Probability
And Statistics
Prepared by Paul CHEGE
African Virtual university
Université Virtuelle Africaine
Universidade Virtual Africana
African Virtual University Notice
This document is published under the conditions of the Creative Commons
http://en.wikipedia.org/wiki/Creative_Commons
Attribution
http://creativecommons.org/licenses/by/2.5/
License (abbreviated “cc-by”), Version 2.5.
African Virtual University Table of Contents
I.
Probability and Statistics_____________________________________ 3
II.
Prerequisite Course or Knowledge_ _____________________________ 3
III. Time_____________________________________________________ 3
IV. Materials__________________________________________________ 3
V.
Module Rationale_ __________________________________________ 3
VI. Content___________________________________________________ 4
6.1 Overview ____________________________________________ 4
6.2 Outline_ _____________________________________________ 5
6.3 Graphic Organizer______________________________________ 6
VII. General Objective(s)_________________________________________ 7
VIII. Specific Learning Activities____________________________________ 7
IX. Teaching and Learning Activities________________________________ 9
X.
Compiled List of all Key Concepts (Glossary)_____________________ 12
XI. Compiled List of Compulsory Readings _________________________ 18
XII. Compiled List of Resources__________________________________ 19
XIII. Compiled List of Useful Links_________________________________ 20
XIV. Learning Activities__________________________________________ 21
XV. Synthesis of the Module____________________________________ 112
XVI. Summative Evaluation_ ____________________________________ 113
XVII. References______________________________________________ 121
XVIII. Student records__________________________________________ 122
XIX. Main Author of the Module_ _________________________________ 123
African Virtual University I. Probability and Statistics
by Paul Chege
II. Prerequisite courses or knowledge
Secondary school statistics and probability.
III. Time
The total time for this module is 120 study hours.
IV. Material
Students should have access to the core readings specified later. Also, they will need
a computer to gain full access to the core readings. Additionally, students should
be able to install the computer software wxMaxima and use it to practice algebraic
concepts.
V. Module Rationale
Probability and Statistics, besides being a key area in the secondary schools’ teaching
syllabuses, it forms an important background to advanced mathematics at tertiary level.
Statistics is a fundamental area of Mathematics that is applied across many academic subjects and is useful in analysis in industrial production. The study of statistics
produces statisticians that analyse raw data collected from the field to provide useful
insights about a population. The statisticians provide governments and organizations
with concrete backgrounds of a situation that helps managers in decision making.
For example, rate of spread of diseases, rumours, bush fires, rainfall patterns, and
population changes.
On the other hand, the study of probability helps decision making in government
agents and organizations based on the theory of chance. For example:- predicting
the male and female children born within a given period and projecting the amount
of rainfall that regions expect to receive based on some historical data on rainfall
patterns. Probability has also been extensively used in the determination of high,
middle and low quality products in industrial production e.g the number of good and
defective parts expected in an industrial manufacturing process.
African Virtual University VI.Content
6.1 Overview
This module consists of three units:
Unit 1: Descriptive Statistics and Probability Distributions
Descriptive statistics in unit one is developed either as an extension of secondary
mathematics or as an introduction to first time learners of statistics. It introduces the
measures of dispersion in statistics. The unit also introduces the concept of probability
and the theoretical treatment of probability.
Unit 2: Random variables and Test Distributions
This unit requires Unit 1 as a prerequisite. It develops from the moment and moment
generating functions, Markov and Chebychev inequalities, special univariate distributions, bivariate probability distributions and analyses conditional probabilities.
The unit gives insights into the analysis of correlation coefficients and distribution
functions of random variables such as the Chi-square, t and F.
Unit 3: Probability Theory
This unit builds up from unit 2. It analyses probability using indicator functions. It
introduces Bonferoni inequality random vectors,, generating functions, characteristic functions and statistical independence random samples. It develops further the
concepts of functions of several random variables and independence of X and S2 in
normal samples order statistics. The unit summarises with the treatment of convergence and limit theorems.
African Virtual University 6.2 Outline: Syllabus
Unit 1 ( 40 hours): Descriptive Statistics and Probability Distributions
Level 1. Priority A. No prerequisite.
Frequency distributions relative and cumulative distributions, various frequency
curves, mean, Mode Median. Quartiles and Percentiles, Standard deviation, symmetrical and skewed distributions. Probability; sample space and events; definition
of probability, properties of probability; random variables; probability distributions,
expected values of random variables; particular distributions; Bernoulli, binomial,
Poisson, geometric, hypergeometric, uniform, exponential and normal. Bivariate
frequency distributions. Joint probability tables and marginal probabilities.
Unit 2 ( 40 hours): Random Variables and Test Distributions
Level 2. Priority B. Statistics 1 is prerequisite.
Moment and moment generating function. Markov and Chebychev inequalities,
special Univariate distributions. Bivariate probability distribution; Joint Marginal
and conditional distributions; Independence; Bivariate expectation Regression and
Correlation; Calculation of regression and correlation coefficient for bivariate data.
Distribution function of random variables, Bivariate normal distribution. Derived
distributions such as Chi-Square. t. and F.
Unit 3 ( 40 hours): Probability Theory
Level 3. Priority C. Statistics 2 is prerequisite.
Probability: Use of indicator functions. Bonferoni inequality Random vectors.
Generating functions. Characteristics functions. Statistical independence Random
samples. Multinomial distribution. Functions of several random variables.
The independence of X and S2 in normal samples Order statistics Multivariate
normal distribution. Convergence and limit theorems. Practical exercises.
