Solution to final exam engineering statistics 2014 2015

Solution to Engineering Statistics Exam 2014-2015 Dam and Water Resources Department
Page 1 of 12
Solution to Final Exam (2014-2015)
Salahaddin University Date 01/06/2015
College of Engineering Time: 3 Hours
Dams & Water Resources Eng. Dept. Subject: Engineering Statistics
Class: 2rd
Year Lecturer: M. Chener S. Qadr
Q1: (28%)
The following series is the minimum monthly flow in m3 .
S-1
over the period 1947 to 2007 of a certain
river.
21 36 4 16 21 21 23 11 46 10
25 12 9 16 10 6 11 12 17 3
11 26 2 6 12 22 13 11 26 20
26 12 8 15 7 10 8 13 19 22
57 37 29 36 38 31 58 63 72 83
86 88 88 83 14 7 9 95 73 69
A) Construct frequency distribution table for the data, including relative and cumulative frequency,
class boundaries and class midpoint.
For the First CLASS: the LOWER CLASS LIMIT = 0
(10%)
B) Calculate Mean, Median, Mode, and Standard deviation for the data (10%)
C) Present the data in form of Ogive, and Pie Chart. (8%)
N.B: More than one solution is acceptable for Q-1- Part A
Solution to Q-1:
Q-1-A
Number of data = 60
Lowest value = 2
Highest Value =95
60, 2, 95
1 3.322log(N)
1 3.322log(60) 6.907 7
95 3
13.28 15
7
,7
N number of data Lowest value Highest value
Number of Classes
Number of Classes use classes
Highest value lowest value
Class width use
N
Use classes with class wi
   
 
  
 
  
 15, lim 0 ( )dth of Lower class it of the First Class GIVEN
The lower Class Limit of Successive Class = Lower Class limit of Proceeding Class + Class Width
The lower Class Limit of 2nd
Class = 0 + 15 =15
The lower Class Limit of 3rd
Class = 15+15=30 and So on.
Lower Class Boundary of Class = (lower Class limt of the same class + Upper Class limit of the
proceeding Class) /2
Lower Class boundary of 2nd
Class = (15+14)/2 = 14.5

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Lowe
r
Class
Limit
Upper
Class
Limit
Freq. Relative
Freq.
Classes Cum.
Frequency
Lower Class
Boundary
Upper Class
Boundary
Class
Midpoi
nt
Fx Fx2
1 0 14 25 41.7 Less than 15 25 -0.5 14.5 7 175 1225
2 15 29 17 28.3 Less than 30 42 14.5 29.5 22 374 8228
3 30 44 5 8.3 Less than 45 47 29.5 44.5 37 185 6845
4 45 59 3 5 Less than 60 50 44.5 59.5 52 156 8112
5 60 74 4 6.7 Less than 75 54 59.5 74.5 67 268 17956
6 75 89 5 8.3 Less than 90 59 74.5 89.5 82 410 33620
7 90 104 1 1.7 Less than 105 60 89.5 104.5 97 97 9409
Sum (N) =60 1665 85395
Q-1-B
a) Mean
60 ( ) 1665
( ) 1665
27.75
•
•
60
f
Mean x
f x
f x
x
f
  
   



b) Median
Lower Class
Limit
Upper Class
Limit
Freq. Relative
Freq.
Cum.
Frequency
Lower Class
Boundary
Upper Class
Boundary
Class
Midpoint
1 0 14 25 41.7 25 -0.5 14.5 7 MODAL CLASS
2 15 29 17 28.3 42 14.5 29.5 22 Median CLASS
The second is the Median Class, since Cumulative frequency of 2nd class = 42 > N/2
60 30 14.5
2
15 60
14.5 25 18.91
2 17 2
N
N l
h n
X l c X
f
  
   
          
   
c) Mode
1 2
1 2
0.5 14.5
25 0 25, 25 17 8
L L
d d
  
     
 
1
1 2 1
1 2
( )
25
( 0.5) 14.5 ( 0.5 10.86
25 8
d
Mode L L L
d d
Mode
  

     

d) Standard Deviation
2
( ) 1664 ( ) 5• 539• 8f x f x  
2 2 2
[ ( )] [ ( )] 60 85395 1665
25.77
( 1) 60 (60 1)
n f x f x
S
n n
      
