1. The question asks to find the probability of a couple having at least 2 boys among 5 children, assuming equal probability of boys and girls and independence between children.
2. The sample space includes outcomes with 0 boys (1 outcome), 1 boy (5 outcomes), and at least 2 boys.
3. The probability of having at least 2 boys is calculated as 1 minus the probability of having less than 2 boys (0 or 1 boy). This gives a probability of 0.324 of having at least 2 boys among 5 children.
Chapter-4: More on Direct Proof and Proof by Contrapositivenszakir
Proofs Involving Divisibility of Integers, Proofs Involving Congruence of Integers, Proofs Involving Real Numbers, Proofs Involving sets, Fundamental Properties of Set Operations, Proofs Involving Cartesian Products of Sets
Chapter-4: More on Direct Proof and Proof by Contrapositivenszakir
Proofs Involving Divisibility of Integers, Proofs Involving Congruence of Integers, Proofs Involving Real Numbers, Proofs Involving sets, Fundamental Properties of Set Operations, Proofs Involving Cartesian Products of Sets
Mathematical Statistics with Applications in R 2nd Edition Ramachandran Solut...Alvaradoree
Full download : http://alibabadownload.com/product/mathematical-statistics-with-applications-in-r-2nd-edition-ramachandran-solutions-manual/ Mathematical Statistics with Applications in R 2nd Edition Ramachandran Solutions Manual
Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data. In applying statistics to a scientific, industrial, or social problem, it is conventional to begin with a statistical population or a statistical model to be studied
For helpful CXC Maths Multiple Choice Videos please click below
These videos are very helpful
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1of 20Use the formula for the sum of the first n terms of a geom.docxhyacinthshackley2629
1of 20
Use the formula for the sum of the first n terms of a geometric sequence to solve the following.
Find the sum of the first 12 terms of the geometric sequence: 2, 6, 18, 54 . . .
A. 531,440
B. 535,450
C. 535,445
D. 431,440
2 of 20
5.0 Points
The following are defined using recursion formulas. Write the first four terms of each sequence.
a1 = 7 and an = an-1 + 5 for n ≥ 2
A. 8, 13, 21, 22
B. 7, 12, 17, 22
C. 6, 14, 18, 21
D. 4, 11, 17, 20
3 of 20
5.0 Points
How large a group is needed to give a 0.5 chance of at least two people having the same birthday?
A. 13 people
B. 23 people
C. 47 people
D. 28 people
4 of 20
5.0 Points
Write the first six terms of the following arithmetic sequence.
an = an-1 + 6, a1 = -9
A. -9, -3, 3, 9, 15, 21
B. -11, -4, 3, 9, 17, 21
C. -8, -3, 3, 9, 16, 22
D. -9, -5, 3, 11, 15, 27
5 of 20
5.0 Points
Write the first four terms of the following sequence whose general term is given.
an = 3n
A. 3, 9, 27, 81
B. 4, 10, 23, 91
C. 5, 9, 17, 31
D. 4, 10, 22, 41
6 of 20
5.0 Points
If 20 people are selected at random, find the probability that at least 2 of them have the same birthday.
A. ≈ 0.31
B. ≈ 0.42
C. ≈ 0.45
D. ≈ 0.41
7 of 20
5.0 Points
Consider the statement "2 is a factor of n2 + 3n."
If n = 1, the statement is "2 is a factor of __________."
If n = 2, the statement is "2 is a factor of __________."
If n = 3, the statement is "2 is a factor of __________."
If n = k + 1, the statement before the algebra is simplified is "2 is a factor of __________."
If n = k + 1, the statement after the algebra is simplified is "2 is a factor of __________."
A.4; 15; 28; (k + 1)2 + 3(k + 1); k2 + 5k + 8
B.4; 20; 28; (k + 1)2 + 3(k + 1); k2 + 5k + 7
C.4; 10; 18; (k + 1)2 + 3(k + 1); k2 + 5k + 4
D.4; 15; 18; (k + 1)2 + 3(k + 1); k2 + 5k + 6
8 of 20
5.0 Points
k2 + 3k + 2 = (k2 + k) + 2 ( __________ )
A. k + 5
B. k + 1
C. k + 3
D. k + 2
9 of 20
5.0 Points
The following are defined using recursion formulas. Write the first four terms of each sequence.
a1 = 3 and an = 4an-1 for n ≥ 2
A. 3, 12, 48, 192
B. 4, 11, 58, 92
C. 3, 14, 79, 123
D. 5, 14, 47, 177
10 of 20
5.0 Points
To win at LOTTO in the state of Florida, one must correctly select 6 numbers from a collection of 53 numbers (1 through 53). The order in which the selection is made does not matter. How many different selections are possible?