African Virtual University 6.3 Graphic Organiser
Variance
& Standard
deviation
Mean,
Mode, and
Median
Indicator
functions
Bonferoni
Inequalities,
random
vectors
Generating
functions,
characteristic
functions &
random samples
Multinomial
distributions,
Functions of
random variables
DATA
Probability
Probability
distributions
Multivariate
distribution,
Convergence &
limit theorems
Frequency
Curves,
Quartiles
Deciles and
Percentiles,
Moment
and
moment
generating
function
Markov and
Chebychev
inequalities
Joint marginal
& conditional
distributions
Univariate
and
Bivariate
distributions
Regression
& correlation
Derived
distributionsChi-square, t
and F
Joint
probability
tables
5
African Virtual University VII. General Objective(s)
By the end of this module, the trainee should be able to compute the various measures
of dispersions in statistics and work out probabilities based on laws of probability
and carry out tests on data using the theories of probability
VIII. Specific Learning Objectives
(Instructional Objectives)
Unit 1: Descriptive Statistics and Probability Distributions ( 40 Hours)
By the end of unit 1, the trainee should be able to:
• Draw various frequency curves
• Work out the mean, mode, median, quartiles, percentiles and standard deviations of discrete and grouped data
• Define and state the properties of probability
• Illustrate random variables, probability distributions, and expected values of
random variables.
• Illustrate Bernoulli, Binomial, Poisson, Geometric, Hypergeometric, Uniform,
Exponential and Normal distributions
• Investigate Bivariate frequency distributions
• Construct joint probability tables and marginal probabilities.
Unit 2: Random Variables and Test Distributions ( 40 Hours)
By the end of unit 2, the trainee should be able to:
• Illustrate moment and moment generating functions
• Analyse Markov and Chebychev inequalities
• Examine special Univariate distributions, bivariate probability distributions,
Joint marginal and conditional distributions.
• Show Independence, Bivariate expectation, regression and correlation
• Calculate regression and correlation coefficient for bivariate data
• Show distribution function of random variables.
• Examine Bivariate normal distribution
• Illustrate derived distributions such as Chi-Square, t, and F.
African Virtual University Unit 3: Probability Theory ( 40 Hours)
By the end of unit 3, the trainee should be able to:
•
•
•
•
•
•
•
•
•
Use indicator functions in probability
Show Bonferoni inequality random vectors
Illustrate generating and characteristic functions
Examine statistical independence random samples and multinomial distribution
Evaluate functions of several random variables
Illustrate the independence of X and S2 in normal samples order statistics
Show multivariate normal distribution
Illustrate convergence and limit theorems.
Work out practical exercises.
African Virtual University IX. Teaching and Learning Activities
9.1 Pre-assessment
Basic mathematics is a pre-requisite for Probability and Statistics.
Questions
1) When a die is rolled, the probability of getting a number greater than 4 is
A.
1
6
B.
1
3
C.
1
2
D. 1
2) A single card is drawn at random from a standard deck of cards. Find the probability that is a queen.
A.
B.
C.
D.
1
13
1
52
4
13
1
2
3) Out of 100 numbers, 20 were 4’s, 40 were 5’s, 30 were 6’s and the remainder
were 7’s. Find the arithmetic mean of the numbers.
A. 0.22
B. 0.53
C. 2.20
D. 5.30
African Virtual University 10
4) Calculate the mean of the following data.
Height (cm)
60 - 62
63 - 65
66 - 68
69 - 71
72 - 74
A.
B.
C.
D.
Class mark (x)
61
64
67
70
73
57.40
62.00
67.45
72.25
5) Find the mode of the following data: 5, 3, 6, 5, 4, 5, 2, 8, 6, 5, 4, 8, 3, 4, 5, 4, 8,
2, 5, and 4.
A.
B.
C.
D.
4
5
6
8
6) The range of the values a probability can assume is
A. From 0 to 1
B. From -1 to +1
C. From 1 to 100
D. From 0 to
1
2
7) Find the median of the following data: 8, 7, 11, 5, 6, 4, 3, 12, 10, 8, 2, 5, 1, 6, 4.
A.
B.
C.
D.
12
5
8
6
8) Find the range of the set of numbers: 7, 4, 10, 9, 15, 12, 7, 9.
A.
B.
C.
D.
9
11
7
8.88
African Virtual University 11
9) When two coins are tossed, the sample space is
A.
B.
C.
D.
H, T and HT
HH, HT, TH, TT
HH, HT, TT
H, T
10) If a letter is selected at random from the word “Mississippi”, find the probability
that it is an “i”
A.
1
8
B.
1
2
C.
D.
3
11
4
11
Answer Key
1. B
2.
A
3. D
4.
C
5.
B
6. A
7.
D
8.
B
9.
B
10.
D
Pedagogical Comment For Learners
This pre-assessment is meant to give the learners an insight into what they can
remember regarding Probability and Statistics. A score of less than 50% in the preassessment indicates the learner needs to revise Probability and Statistics covered in
secondary mathematics. The pre-assessment covers basic concepts that trainees need
to be familiar with before progressing with this module. Please revise Probability
and Statistics covered in secondary mathematics to master the basics if you have
problems with this pre-assessment.
African Virtual University 12
X. Key Concepts ( Glossary)
Mutually Exclusive: Two events are mutually exclusive if they cannot occur at the
same time.
Variance of a set of data is defined as the square of the standard deviation i.e variance = s2.