  
  

Page 3 of 12
Q-1-C
360 360
Component Part Frequencyof theclass
Angle
Whole part Total Frequency
   
Class 1 2 3 4 5 6 7
Frequency 25 17 5 3 4 5 1
Angle 150 102 30 18 24 6 30
0
10
20
30
40
50
60
70
0 20 40 60 80 100 120
CumulativeFrequency
Class Midpoint
OGIVE CHART
1st Class, 150, 42%
2nd Class, 102, 28%
3rd Class, 30, 8%
4th Class, 18, 5%
5th Class, 24, 7%
6th Class, 30, 8%
7th Class, 6, 2%
PIE chart
Class 1st Class 2nd Class 3rd Class 4th Class 5th Class 6th Class 7th Class

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Q2: (8%): The following data represent score fo 20 student in a certain test out of 30
{2, 7, 5, 8, 13, 14, 18, 13, 12, 17, 15, 12, 19, 11, 14, 14, 18, 22, 23, 29}
A) Find the percentile rank of a score of 19 (2%)
B) Make a box-and-whisker plot of the data. Find the interquartile range (3%)
C) Find the value corresponding to the 30th
and 91th
(3%)
Solution Q-2:
Sort the data from lowest to highest
{2, 7, 5, 8, 11, 12, 12, 13, 13, 14, 14, 14, 15, 17, 18, 18, 19, 22, 23, 29}
a) Percentile rank of a score of 19
0.5
100
16 0.5
19 100 82.5%
20
Number of valueslessthan X
Percentile of Value X
Total number of values
Percentile of Score

 

  
b) Make a box-and-whisker plot of the data. Find the interquartile range
 1
11 12
Q , 11.5
2
median Low MD

  
 2
14 14
Q , 14
2
median Low High

  
 3
18 18
Q , 18
2
median MD High

  
Now you can draw box and whisker plot from the data
c) Find the value corresponding to the 30th and 91th
{2, 7, 5, 8, 11, 12, 12, 13, 13, 14, 14, 14, 15, 17, 18, 18, 19, 22, 23, 29}
6th value, 7th value
30
30
20 6
100 100
1 6 12 7 12
   
     . ,
k
L n L
L L L L L
∴ L=12
A score of 12 correspond to 30th percentile. In another meaning A Student with a score of
12, did better than 30 percent of the students,
91
91
20 18 2 19
100 100
      .
k
L n L next whole number
L19 = 23
A score of 23 correspond to 91th percentile. In another meaning A Student with a score of 23, did
better than 91 percent of the students.

Page 5 of 12
Q3: (8%)
A developer of a new subdivision offers prospective home buyers a choice of Tudor, rustic, colonial, and
traditional exterior styling in ranch, two-story, and split-level floor plans.
A) Prepare Tree Diagram (4%)
B) In how many different ways can a buyer order one of these homes? (4%)
Solution to Q-3
A) Tree Diagram
B) In how many different ways can a buyer order one of these homes?
, , , 41 Tudor rustic colonial and traditin Exterior st onal exteryling ior 
2 , , 3ranch two story and spn lFloor P it levellan   
1 2 4 3 12
in12wayscan a buyer order one of thesehomes
n n   


of 12
Q4: (6%)
If a two dice is tossed in a balanced dice.
A) What is the probability of getting a total of 7 or 11 when pair of fair dice is tossed? (3%)
B) Fair dice are rolled. What is the probability of getting a sum less than 7 or a sum equal to 10? (3%)
SOLUTION Q-4
Q-4-A
Let A = Probability of getting a sum of 7
For getting a sum of 7,
(1,6) (2,5) (3,4) (4,3) (5,2) and (6,1) are favourable outcomes
2
Samplespace=Totalnumberof possibleoutcomes 6 36
Probability of getting a sum of 7
6

36
N
n
PA
N
  
  
For getting a sum of 11,
Let B = Probability of getting a sum of 11
(5,6) and (6,5) are favourable outcomes
Probability of getting a sum of 11
2
36
n
PB
N
  