A. 32,957,326 selections
B. 22,957,480 selections
C. 28,957,680 selections
D. 225,857,480 selections
11 of 20
5.0 Points
Write the first six terms of the following arithmetic sequence.
an = an-1 - 0.4, a1 = 1.6
A. 1.6, 1.2, 0.8, 0.4, 0, -0.4
B. 1.6, 1.4, 0.9, 0.3, 0, -0.3
C. 1.6, 2.2, 1.8, 1.4, 0, -1.4
D. 1.3, 1.5, 0.8, 0.6, 0, -0.6
12 of 20
5.0 Points
You volunteer to help drive children at a charity event to the zoo, but you can fit only 8 of the 17 children present in your van. How many different groups of 8 children can you drive?
A. 32.
QuizTop of FormQuestion 1 (24 points)Question 1True or Fa.docxsleeperharwell
Quiz
Top of Form
Question 1 (24 points)
Question 1:
True or False.
Enter the answer to each of the the questions with:
T for True
F for False
(a) If all the observations in a data set are identical, then the variance for this data set is zero.
[removed]
(b) If P(A) = 0.4 and P(B) = 0.5, then P(A AND B) = 0.2.
[removed]
c. The mean is always equal to the median for a normal distribution.
[removed]
(
d) A 95% confidence interval is wider than a 90% confidence interval of the same parameter
.
[removed]
(e) In a two-tailed hypothesis testing at significance level α of 0.05, the test statistic is calculated as 2. If P(X >2) = 0.03, then we have sufficient evidence to reject the null hypothesis.
[removed]
Bottom of Form
Question 2 (5 points)
Question 2 options:
Refer to the following frequency distribution for Questions 2, 3, 4, and 5.
Show all work. Just the answer, without supporting work, will receive no credit.
A random sample of 100 students was chosen from UMUC STAT 200 classes. The frequency distribution below shows the distribution for study time each week (in hours).
Checkout Time (in minutes)
Frequency
Relative Frequency
0.0 - 4.9
5
A
5.0 - 9.9
13
B
10.0 - 14.9
C
22
15.0 - 19.9
42
D
20.0 - 24.9
E
F
Total
100
G
Complete the Frequency Table with the missing frequency and relative frequency numbers.
Enter answer for "A" as a decimal with 2 decimal places with a zero to the left of the decimal point.
[removed]
Enter answer for "B" as a decimal with 2 decimal places with a zero to the left of the decimal point.
[removed]
Enter answer for "C" as an Integer.
[removed]
Enter answer for "D" as a decimal with 2 decimal places with a zero to the left of the decimal point.
[removed]
Enter answer for "E" as an integer.
[removed]
Enter answer for "F" as a decimal with 2 decimal places with a zero to the left of the decimal point.
[removed]
Question 3 (5 points)
Question 3 options:
A random sample of 100 students was chosen from UMUC STAT 200 classes. The frequency distribution below shows the distribution for study time each week (in hours).
This is the same distribution table from Question 2
Checkout Time (in minutes)
Frequency
Relative Frequency
0.0 - 4.9
5
A
5.0 - 9.9
13
B
10.0 - 14.9
C
22
15.0 - 19.9
42
D
20.0 - 24.9
E
F
Total
100
G
What percentage of the study times was at least 15 hours?
Enter answer as a percent without the percent sign to 0 decimal places.
[removed]
Question 4 (5 points)
Question 4 options:
A random sample of 100 students was chosen from UMUC STAT 200 classes. The frequency distribution below shows the distribution for study time each week (in hours).
This is the same distribution table from Question 2
Checkout Time (in minutes)
Frequency
Relative Frequency
0.0 - 4.9
5
A
5.0 - 9.9
13
B
10.0 - 14.9
C
22
15.0 - 19.9
42
D
20.0 - 24.9
E
F
Total
100
G
In what class interval must the median lie? Explain your answer.
.
This is a Question Papers of Mumbai University for B.Sc.IT Student of Semester - IV [Quantitative Technology] (Revised Course). [Year - September / 2013] . . . Solution Set of this Paper is Coming soon . . .
Distinguish between Parameter and Statistic.
Calculate sample variance and sample standard deviation.
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QuizQuestion 1 (24 points)Question 1True or Fal.docxsleeperharwell
Quiz
Question 1 (24 points)
Question 1:
True or False.
Enter the answer to each of the the questions with:
T for True
F for False
(a) If all the observations in a data set are identical, then the variance for this data set is zero.