A trial: This refers to an activity of carrying out an experiment like picking a card
from a deck of cards or rolling a die or dices
Sample space: This refers to all possible outcomes of a probability experiment. e.g.
in tossing a coin, the outcomes are either Head(H) or tail(T)
A random variable: is a function that assigns a real number to every possible result
of a random experiment.
Random sample is one chosen by a method involving an unpredictable component.
Bernoulli distribution: is a discrete probability distribution, which takes value 1
with success probability p and value 0 with failure probability q = 1 − p.
Binomial distribution is the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields
success with probability .p
Hypergeometric distribution: is a discrete probability distribution that describes the
number of successes in a sequence of n draws from a finite population without
replacement.
Poisson distribution: is a discrete probability distribution that expresses the probability of a number of events occurring in a fixed period of time if these events
occur with a known average rate, and are independent of the time since the last
event
Correlation: is a measure of association between two variables.
Regression: is a measure used to examine the relationship between one dependent
and one independent variable.
Chi-square test is any statistical hypothesis test in which the test statistic has a
chi-square distribution when the null hypothesis is true, or any in which the
probability distribution of the test statistic (assuming the null hypothesis is true)
can be made to approximate a chi-square distribution as closely as desired by
making the sample size large enough.
Multivariate normal distribution is a specific probability distribution, which can
be thought of as a generalization to higher dimensions of the one-dimensional
normal distribution.
t -test is any statistical hypothesis test for two groups in which the test statistic has
a Student’s t distribution if the null hypothesis is true
African Virtual University 13
Statistical Terms
1. Raw data: Data that has not been organised numerically.
2. Arrays: An arrangement of raw data numerical data in ascending order of magnitude.
3. Range: the difference between the largest and the smallest numbers in a data.
4. Class intervals: In a range of grouped data e.g 21-30, 31-40 etc, then 21-30 l is
called the class interval.
5. Class limits: In a class interval of 21-30, then 21 and 30 are called class limits.
6. Lower class limits (l.c.l) : In the class interval 21-30, the lower class limit is 21
7. Upper class limit (u.c.l): in the class interval 21-30, the upper class limit is 30
8. Lower and upper class boundaries: In the class interval 21-30, the lower
class boundary is 20.5 and the upper class boundary is 30.5. These boundaries
assume that theoretically measurements for a class interval 21-30 includes all
the numbers from 20.5 to 30.5
9. Class Interval: In a class 21-30, then the class interval is the difference between
the upper class limit and the lower class limit i.e. 30.5-20.5 = 10. The class interval is also known as class width or class size.
10. Class Mark or Mid-point: In a class interval 21-30, the class mark is the average
21 + 30
= 25 .5
2
11. Frequency Distributions: large masses of raw data maybe arranged in classes
in tabular form with their corresponding frequencies. e.g.
of 21 and 30 i.e
Mass (kg)
Number of pupils (f)
10-19
5
20-29
7
30-39
10
40-49
6
This tabular arrangement is called a frequency distribution or frequency table.
12. Cumulative Frequency: For the following frequency distribution, the cumulative
frequencies are calculated as additions of individual frequencies
Mass ( X)
Frequency (f)
Cumulative
Frequency( C.F)
20-24
4
4
25-29
10
4+10=14
30-34
16
14=16=30
35-39
8
30+8=38
40-44
2
38+2=40
African Virtual University 14
Hence the cumulative frequency of a value is its frequency plus frequencies of all
smaller values.
The above table is called a Cumulative Frequency table.
13. Relative – Frequency Distributions: In a frequency distribution
Mass ( X)
Frequency (f)
20-24
4
25-29
10
30-34
16
35-39
8
40-44
2
∑ f = 40
The relative frequency of a class 25-29 is the frequency of the class divided by the
total frequency of all classes (cumulative frequency) and generally expressed as a
percentage.
Example:
The relative frequency of the class 25-29 =
f
∑f
× 100% =
10
× 100 = 25%
40
Note: the sum of relative frequencies is 100% or 1.
14. Cumulative Frequency Curve ( Ogive)
Mass ( X)
Frequency (f)
Cumulative
Frequency( C.F)
20-24
4
4
25-29
30-34
35-39
40-44
10
16
8
2
4+10=14 14=16=30 30+8=38 38+2=40
African Virtual University 15
From the above cumulative frequency table, we can draw a graph of cumulative
frequency verses the upper class boundaries.
Cumulative frequency
Upper class
boundaries
Cumulative
frequencies
24.5
29.5
34.5
39.5
44.5
3
14
30
38
40
Ogive
45
40
35
30
25
20
15
10
5
0
20
25
30
35
40
45
Upper class limit
Note: From the cumulative frequency data, the first plotting point is ( 24.5, 3). If
we started our graph at this point, it would remain hanging on the y-axis. We create
another point (19.5, 0) as a starting point. 19.5 is the projected upper class boundary
of the preceding class.
African Virtual University 16
Shapes of Frequency Curves
Symmetrical or bell-shaped.
Skewed to the right ( positive skewness)
Has equal frequency to the left and right
of the central maximum e.g. normal curve
Skewed to the left ( Negative skewness)
Has the maximum towards the right of
the and the longer tail to the left
Has the maximum towards the left and
the longer tail to the right
J –Shaped
Has the maximum occurring at the right
end
3
African Virtual University 17
Reverse J-Shaped
Has the maximum occurring at the left
end
Bimodal
Has two maxima
U- shaped
Has maxima at both ends
Multimodal
Has more than two maxima.