Probability of getting a sum of 7 or 11 =
6 2 8
36 36 36
PAOR B PA PB PA PB      
Q-4-B
For getting a sum of less than 7,
Let C = Probability of getting a sum of less than 7
Sums less than 7= 2, 3, 4, 5, 6
∴ Favourable outcomes are:
Outcomes that give a sum of 2: 1-1
Outcomes that give a sum of 3: 1-2, 2-1
Outcomes that give a sum of 4: 1-3, 2-2, 3-1
Outcomes that give a sum of 5: 1-4, 2-3, 3-2, 4-1
Outcomes that give a sum of 6: 1-5, 2-4, 3-3, 4-2, 5-1
Total number of outcomes in our EVENT: 15
15
3
7
6
Probability of getting a su
n
PBm of less tha
N
n   
For getting a sum equal to 10,
Let D = Probability of getting a sum of 10
(4,6) (5,5) (6,4) are favourable outcomes
1
6
0
3
3
Probability of getting
n
PBa sum of
N
   
Probability of getting a sum of less than 7 or a sum equal to 10
18 3 21
36 36 36
PC or D PC PD PC PD      

Page 7 of 12
Q5: (7%)
Find the probability of a couple having at least 2 boy among 5 children. Assume that boys and girls are
equally likely and that the gender of a child is independent of any other child.
Solution to Q5:
Sample Space:
Sample space for No boy: (G, G, G, G, G)
Sample space for One boy:
(B, G, G, G, G) (G, B, G, G, G) (G, G, B, G, G) (G, G, G, B, G) (G, G, G, G, B)
Let A=havingat least twoboys
1
(at least one)
1
(at least twoboys)
having lessthan twoboys
P P
none
P PA P
none
P P P
none Noboy One boy
 
  
  
5
( )( )
1
P
2
Sample Spacefor one child ( , ) 2
Sample Spacefor 5 child 2 32
girlboy
P
Boy Girl
 
 
 
1
( )
1
(at least twoboys)
having lessthan twoboy
P P
at least one none
P PA P
none
P s P P
none Noboy One boy
 
  
  
Girl and Girl andGirl and Girl and Girl
1 5
,
32 32
1 5 6
32 32 32
P
Noboy
P P
Noboy one boy
P none P P
Noboy one boy

 
    
1
( )
6 26
1
( ) 32 32
P P
at least one none
P PA
at least twoboys
 
   

of 12
Q6: (8%)
A town has two fire engines operating independently. The probability that a specific engine is available
when needed is 0.96
A) What is the probability that neither is available when needed? (4%)
B) What is the probability that a fire engine is available when needed?
SOLUTION Q6: (8%)
(4%)
Or Probability that a fire engine is available when need needed = One or both fire engine available
= Probability of at least one
atleat one fireengineisavailble
let
P PA
1
0.016 ( )
PA Pnone
Pnone both fire enginesarenot available from part A
 
 
1
0.0016 ( )
1 1 0.0016 0.9984
PA Pnone
Pnone both fire engines arenot available from part A
PA Pnone
 
 

    

Page 9 of 12
Q7: (15%)
In 1970, 11% of Americans completed four years of college; 43% of them were women. In 1990, 22% of
Americans completed four years of college; 53% of them were women.
A) What is the probability that a woman finished four years of college in 1990? (3%)
B) What is the probability that a man had not finished college in 1990? (4%)
C) Given that a person was man, what is the probability that a person completed four years of
college in 1990?
(4%)
D) Given that a person completed four years of college in 1970, what is the probability that
the person was a woman?
(4%)
Illustrative Figure
1970
11% Completed four years of college
89% DID NOT completed four years of
college
57% Man
43% Woman
43% Man
57% Woman
1990
22% Completed four years of college
78% DID NOT completed four years of college
53% Woman
P(𝐴̅∩C)=0.78×0.53
47% Man
P A and B =0.22×0.53
53% Man
47% Woman

of 12
Solution Q-7
Q-7-A
What is the probability that a woman finished four years of college in 1990?
let
F= American finished four years of college in 1990
W= American woman finished four years of college in 1990
M= American man finished four years of college in 1990
( ) 0.22 0.53 0.1166 11.66%P F W    
Q-7-B
What is the probability that a man had not finished college in 1990?
F