[removed]
(b) If P(A) = 0.4 and P(B) = 0.5, then P(A AND B) = 0.2.
[removed]
c. The mean is always equal to the median for a normal distribution.
[removed]
(
d) A 95% confidence interval is wider than a 90% confidence interval of the same parameter
.
[removed]
(e) In a two-tailed hypothesis testing at significance level α of 0.05, the test statistic is calculated as 2. If P(X >2) = 0.03, then we have sufficient evidence to reject the null hypothesis.
[removed]
Question 2 (5 points)
Question 2 options:
Refer to the following frequency distribution for Questions 2, 3, 4, and 5.
Show all work. Just the answer, without supporting work, will receive no credit.
A random sample of 100 students was chosen from UMUC STAT 200 classes. The frequency distribution below shows the distribution for study time each week (in hours).
Checkout Time (in minutes)
Frequency
Relative Frequency
0.0 - 4.9
5
A
5.0 - 9.9
13
B
10.0 - 14.9
C
22
15.0 - 19.9
42
D
20.0 - 24.9
E
F
Total
100
G
Complete the Frequency Table with the missing frequency and relative frequency numbers.
Enter answer for "A" as a decimal with 2 decimal places with a zero to the left of the decimal point.
[removed]
Enter answer for "B" as a decimal with 2 decimal places with a zero to the left of the decimal point.
[removed]
Enter answer for "C" as an Integer.
[removed]
Enter answer for "D" as a decimal with 2 decimal places with a zero to the left of the decimal point.
[removed]
Enter answer for "E" as an integer.
[removed]
Enter answer for "F" as a decimal with 2 decimal places with a zero to the left of the decimal point.
[removed]
Question 3 (5 points)
Question 3 options:
A random sample of 100 students was chosen from UMUC STAT 200 classes. The frequency distribution below shows the distribution for study time each week (in hours).
This is the same distribution table from Question 2
Checkout Time (in minutes)
Frequency
Relative Frequency
0.0 - 4.9
5
A
5.0 - 9.9
13
B
10.0 - 14.9
C
22
15.0 - 19.9
42
D
20.0 - 24.9
E
F
Total
100
G
What percentage of the study times was at least 15 hours?
Enter answer as a percent without the percent sign to 0 decimal places.
[removed]
Question 4 (5 points)
Question 4 options:
A random sample of 100 students was chosen from UMUC STAT 200 classes. The frequency distribution below shows the distribution for study time each week (in hours).
This is the same distribution table from Question 2
Checkout Time (in minutes)
Frequency
Relative Frequency
0.0 - 4.9
5
A
5.0 - 9.9
13
B
10.0 - 14.9
C
22
15.0 - 19.9
42
D
20.0 - 24.9
E
F
Total
100
G
In what class interval must the median lie? Explain your answer.
Enter answer by selecting the.
6th International Conference on Machine Learning & Applications (CMLA 2024)ClaraZara1
6th International Conference on Machine Learning & Applications (CMLA 2024) will provide an excellent international forum for sharing knowledge and results in theory, methodology and applications of on Machine Learning & Applications.
Hierarchical Digital Twin of a Naval Power SystemKerry Sado
A hierarchical digital twin of a Naval DC power system has been developed and experimentally verified. Similar to other state-of-the-art digital twins, this technology creates a digital replica of the physical system executed in real-time or faster, which can modify hardware controls. However, its advantage stems from distributing computational efforts by utilizing a hierarchical structure composed of lower-level digital twin blocks and a higher-level system digital twin. Each digital twin block is associated with a physical subsystem of the hardware and communicates with a singular system digital twin, which creates a system-level response. By extracting information from each level of the hierarchy, power system controls of the hardware were reconfigured autonomously. This hierarchical digital twin development offers several advantages over other digital twins, particularly in the field of naval power systems. The hierarchical structure allows for greater computational efficiency and scalability while the ability to autonomously reconfigure hardware controls offers increased flexibility and responsiveness. The hierarchical decomposition and models utilized were well aligned with the physical twin, as indicated by the maximum deviations between the developed digital twin hierarchy and the hardware.
HEAP SORT ILLUSTRATED WITH HEAPIFY, BUILD HEAP FOR DYNAMIC ARRAYS.
Heap sort is a comparison-based sorting technique based on Binary Heap data structure. It is similar to the selection sort where we first find the minimum element and place the minimum element at the beginning. Repeat the same process for the remaining elements.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
Industrial Training at Shahjalal Fertilizer Company Limited (SFCL)MdTanvirMahtab2
This presentation is about the working procedure of Shahjalal Fertilizer Company Limited (SFCL). A Govt. owned Company of Bangladesh Chemical Industries Corporation under Ministry of Industries.