4
African Virtual University 18
XI. Compiled List of Compulsory Readings
Reading # 1: Wolfram MathWorld (visited 06.05.07)
Complete reference : http://mathworld.wolfram.com/Probabilty
Abstract : This reference gives the much needed reading material in probability
and statistics. The reference has a number of illustrations that empower the learner
through different approach methodology. Wolfram MathWorld is a specialised online mathematical encyclopaedia.
Rationale: It provides the most detailed references to any mathematical topic.
Students should start by using the search facility for the module title. At any point
students should search for key words that they need to understand. The entry should
be studied carefully and thoroughly.
Reading # 2: Wikipedia (visited 06.05.07)
Complete reference : http://en.wikipedia.org/wiki/statistics
Abstract : Wikipedia is an on-line encyclopaedia. It is written by its own readers.
It is extremely up-to-date as entries are continually revised. Also, it has proved to be
extremely accurate. The mathematics entries are very detailed.
Rationale: It gives definitions, explanations, and examples that learners cannot access
in other resources. The fact that wikipedia is frequently updated gives the learner
the latest approaches, abstract arguments, illustrations and refers to other sources to
enable the learner acquire other proposed approaches in Probability and Statistics.
Reading # 3: MacTutor History of Mathematics (visited 03.05.07)
Complete reference : http://www-history.mcs.standrews.ac.uk/Indexes
Abstract : The MacTutor Archive is the most comprehensive history of mathematics
on the internet. The resources are rganised by historical characters and by historical
themes.
Rationale: Students should search the MacTutor archive for key words in the topics
they are studying (or by the module title itself). It is important to get an overview
of where the mathematics being studied fits in to the history of mathematics. When
the student completes the course and is teaching high school mathematics, the characters in the history of mathematics will bring the subject to life for their students.
Particularly, the role of women in the history of mathematics should be studied to
help students understand the difficulties women have faced while still making an
important contribution.. Equally, the role of the African continent should be studied
to share with students in schools: notably the earliest number counting devices (e.g.
the Ishango bone) and the role of Egyptian mathematics should be studied.
African Virtual University 19
XII. Compiled list of Compulsory Resources
Resource #1 Maxima.
Complete reference : Copy of Maxima on a disc is accompanying this course
Abstract : The distance learners are occasionally confronted by difficult mathematics without resources to handle them. The absence of face to face daily lessons with
teachers means that learners can become totally handicapped if not well equipped
with resources to solve their mathematical problems. This handicap is solved by use
of accompanying resource: Maxima.
Rationale: Maxima is an open-source software that can enable learners to solve linear
and quadratic equations, simultaneous equations, integration and differentiation,
perform algebraic manipulations: factorisation, simplification, expansion, etc This
resource is compulsory for learners taking distance learning as it enables them learn
faster using the ICT skills already learnt.
Resource #2 Graph
Complete reference : Copy of Graph on a disc is accompanying this course
Abstract : It is difficult to draw graphs of functions, especially complicated functions,
most especially functions in 3 dimensions. The learners, being distance learners, will
inevitably encounter situations that will need mathematical graphing. This course
is accompanied by a software called Graph to help learners in graphing. Learners
however need to familiarise with the Graph software to be able to use it.
Rationale: Graph is an open-source dynamic graphing software that learners can
access on the given CD. It helps all mathematics learners to graph what would otherwise be a nightmare for them. It is simple to use once a learner invests time to learn
how to use it. Learners should take advantage of the Graph software because it can
assist the learners in graphing in other subjects during the course and after. Learners
will find it extremely useful when teaching mathematics at secondary school level.
African Virtual University 20
XIII. Compiled List of Useful Links
Useful Link #1
Title : Wikipedia
URL : http://en.wikipedia.org/wiki/Statistics
Description: Wikipedia is every mathematician’s dictionary. It is an open-resource
that is frequently updated. Most learners will encounter problems of reference materials from time to time. Most of the books available cover only parts or sections
of Probability and Statistics. This shortage of reference materials can be overcome
through the use of Wikipedia. It’s easy to access through “Google search”
Rationale: The availability of Wikipedia solves the problem of crucial learning
materials in all branches of mathematics. Learners should have first hand experience
of Wekipedia to help them in their learning. It is a very useful free resource that not
only solves student’s problems of reference materials but also directs learners to other
related useful websites by clicking on given icons. Its usefulness is unparalleled.
Useful Link #2
Title : Mathsguru
URL : http://en.wikipedia.org/wiki/Probability
Description: Mathsguru is a website that helps learners to understand various branches
of number theory module. It is easy to access through Google search and provides
very detailed information on various probability questions. It offers explanations and
examples that learners can understand easily.
Rationale: Mathsguru gives alternative ways of accessing other subject related topics,
hints and solutions that can be quite handy to learners who encounter frustrations of
getting relevant books that help solve learners’ problems in Probability. It gives a
helpful approach in computation of probabilities by looking at the various branches
of the probability module.
Useful Link #3
Title : Mathworld Wolfram
URL : http://mathworld.wolfram.com/Probability
Description: Mathworld Wolfram is a distinctive website full of Probability solutions. Learners’ should access this website quite easily through Google search for
easy reference. Wolfram also leads learners to other useful websites that cover the
same topic to enhance the understanding of the learners.
Rationale: Wolfram is a useful site that provides insights in number theory while
providing new challenges and methodology in number theory. The site comes handy
in mathematics modelling and is highly recommended for learners who wish to study
number theory and other branches of mathematics. It gives aid in linking other webs
thereby furnishing learners with a vast amount of information that they need to comprehend in Probability and Statistics.
African Virtual University 21
XIV. Learning Activity
Unit 1
40 Hours
Descriptive Statistics And Probability Distributions
A curious farmer undertakes the following activities in her farm.
