswick

ADM 2623 Midterm I (11-02-2016)
Instructor : Donglei Du
Duration: 80 minutes
Instructions:
1. Please justify your answers by providing the methods and/or the procedures for your final results.
You will not get any credit for a single answer without proper process.
2. The exam is closed book and notes.
3. Duration: 80 minutes
4. Formulas are provided along with the exam papers (the last page).
5. There are 5 questions in total.
6. The total mark is 30.
Student Name:_______________________
Student ID________
________________
1. Problem 1. Short Explanations (8 marks).
a. (2 points) What is Descriptive Statistics?
Answer: It is collecting, organizing, and representing data in such a way
that the characteristics and patterns of the data can be easily captured.
b. (2 points) What is the difference between the general and special rule of addition?
Answer: general rule of addition: valid for any events
special rule of addition: valid for mutually exclusive events.
c. (2 points) What is Sample Space?
Answer: Sample Space: is the collection or set of all the possible outcomes of
a random experiment
d. (2 points) What is the difference between the Interval and the Ratio level of
measurement?
Answer: Interval data don’t true starting point (absolute zero).
2. Problem 2. (4 marks). A survey was taken of the amount of money spent at a farmer’s arket by 100
shoppers on a Saturday:
20
23
33
35
39
40
44
44
52
53
Generate a frequency table for this raw data
27
35
42
47
56
28
36
42
48
61
31
38
43
49
69
i. (1 point) Calculate the number of classes:
Solution: There are 25 numbers and hence according to the 2 to kth rule: k = 5
since 2^5 =32 >25, but 2^4 =16 <25.
ii. (1 point) Calculate the class width
Solution: w = (69-20)/5=10. and rounded to 10.
iii. (1 point) Calculate the lower limit of the first class
Solution: L= 20-(5(10)-49)/2=20 So choose 20.
iv. (1 point) draw a frequency distribution table below:
Solution:
class
freq
[20,30)
4
[30,40)
7
[40,50)
9
[50,60)
3
[60,70)
2
3. Problem 3. (6 marks). A real estate survey company selects a sample of 100 Condo sales in
New Brunswick for real estate research in 2016. The prices are grouped and shown in the table.
Weights (kg)
50 up to 75
75 up to 100
100 up to 125
125 up to 150
150 up to 175
Frequencies
x
fx
fx2
5
15
20
35
25
62.5
87.5
112.5
137.5
162.5
312.5
1312.5
2250
4812.5
4062.5
12750
19531.25
114843.75
253125
661718.75
660156.25
1709375
100
CF
5
20
40
75
100
a. (2 points) From the grouped data of this survey, calculate the mean for this grouped
data to estimate average condo price in New Brunswick.
Solution: mean = 12750/100 = 127.5(K)
b. (2 points) Calculate the median for this grouped data.
Solution: The median class is [125,150). So
Median = 125 +(100/2 - 40)*25/35 =132.14(K)
c. (2 points) Calculate the standard deviation for this grouped data.
Solution: Variance = (1709375 – (12750)^2/100)/99 = 845.96. So SD = 29.09K
Problem 4. (4 marks) In a study of pleas and prison sentences, it is found that 45% of the
subjects studied were sent to prison. Among those sent to prison, 25% chose to plead guilty.
Among those not sent to prison, 80% chose to plead guilty. If a study subject is randomly
selected and it is then found that the subject entered a guilty plea, find the probability that this
person was not sent to prison. If a study subject is randomly selected and it is then found that the
subject entered without guilty plea, find the probability that this person was not sent to prison. If
a person doesn’t want to be sent to prison, what decision he/she would like to make according to
the above information?
Solution: Let A1 be the event that sent to prison
Let A2 be the event that do not sent to prison.
Then P(A1) = 0.45 and P(A2) = 0.55.
Let B be the event that choose to plead guilty.
Let 𝐵̅ be the event that choose to do not plead guilty.
Then P(B|A1) = 0.25 and P(B|A2) = 0.8.
Then P(𝐵̅ |A1) = 0.75 and P(𝐵̅ |A2) = 0.2.
We are looking for the following conditional probabilities P(A2|B) and P(A2|𝐵̅), the one with
higher probability is more likely.
According to Bayes’ Theorem, we have
P(A2|B) = P(B|A2) P(A2)/( P(B|A2) P(A2)+ P(B|A1) P(A1))
=0.8* 0.55 /(0.55*0.8+ 0.45*0.25)= 0.44/0.5525= 0.79638009≈79.64%
P(A2|𝐵̅) = P(𝐵̅ |A2) P(A2)/( P(𝐵̅ |A1) P(A1)+ P(𝐵̅ |A2) P(A2))
=0.2*0.55/(.75*0.45+ 0.55*0.20)= 0.11/(0.3375+0.11)= 0.11/ 0.4475= 0.245810056
≈24.58%
79.64% > 24.58%
So plead guilty would be better choice according to this study.
Problem 5. (8 marks) Short calculations
a. (2 points). Listed below is the percentage increase in sales for the MG Corporation
over the last four years.
8.4% 12.3% -5.2% 8.6%
Determine the average annual increase (Geometric Mean) in the sales over this
four-year period.
Solution:
Average annual increase= (1.084*1.123*0.948*1.086)^0.25 -1=1.058-1=0.058
(5.81%)
b.
(2 points) A sample of teachers had a mean annual income of $45 000 with a
standard deviation of $2000. Using Chebyshev’s Theorem, what is the proportion of
faculty that earn more than $40,000 but less than $50 000?
Solution: mean =45K and SD = 2K. Note that k = 2.5. So according to the
Chebyshev's theorem, there is at least 1-1/2.5^2=0.84.
c. (2 points) Suppose in a student survey, 84% of students use Google for search engine
and 32% use Bing for search engine. 23% of person use both. What is the probability that
a person either use Google or Bing as search engine? What is the probability that a person
neither use Google nor Bing as search engine?
Solution: P (G ∪ B ) = P (G) + P (B ) − P (G ∩ B ) = 0.84 + 0.32 − 0.23 = 0.93
P (Not G ∩ Not B )= 1- P (G ∪ B )=0.07
d. (2 points) Suppose a financial analyst is doing research on two bonds X and Y. Consider the
following joint distribution of the rate of return for these two bonds during a randomly chosen
year
X
Y
Total
-1%
-5%
0.15
2%
0.1
5%
0.05
3%
0.05
0.15
0.05
8%
0.05
0.25
0.10
0.35
0.30
0.40
Total
0.30
0.25
0.45
1
(a) (1/35 marks) Are the rate of return for Bond X and Y independent? Why?
Solution: p(x=-1)*p(y=-5)=(0.25)(0.3)=0.075
p(x=-2 and y=-5)=0.15  0.075
x and y are not independent.
(b) (1/35 marks) Compute E(X) and E(Y)
Solution: E(x)= xp(x) = 2.85%

E(y)=
 yp( y ) =2.45%
Formula
22. E[ X ] 

 x P( x )
i 1
i
i
 

23. V [ X ]   x P( xi )    xi P( xi ) 
i 1
 i 1


2
i
2