Harnessing WebAssembly for Real-time Stateless Streaming PipelinesChristina Lin
Traditionally, dealing with real-time data pipelines has involved significant overhead, even for straightforward tasks like data transformation or masking. However, in this talk, we’ll venture into the dynamic realm of WebAssembly (WASM) and discover how it can revolutionize the creation of stateless streaming pipelines within a Kafka (Redpanda) broker. These pipelines are adept at managing low-latency, high-data-volume scenarios.
Student information management system project report ii.pdfKamal Acharya
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Dr.Costas Sachpazis
Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
Saudi Arabia stands as a titan in the global energy landscape, renowned for its abundant oil and gas resources. It's the largest exporter of petroleum and holds some of the world's most significant reserves. Let's delve into the top 10 oil and gas projects shaping Saudi Arabia's energy future in 2024.
An Approach to Detecting Writing Styles Based on Clustering Techniquesambekarshweta25
An Approach to Detecting Writing Styles Based on Clustering Techniques
Authors:
-Devkinandan Jagtap
-Shweta Ambekar
-Harshit Singh
-Nakul Sharma (Assistant Professor)
Institution:
VIIT Pune, India
Abstract:
This paper proposes a system to differentiate between human-generated and AI-generated texts using stylometric analysis. The system analyzes text files and classifies writing styles by employing various clustering algorithms, such as k-means, k-means++, hierarchical, and DBSCAN. The effectiveness of these algorithms is measured using silhouette scores. The system successfully identifies distinct writing styles within documents, demonstrating its potential for plagiarism detection.
Introduction:
Stylometry, the study of linguistic and structural features in texts, is used for tasks like plagiarism detection, genre separation, and author verification. This paper leverages stylometric analysis to identify different writing styles and improve plagiarism detection methods.
Methodology:
The system includes data collection, preprocessing, feature extraction, dimensional reduction, machine learning models for clustering, and performance comparison using silhouette scores. Feature extraction focuses on lexical features, vocabulary richness, and readability scores. The study uses a small dataset of texts from various authors and employs algorithms like k-means, k-means++, hierarchical clustering, and DBSCAN for clustering.
Results:
Experiments show that the system effectively identifies writing styles, with silhouette scores indicating reasonable to strong clustering when k=2. As the number of clusters increases, the silhouette scores decrease, indicating a drop in accuracy. K-means and k-means++ perform similarly, while hierarchical clustering is less optimized.
Conclusion and Future Work:
The system works well for distinguishing writing styles with two clusters but becomes less accurate as the number of clusters increases. Future research could focus on adding more parameters and optimizing the methodology to improve accuracy with higher cluster values. This system can enhance existing plagiarism detection tools, especially in academic settings.
Literature Review Basics and Understanding Reference Management.pptxDr Ramhari Poudyal
Three-day training on academic research focuses on analytical tools at United Technical College, supported by the University Grant Commission, Nepal. 24-26 May 2024
Cosmetic shop management system project report.pdfKamal Acharya
Buying new cosmetic products is difficult. It can even be scary for those who have sensitive skin and are prone to skin trouble. The information needed to alleviate this problem is on the back of each product, but it's thought to interpret those ingredient lists unless you have a background in chemistry.
Instead of buying and hoping for the best, we can use data science to help us predict which products may be good fits for us. It includes various function programs to do the above mentioned tasks.
Data file handling has been effectively used in the program.
The automated cosmetic shop management system should deal with the automation of general workflow and administration process of the shop. The main processes of the system focus on customer's request where the system is able to search the most appropriate products and deliver it to the customers. It should help the employees to quickly identify the list of cosmetic product that have reached the minimum quantity and also keep a track of expired date for each cosmetic product. It should help the employees to find the rack number in which the product is placed.It is also Faster and more efficient way.
Solution to final exam engineering statistics 2014 2015
1. 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
2. Page 2 of 12
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
3. Solution to Engineering Statistics Exam 2014-2015 Dam and Water Resources Department
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
4. Page 4 of 12
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.
5. Solution to Engineering Statistics Exam 2014-2015 Dam and Water Resources Department
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
6. Page 6 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
7. Solution to Engineering Statistics Exam 2014-2015 Dam and Water Resources Department
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
8. Page 8 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
9. Solution to Engineering Statistics Exam 2014-2015 Dam and Water Resources Department
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
10. Page 10 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
11. Solution to Engineering Statistics Exam 2014-2015 Dam and Water Resources Department
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
12. Page 12 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