1. She plants 80 tree seedlings on 1st March. She measures the heights of the trees on 1st December.
2. She weighs all the 40 cows in her farm and records the weights in her diary.
3. She records the daily production of eggs from the poultry section.
4. She records the time taken to deliver the milk to the processing plant.
The records are kept as below.
1. Heights of plants in cm
77
74
76
85
62
71
85
53
63
78
68
60
82
81
67
80
75
88
68
73
75
53
95
71
85
74
73
62
75
61
71
68
69
83
95
94
87
78
82
66
60
83
60
68
77
75
75
78
89
96
72
71
76
63
62
78
61
65
67
79
75
53
62
85
93
88
97
79
73
65
93
85
76
76
90
72
57
84
73
86
2. Weights of goats in kg
Weight
(kg)
No. of
goats
118-126
127-135
136-144 145-153 154-162 163-171 172-180
3
5
9
12
5
4
2
African Virtual University 22
3. Number of laid eggs
Eggs
No of
days
462
98
480
75
498
56
516
42
534
30
552
21
570
15
588
11
606
6
624
2
4. Delivery time of milk to processing plant
Time in minutes
No. of days
90-100
9
80-89
32
70-79
43
60-69
21
50-59
11
40-49
3
30-39
1
CASE 1:
A local firm dealing with agriculture extension services visits the farmer. She proudly
produces her records. The agricultural officer is very impressed by her good records
but clearly realises that the farmer needs some skills in data management to enable
her make informed decisions based on her farm outputs.
The agricultural officer designs a short course on data processing for all the rural
farmers.
During the course planning stage, the following terms are defined and designed for
a lesson one to the farmers.
a)
b)
c)
d)
e)
Data : The result of observation e.g. height of tree seedlings
Frequency: Rate of occurrence e.g. number of goats weighed.
Mean: The average of a data
Mode: The highest occurring in a data.
Median: In an ascending data, the median is the term occurring at the middle
of the data.
f) Range: the difference between the highest and the lowest in the data.
Lesson One: Measures Of Dispersion
Introduction to Statistics
Descriptive statistics is used to denote any of the many techniques used to summarize a set of data. In a sense, we are using the data on members of a set to describe
the set. The techniques are commonly classified as:
1. Graphical description in which we use graphs to summarize data.
2. Tabular description in which we use tables to summarize data.
3. Parametric description in which we estimate the values of certain parameters
which we assume to complete the description of the set of data.
In general, statistical data can be described as a list of subjects or units and the data
associated with each of them. We have two objectives for our summary:
African Virtual University 23
1. We want to choose a statistic that shows how different units seem similar.
Statistical textbooks call the solution to this objective, a measure of central
tendency.
2. We want to choose another statistic that shows how they differ. This kind of
statistic is often called a measure of statistical variability.
When we are summarizing a quantity like length or weight or age, it is common to
answer the first question with the arithmetic mean, the median, or the mode. Sometimes, we choose specific values from the cumulative distribution function called
quartiles.
The most common measures of variability for quantitative data are the variance; its
square root, the standard deviation; the statistical range; interquartile range; and the
absolute deviation.
Farmers lessons
The farmers are taught how to compute the
a) Mean or Average of a data as follows:
Average of a data= Sum total of the data divided by number of items in data.
Example:
Calculate the mean of the following data:
1) 1,3,4,4,5,6,3,7,
Solution: Mean =
1 + 3 + 4 + 4 + 5 + 6 + 3 + 7 33
=
= 4.125
8
8
2) 650,675, 700, 725, 800, 900, 1050, 1125, 1200, 575
Solution:
Mean =
650 + 675 + 700 + 725 + 800 + 900 + 1050 + 1125 + 1200 + 575
10
= = 840
8400
10
African Virtual University 24
Lesson Two
Mean Of Discrete Data
Example:
1) Find the mean of the following data:
X 22
f 5
24
7
25
8
33
4
36
6
37
9
41
11
Solution:
Mean
22(5) + 24(7) + 25(8) + 33(4) + 36(6) + 37(9) + 41(11) 1628
=
= 32.56
5 + 7 + 8 + 4 + 6 + 9 + 11
50
2) Find the mean wage of the workers:
Wage in $
No. of Workers
220
12
250
15
300
18
350
20
375
5
Solution:
Mean =
220(12) + 250(15) + 300(18) + 350(20) + 375(5) 20665
=
12 + 15 + 18 + 20 + 5
70
= $ 295.214
Frequency Tables And Mean Of Grouped Data
Example:
The weights of milk deliveries to a processing plant are shown below:
45
48
56
39
49
45
36
47
50
45
42
46
46
41
39
45
48
46
52
35
42
37
46
44
39
46
43
45
47
47
51
46
42
43
46
40
51
33
54
47
a) Using class intervals of 5, tabulate this data in a frequency table
b) Calculate the mean mass of the milk delivered.
African Virtual University 25
Solution
Frequency / Tally table
Class
Tally
Frequency
33- 37
37-42
43-47
48-52
53-57
////
///// ///
//// //// //// ///
//// //
//
Total
4
8
19
7
2
40
c) Mean of a grouped data
Class
Tally
Frequency(f)
33- 37
////
4
37-42
43-47
48-52
53-57
///// ///
//// //// //// ///
//// //
//
Total
8
19
7
2
40
Mean =
∑ fx = 1775 = 44.375
∑ f 40
Mid-point (x)
fx
33 + 37
= 35
2
4 ×35 = 140
40
45
50
55
320
855
350
110
1775
African Virtual University 26
DO THIS
Work out the mean of;
1). 63, 65, 67, 68, 69
2).
x
f(x)
1
11
2
10
3
5
4
3
5
1
3).
