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Last Name _________________________ First Name _________________________ ID _________________________
Treatment of Experimental Data 85-222 Winter 2005
Faculty of Engineering
University of Windsor
Midterm Exam I Solution
Friday, February 11, 11:30 am – 1:30 pm, Ambassador Auditorium
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and text (no notes).
Time available: 2 hour
Instructions:
 This solution has 9 pages.
 Please be sure to put your name and student ID on each odd-numbered page.
 State answers upto four decimal places.
 Show your work.
Grading:
Question
Marks:
1
/10
2
/7
3
/9
4
/5
5
/8
6
/5
7
/4
8
/8
9
/9
Total:
/65
Name:_________________________________________________
ID:_________________________
Question 1: (10 points) Circle the most appropriate answer
1.1 Descriptive statistics involves data
a. collection
b. organizing or summarizing
c. presentation
d. all of the above
1.2 Nominal data
a. allow all arithmetic operations
b. convey ranking
c. represents arbitrary codes
d. represents quantity or amounts of something such as length, weight, etc.
1.3 Which of the following is an advantage of sampling?
a. Accuracy of information
b. Cost of data collection
c. Both
d. None
1.4 A bar chart
a. emphasizes trend, if any
b. emphasizes relative values e.g., frequencies
c. can show order of categories
d. b and c
1.5 The following is the most suitable measure of central tendency for ranked data
a. Mean
b. Median
c. Mode
d. b and c
1.6 Defining sample variance as the mean squared deviation from the sample mean tends to
a. Underestimate the population variance
b. Overestimate the population variance
c. Accurately estimate the population variance
d. b and c
1.7 For positively skewed data
a. mean > median
b. mean = median
c. mean < median
d. there are usually two modes
1.8 If A and B are two mutually exclusive events
a. P(A)+P(B) = 1
b. P(A and B) = P(A)×P(B)
c. P(A or B) = P(A)+P(B)
d. P(A|B) = P(A)
2
Name:_________________________________________________
ID:_________________________
1.9 The reliability increases
a. if the components are in parallel and the number of components increases
b. if the components are in series and the number of components increases
c. if the components are in parallel and the number of components decreases
d. b and c
1.10
a.
b.
c.
d.
An example of a continuous random variable is the one that assumes value of
number of defective parts in a production lot
time between two customers arriving in a bank
number of accidents per month
b and c
Question 2: (7 points) Descriptive statistics
A major airline wanted some information on those enrolled in their “frequent flyer” program. A sample
of 20 members resulted in the following number of miles flown, to the nearest 1000 miles, by each
participant.
21
19
23
22
20
19
17
22
14
20
19
11
16
16
13
18
12
15
23
24
a. (2 points) Construct a frequency distribution table for the data, using five class intervals and the
value 10 as the lower limit for the first class.
Class
largest number - smallest number
Class width 
Interval
Frequency
Number of classes
10
to
13
2
24  10

 2.8  3
13 to 16
3
5
16 to 19
4
19 to 22
6
22 to 25
5
b. (4 points) Construct a relative frequency histogram for the data, using five class intervals and the
value 10 as the lower limit for the first class.
Relative Frequency Histogram
0.35
Frequency
2
3
4
6
5
Relative
Frequency
0.10
0.15
0.20
0.30
0.25
Relative Frequency
Class
Interval
10 to 13
13 to 16
16 to 19
19 to 22
22 to 25
0.3
0.25
0.2
0.15
0.1
0.05
0
13
16
19
22
25
Miles flown (in 1000)
c. (1 point) Comment if the data is symmetric, positively skewed, or negatively skewed. Justify your
answer in brief.
The data is negatively skewed. The mean is less than the median. There are more larger
numbers than smaller numbers.
3
Name:_________________________________________________
ID:_________________________
Question 3: (9 points) Central location and box plot
The following are numbers of twists that were required to break 11 forged alloy bars: 24, 39, 48, 26,
35, 38, 54, 23, 34, 29, and 37. Find:
a. (1 point) Mean
n
X 