= American HAD NOT finished four years of college in 1990
∴22% of American finished 4 year college, then 78% had not finished college
( ) 0.78 0.53 0.4134 41.34%P F M

    
Q-7-C
Given that a person was man, what is the probability that a person completed four years of college
in 1990?
( / ) 0.47P F M 
Q-7-D
Given that a person completed four years of college in 1970, what is the probability that the person
was a woman?
( / ) 0.43P W F 

Page 11 of 12
Q8: (20%)
The demand for a particular type of pump at an isolated mine is random and independent of previous
occurrences, but the average demand in a week (7 days) is for 2.8pumps. Further supplies are ordered
each Tuesday morning and arrive on the weekly plane on Friday morning. Last Tuesday morning only
one pump was in stock, so the stores man ordered six more to come in Friday morning.
A) Find the probability that one pump will still be in stock on Friday morning when new
stock arrives.
(6%)
B) Find the probability that stock will be exhausted and there will be unsatisfied demand for
at least one pump by Friday morning.
(6%)
C) Find the probability that one pump will still be in stock this Friday morning and at least
five will be in stock next Tuesday morning.
(8%)
Solution to Q-8
First we have to recognize that the Poisson distribution will apply.
2.8
0.4 /
7
day  
Q-8-A
(3 ) 0.4 3 1.2day   
X = Demand of pump per day.
If one pump will still be in stock in Friday,
It means that the demand on Pump in three days was zero ∴ X=0
1.2 0
( 0)
1.2
(No demand in 3days) 0.3012
! 0!
x
x
e e
P P
x

 
   
(No demand in 3days) ( ) 0.3012P P one pump will be in stock at Fridaymorning 
Q-8-B
If stock will be exhausted and there will be unsatisfied demand for at least one pump by Friday
morning
∴ X= ( one pump in stock + unsatisfied demand for at least one)
∴ X= Demand at least two
( 3days) ( 2) 1Demand of at least two pump x noneP P P  
(Demand for zero or onepumpinthree days)noneP P
( 0) ( 1)
1.2 0 1.2 1
1.2 1.2
0.3012 0.3614 0.6626
0! 1!
none x x
none
P P P
e e
P
 
 
 
    

of 12
( 2)
( 2)
(Demandof at least two pump 3days)
(Demandof at least two pump 3days)
1
1 0.6626 0.3374
x none
x
P P P
P P


  
   
( 2)
(unsatisfied demand for a
(Demandof at least twopump 3day
t least one pump by Friday mo
s)
rning)
0.3374
0.3374
xP P
P
 
 
Q-8-C
Find the probability that one pump will still be in stock this Friday morning and at least five will be
in stock next Tuesday morning.
at least five will be in stock next Tuesday morning.
Pr(one pump in stock this Friday and at least five
let A=Onepump beinstock this
will be in stock next Tue
fridaymornin
sday)=P(A
g
B=
andB)
It is a dependant multiplication rule,
P(A ) P(A B) PA PB/ Aand B    
from Part A,
0 0.3012PA Px  
From Friday Morning to Tuesday morning, there is 4 days: Saturday, Sunday Monday, Tuesday =4
0.4 4 1.6
(4 )day
   
After the new stock arrives, we will have 1+ 6 = 7 pump in stock Friday morning.
If we have at least five in stock Tuesday morning, then the demand in four days most be ≤ 2 pump.
( 3days) ( 2) ( 0) ( 0) ( 2)Demand of at mosttwo pump x x x xP P P P P      
( 4days) ( 2) ( 0) ( 0) ( 2)
1.6 0 1.6 1 1.6 2
( 2)
1.2 1.2 1.2
0.7834
0! 1! 2!
Demand of at mosttwo pump x x x x
x
P P P P P
e e e
P
   
  

   
   
( 4days) ( 2)
(at leastfivewill be in stock nextTuesdaygiven that thereisone stockin friday)
0.7834
/ 0.7834
Demand of at most two pump xP P
P PB A
 
  
P(A ) P(A B) PA PB/ A
P(A B) 0.3012 0.7834 0.236
and B    
   

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