Weight (x)
Frequency
4-8
2
9-13
4
14-18
7
19-23
14
24-28
8
29-33
5
4). 91,78, 82,73,84
5).
Height (x)
Frequency
61
5
64
18
67
42
70
27
73
8
6).
Weight (x)
Frequency
30.5-36.5
4
36.5-42.5
10
42.5-48.5
14
Answer Key
1). 66.4
2) 2.1
3). 20.6
4) 80
5) 76.45
6) 51.44
48.5-54.5
27
54.5-60.5
45
African Virtual University 27
Lesson Three
Mode
Example
1) Find the mode of the following data: 1,3,4,4,5,6,1,3,3,2,2,3,3,5
Solution:
The mode of a data is the item that appears most times. In this data, 3 occurs most
times or most frequently i.e. 5 times. Therefore the mode is 3.
2) Find the mode of the following data: 22, 24, 25,22, 27, 22, 25, 30, 25, 31
Solution
22 and 25 occur three times each. Therefore the modes are 22 and 25. this is called
a bimodal data.
3) Find the mode of the data:
Observation ( X)
0
1
2
3
4
Frequency ( f)
3
7
10
16
11
Solution
The most occurring observation is 3 i.e. 3 occurs 16 times.
4) Find the modal class of the following data
Weight ( X)
Frequency ( f)
50 – 54
3
55-59
6
60-64
8
65-69
5
70-74
15
75-79
9
Solution
The modal class is 70-74 because it has the highest frequency of occurrence.
80-84
13
African Virtual University 28
DO THIS
Work out the modes or modal classes of the following data;
1) 6, 8, 3,5,2,6,5,9,5
2) 20.4, 20.8, 22.1, 23.4, 19.7, 31.2, 23.4, 20.8, 25.5,23.4
3)
Weight (x)
Frequency
4-8
2
9-13
4
14-18
7
19-23
14
24-28
8
29-33
5
4)
Weight (x)
Frequency
30.5-36.5
4
36.5-42.5
10
42.5-48.5
14
48.5-54.5
27
Answer key
1) 5
2) 23.4
3) 19-23
4) 54.5-60.5
54.5-60.5
45
African Virtual University 29
Lesson Four
Median
The median is the value in the middle of a distribution e.g. in 1, 2,3,4,5, the median is 3
i.e it comes at exactly in the middle of the distribution. For the data 1,2,2,3,4,5,6,7,7,8;
there are 10 terms and no middle number. In such a case, the median is the average
of the two numbers bordering the centre line
Eg 1,2,2,3,
5
4
Therefore the median
6,7,7, 8
4+5
= 4.5
2
Median of a Grouped Data
Example
Find the median of the following grouped data
Mass ( X)
Frequency (f)
20-24
4
25-29
10
30-34
16
35-39
8
40-44
2
Solution
∑ f = 40 Therefore the median is the average of the 20
th
20 + 21
= 10.5th term
2
and 21st terms
African Virtual University 30
Definition: Lower and Upper Limits of a Class.
The Lower Class Limit ( L.C.L) or lower class boundary and the Upper Class Limits
(U.C.L) or upper class boundary are the lower and upper bounds of a class interval
e.g the lower and upper limits of the class interval 20-24 are 19.5 and 20.5 and the
L.C.L and U.C.L of the class interval 35-39 are 34.5 and 39.5.
Mass ( X)
Frequency (f)
Cumulative
20-24
4
4
25-29
10
4+10=14
30- 34
16
14 + 16 = 30
35-39
8
30+8=38
40-44
2
39+2 =40
Frequency
Procedure for Calculation of the Median
Step 1: The median occurs in the class interval 30-34
Step 2: L.C.L and U.C.L of 30-34 are 29.5 and 34.5
Step 3: Work out the Cumulative Frequency ( C.F)
Step 4: Work out the class interval as U.C.L – L.C.L
Step 5: To get the 10.5th term.
Summation difference
10.5th term = L.C.L of class with median + x Class Interval
Class frequency
i.e Summation difference 20.5 – 14 = 6.5 where 14 is the C.F of the class interval
25-29.
Step 6: The median = 29.5 +
6.5
× 5 = 31.53125.
16
Note that the denominator 16 is the class frequency in the class interval 30-34.
Range of a Data
The range of a data is simply the difference between the highest and the lowest score
in a data
Example: 23,26,34, 47,63 the range is 63-23=40 and in 121, 65, 78, 203, 298, 174
the range is 298 – 65= 233.
African Virtual University 31
Lesson Five: Measures Of Dispersion
1) Quartiles
Data arranged in order of magnitude can be subdivided into four equal portions i.e.
25% each. The first portion is the lower quartile occurring at 25%. The middle or
centre occurring at 50% is called the median while the third quarter occurring at
75% is called the upper quartile. The three points are normally referenced as Q1, Q2 ,
Q3 respectively.
2) Semi –interquartile Range
The semi-interquartile range or the quartile deviation of a data is defined as
Q=
Q3 − Q1
2
3) Deciles
If data arranged in order of magnitude is sub-divided into 10 equal portions ( 10%
each), then each portion constitutes a decile. The deciles are denoted by D1, D2,
D3,……D9
4) Percentiles
If data divided arranged in order of magnitude is subdivided into 100 equal portions
(1%each), then the portion constitutes a percentile. Percentiles are denoted as P1,
P2, P3…, P99
The Mean Deviation
The mean deviation (average deviation), of a set of N numbers X1 ,X2, X3, X4, X5,……,
XN is defined by
N
∑ X −X
j
j =1
Mean deviation (MD) =
=
N
∑X −X
N
= X − X , where X is the
arithmetic mean of the numbers and X − X is the absolute value of the deviation
of X
j
from X .