x
i 1
i
n

24  39  48  26  35  38  54  23  34  29  37
11
387
11
=35.1818
b. (2 point) The 30th percentile
First, sort the data: 23, 24, 26, 29, 34, 35, 37, 38, 39, 48, 54
d  0.30, k  n  1d  11  10.30  3.6, Hence, k  3
Q0.30  xk  n  1d  k xk 1  xk   x3  3.6  3x4  x3   26  3.6  329  26  27.8
c. (3 points) The first, second and third quartiles.
The 1st quartile: d  0.25, k  n  1d  11  10.25  3, whole number, so Q0.25  xk  x3  26
The 2nd quartile: d  0.50, k  n  1d  11  10.50  6, whole number, so Q0.50  xk  x6  35
The 3rd quartile: d  0.75, k  n  1d  11  10.75  9, whole number, so Q0.75  xk  x9  39
d. (3 points) Construct a box plot.
20
30
40
Number of Twists
4
50
60
Name:_________________________________________________
ID:_________________________
Question 4: (5 points) Variation
A. A. Michelson (1852-1931) made many series of measurements of the speed of light. Using a
revolving mirror technique, he obtained
12, 30, 30, 27, 30, 39, 18, 27, 48, 24, 18
for the differences (velocity of light in air) – (299,700) km/s. Find:
a. (1 point) Range
Range = Largest value – smallest value = 48 –12 = 36
b. (3 points) Variance
X 
12  30  30  27  30  39  18  27  48  24  18 303

 27.5454
11
11
 X
n
s2 
i 1
i
X

2
n 1
2
2
2
2
1 12  27.5454  330  27.5454  227  27.5454  39  27.5454 



11  1  218  27.54542  48  27.54542  24  27.54542

1004.7272
10
 100.4727

Marking note: The sample variance must be computed for experimental data. It’s incorrect to
compute population variance. 1 point is taken off if denominator shows n instead of n  1.
c. (1 point) Standard deviation
s  s 2  100.4727  10.0236
5
Name:_________________________________________________
ID:_________________________
Question 5: (8 points) Probability laws
Consider the following information about 500 machine parts which are inspected before shipping:
The machine part is
improperly assembled
The machine part is
properly assembled
(A)
(AC)
10
5
15
20
465
485
30
470
500
The machine part
contains one or more
defective components
Total
(D)
The machine part
contains no defective
component
(DC)
Total
Find:
a. (2 points) P A
P  A 
30
 0.06
500
 
b. (2 points) P D C
 
P DC 
485
 0.97
500
c. (2 points) P A or D
P( A or D) 
10  5  20 35

 0.07
500
500
Alternately, P( A or D)  P A  PD   P A and D  
30
15
10


 0.07
500 500 500
Marking note: Since, there are some cases (10) when the events A and D occur simultaneously,
the events are not mutually exclusive. If addition law is used to solve the problem, the general
addition law must be used. It’s incorrect to use the addition for mutually exclusive events. So, 1
point is taken off for missing  P A and D .
d. (2 points) P A | D
P A | D  
10 2
  0.6667
15 3
Alternately,
10
P A and D  500 10 2
P A | D  


  0.6667
15 15 3
P D 
500
6
Name:_________________________________________________
ID:_________________________
Question 6: (5 points) Probability trees
The Olive Construction Co. is determining whether it should submit a bid for the construction of a
new shopping center. In the past, Olive’s main competitor, Base Construction Co., has submitted
bids 60% of the time. If Base Construction Co. does not bid on a job, the probability that the Olive
Construction Co. will get the job is 0.80; if Base Construction Co. does bid on a job, the probability
that the Olive Construction Co. will get the job is 0.25.
a. (4 points) Construct a probability tree showing all the probabilities, simple events and joint
probabilities.
Simple Joint
Stage 1
Define the following events:
Stage 2
B : Base construction bids on the job
=0.25
P(O|B)
B
0.
6
O : Olive Construction gets the job
P(
B)
=
P(O c|B
)=
0.75
c
B
P(
40
0.
)=
B
c )=0.50
P(O|B
c
P(O c|B c
)
=0.50
Events Probabilities
O
BO
P(BO)=(0.6)(0.25)=0.15
Oc
BOc
P(BOc)=(0.6)(0.75)=0.45
O
BcO
P(BcO)=(0.4)(0.80)=0.32
Oc
BcOc
P(BcOc)=(0.4)(0.20)=0.08
b. (1 point) What is the probability that the Olive Construction Co. will get the job?