African Virtual University 32
Example
Find the mean deviation of the set
3, 4, 6, 8, 9.
Solution
Arithmetic mean =
3 + 4 + 6 + 8 + 9 30
=
=6
5
5
The mean deviation ( X ) =
−3 + −2 + 0 + 2 + 3
=
5
3− 6 + 4 − 6 6 − 6 + 8 − 6 + 9 − 6
=
5
3 + 2 + 0 + 2 + 3 10
=
=5
5
2
The Mean Deviation of a Grouped Data
For the data
Values
Frequencies
X1
f1
X2
f2
X3
f3
……
….
XN
Fm
The mean deviation can be computed as
m
∑ f X −X
j j
∑f X−X
j =1
Mean deviation =
=
= X−X
N
N
African Virtual University 33
The Standard Deviation
The Standard deviation of a set of N numbers X1 ,X2, X3, X4, X5,……, XN is denoted
by s and is defined by:
N
∑ (X − X )2
j
j =1
=
N
s=
∑ (X − X )2
N
=
∑ x2 =
where x represents the deviations of the numbers X
N
j
(X − X )2
from the mean X .
It follows that the standard deviation is the root mean square of the deviations
from the mean.
The Standard Deviation Of A Grouped Data
Values
Frequencies
X1
f1
X2
f2
X3
f3
……
….
XN
Fm
The standard deviation is calculated as:
s=
m
∑ f (X − X )2
j
j =1
=
N
∑ f (X − X )2
=
N
∑ fx 2
= (X − X )2
N
m
where N= ∑ f = ∑ f .
j =1
j
The Variance
The variance of a set of data is defined as the square of the standard deviation i.e
variance = s2. We sometimes use s to denote the standard deviation of a sample of a
population and σ ( Greek letter sigma ) to denote the standard deviation of a po-
pulation population. Thus σ 2 can represent the variance of a population and s2 the
variance of sample of a population.
African Virtual University 34
Examples
Find the Mean and Range of the following data: 5,5,4,4,4,2,2,2
Solutions
∑n
Mean = m
N
x =
5+5+4+4+4+4+2+2+2
= 3.56
9
= 3.56
Range 5 – 2 =3.
Median (Middle )Observation
Example
Given 13 observations
1,1,2,3,4,4,5,6,8,10,14,15,17
The median =
The value
n + 1 14
7
=
= 607
2
2
14
= 7th position. The median is 5
2
If n is odd the Median is the value in position
n+1
2
But if it is even, we consider the average of the two middle terms.
African Virtual University 35
10) Example
1,1,2,2,3,4,4,5,6,8,10,14,15,17
The median = Average of the Middle two terms
4+5
= 4.5
2
=
Median of Grouped Data
When data are grouped the median c 2 is the value at or below 50% of the observation fall.
DO THIS
Find the median of the following data
1. 1,1,2,2,3,4,5,7,7,7,9
2. 7,8,1,1,9,19,11,2,3,4,8
q
Group Work
Study the computation
of the variance and standard
from the following example.
Definition
The mean squared deviation from the mean is called variance:
−
Σ h (x − x )2
s =
N
2
−
Where: x − x is deviation from the mean, N is number of observations
s 2 is variance and
s 2 is standard deviation.
African Virtual University 36
Example
Given the data 2,4,5,8,11. Find the variance and the standard deviation.
X
−
−
(x − x )2
16
4
1
4
25
x−x
2
4
5
8
11
-4
-2
-1
2
5
∑ x =5
∑
−
(x − x )2 =50
−
So x =
30
=6
5
Variance= s 2 =
52 =
50
= 10
5
50
= 10
5
Standard deviation = √10.
DO THIS
1) Calculate range of the data:
1,1,1,2,2,3,3,3,4,5
10) Calculate the variance and the standard deviation: 1,2,3,4,5
Skewness
Definition: Skewness is the degree of departure from symmetry of a distribution.
( Check positive and negative skewness above)
For skewed distributions, the mean tends to lie on the same side of the mode as the
longer tail.
African Virtual University 37
Pearson’s First Coefficient of Skewness
This coefficient is defined as
Skewness=
mean − mod e
X − mod e
=
s tan dard deviation
s
Pearson’s Second Coefficient of Skewness
This coefficient is defined as:
Skewness=
3(mean − median)
3(X − median)
=
s tan dard deviation
s
Quartile Coefficient of Skewness
This is defined as:
Quartile coefficient of skewness =
(Q − Q ) − (Q − Q ) Q − 2Q + Q
3
2
2
1 = 3
2
1
Q −Q
Q −Q
3
1
3
1
10-90 Percentile of Skewness
This is defined as:
10-90 percentile of skewness =
(P − P ) − (P − P ) P − 2P + P
90
50
50 10 = 90
50 10
P −P
P −P
90 10
90 10
African Virtual University 38
(n + 1)x0.25
9(.25) = 22.5( percentile)
Example: Find 25th percentile of the data 1, 2, 3, 4, 5, 6, 7, 9
25th percentile = (n+1)x0.25 = 9(.25) = 2.25 (percentile)
2nd = 2
3rd = 3
2.25 ⇒ 0.25(1) + 2 = 2.25
Find 50th percentile
50th percentile: (8 + 1)x.50 = 9(.5) = 4.5 percentile
4th = 4
5th = 5
0.5(5) = 0.5 + 4 = 4.5
The (1) is the range 5 − 4 = 1
q
Group Work
1. Study the computation
of percentiles and
attempt the following
question..