POlive gets job   P BO or B C O  PBO   P B C O  0.15  0.32  0.47
Question 7: (4 points) Reliability
Compute the reliability of the following system:
0.80
0.90
0.60
0.85
0.60
0.50
0.95
0.75
Rsystem
 0.901  1  0.801  0.60  0.60 1  0.50 1  1  0.851  0.750.95
 0.901  1  0.801  0.361  0.50 1  1  0.851  0.750.95
 0.901  0.20  0.64  0.501  0.15  0.250.95
 0.90  0.936  0.9625  0.95
 0.77027
7
Name:_________________________________________________
ID:_________________________
Question 8: (8 points) Expected value and variance
Let X be a random variable with the following probability distribution:
-10
-5
0
5
x
0.10
0.15
0.25 0.20
0.30
p(x)
Compute
a. (2 points) E(X)
10
0.25
  E X    xpx   100.10   50.15  00.20  50.30  100.25  2.25
b. (2 points) Var(X)
Var(X)
  x    px 
2
  10  2.25 0.10   5  2.25 0.15  0  2.25 0.20  5  2.25 0.30  10  2.25 0.25
 15.00625  7.8844  1.0125  2.2688  15.015625
2
2
 41.1876
2
2
2
 
Alternately, Var(X) = E X 2   2
E X 2    x 2 px    10 0.10   5 0.15  0 0.20  5 0.30  10 0.25  46.25
2
2
2
2
Var(X) = E X 2    2  46.2500  2.25  46.2500  5.0625  41.1875
2
c. (1 point) E(3X+2)
E3X  2  3E X   2  32.25  2  8.75
d. (1 point) Var(3X+2)
Var(3X+2) = 3 Var  X   941.1876  370.6884
2
e. (2 points) E(3X2+4)


 
E 3 X 2  4  3E X 2  4  346.25  4  142.7503
Alternately,

E 3X 2  4
  4
 3E X


2

 3 Var X    2  4


 3 41.1876  2.25  4
 346.2501  4
2
 138.7503  4
 142.7503
8
2
Name:_________________________________________________
ID:_________________________
Question 9: (9 points) Binomial distribution
A study of 5-year trends in the logistics information systems of industries found that the greatest
computerization advances were in transportation (Industrial Engineering, July 1990). Currently, 40%
of all industries contain shipping open order files in their computerized data base. In a random
sample of 10 industries, let y equal the number that include shipping open order files in their
computerized data base. Note that the probability distribution of y can be modelled using the
binomial distribution. Find:
a. (3 points) P y  1 using Binomial distribution formula
n  10,   0.40
P y  1  b1; n,    b1;10,0.4   C110 0.40 1  0.40
101
1
10!
0.401 0.609
1!10  1!
 100.400.01
 0.0403

b. (3 points) P y  1 using Binomial distribution formula
P y  1  B1; n,    b0; n,    b1; n,    b0;10,0.4  0.0403  C010 0.40  1  0.40
0
10!
0.400 0.6010  0.0403  10! 10.6010  0.0403
0!10  0!
110!
 110.0060  0.0403
 0.0463

c. (1 point) P y  7 using Binomial distribution table
P y  7 = 0.9877 from Table A, Appendinx A
d. (1 point) P y  5 using Binomial distribution table
P y  5  1  P y  4  1  0.6331 (from Table A, Appendinx A)
 0.3669
e. (1 point) P5  y  7 using Binomial distribution table
P5  y  7  P y  7  P y  4  0.9877  0.6331 (from Table A, Appendinx A)
 0.3546
9
100
 0.0403