African Virtual University 39
DO THIS
Find the 25th percentile, the 50th percentile, and 90th percentile
46,21,89,42,35,36,67,53,42,75,42,75,47,85,40,73,48,32,41,20,75,48,48,32,52,61,
49,50,69,59,30,40,31,25,43,52,62,50
Answer Key
a) 36
b) 48
c) 73
Kurtosis
Definition: Kurtosis is the degree of peakedness of a distribution, as compared to the
normal distribution.
Eamples
1) Leptokurtic Distribution
A distribution having a relatively high peak
2) Platykurtic Distribution
A distribution having a relatively flat top
African Virtual University 40
3). Mesokurtic Distribution
A Normal Distribution – not very peaked or flat topped
DO THIS
Find the mode for the data collection:
1) 1,3,4,4,2,3,5,1,3,3,5,4,2,2,2,3,3,4,4,5
2) Number of marriage per 1000 persons in Africa population for years 1965
– 1975
Year
1965
1966
1967
1968
1969
1970
1971
1972
1973
1974
1975
Rate
9.3
9.5
9.7
10.4
10.6
10.6
10.6
10.9
10.8
10.5
10.0
African Virtual University 41
3) Number of deaths per 1000 years for years 1960 and 1965 – 1975
1960
1965
1966
1967
1968
1969
1970
1971
1972
1973
1974
1975
9.5
9.4
9.5
9.4
9.7
9.5
9.5
9.3
9.4
9.3
9.1
8.8
Solutions
1. 3
2. 10.6
3. 9.5
READ:
1) An Introduction to Probability by
Charles M. Grinstead
pages 247 -263
• Exercise on pg 263-267 Nos.
4,7,8,9
African Virtual University 42
Probability
1) Sample Space and Events
Terminology
a) A Probability experiment
When you toss a coin or pick a card from a deck of playing cards or roll a dice, the
act constitutes a probability experiment. In a probability experiment, the chances
are well defined with equal chances of occurrence e.g. there are only two possible
chances of occurrence in tossing a coin. You either get a head or tail. The head and
the tail have equal chances of occurrence.
b) An Outcome
This is defined as the result of a single trial of a probability experiment e.g. When
you toss a coin once, you either get head or tail.
c) A trial
This refers to an activity of carrying out an experiment like picking a card from a
deck of cards or rolling a die or dices.
d) Sample Space
This refers to all possible outcomes of a probability experiment. e.g. in tossing a coin,
the outcomes are either Head(H) or tail(T) i.e there are only two possible outcomes
in tossing a coin. The chances of obtaining a head or a tail are equal.
e) A Simple and Compound Events
In an experimental probability, an event with only one outcome is called a simple
event. If an event has two or more outcomes, it is called a compound event.
2) Definition of Probability
Probability can be defined as the mathematics of chance. There are mainly four
approaches to probability;
1)
2)
3)
4)
The classical or priori approach
The relative frequency or empirical approach
The axiomatic approach
The personalistic approach
African Virtual University 43
The Classical or A Priori Approach
Probability is the ratio of the number of favourable cases as compared to the total
likely cases. Suppose an event can occur in N ways out of a total of M possible ways.
Then the probability of occurrence of the event is denoted by
N
p=Pr(N)=
. Probability refers to the ratio of possible outcomes to all possible
outcomes. M
The probability of non-occurrence of the same event is given by {1-p(occurrence)}.
The probability of occurrence plus non-occurrence is equal to one.
If probability occurrence; p(O) and probability of non-occurrence (O’), then
p(O)+p(O’)=1.
Empirical Probability ( Relative Frequency Probability)
Empirical probability arises when frequency distributions are used.
For example:
Observation ( X)
0
1
2
3
4
Frequency ( f)
3
7
10
16
11
The probability of observation (X) occurring 2 times is given by the formulae
P(2)=
freuency of 2
f (2)
10
10
=
=
=
sum of frequencies ∑ f 3 + 7 + 10 + 16 + 11 47
3) Properties of Probability
a) Probability of any event lies between 0 and 1 i.e. 0 p(O) 1. It follows that probability cannot be negative nor greater than 1.
b) Probability of an impossible event ( an event that cannot occur ) is always
zero(0)
c) Probability of an event that will certainly occur is 1.
d) The total sum of probabilities of all the possible outcomes in a sample space
is always equal to one(1).
e) If the probability of occurrence is p(o)= A, then the probability of non-occurrence is 1-A.
African Virtual University 44
Counting Rules
1) Factorials
Definition: Factorial 4 ! = 4 x 3 x 2 x 1 and 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1
2) Permutation Rules
Definition:
P
n r
=
n !
(n − r ) !
Examples
5!
5x4x3x2x1
=
= 5x4x3 = 60
(5 − 3)!
2x1
8!
8! 8x7x6x5x4x3x2x1
• 8P5 =
= =
= 8x7x6x5x4 = 6720
(8 − 5)! 3!
3x2x1
• 5P3 =
3) Combinations
Definition: nCr =
n !
(n − r ) ! r !
Examples
C2 =
5!
5x4x3x2x1 5x4
=
=
= 10
(5 − 2)!2!
3! 2!
2x1
C6 =
10!
10!
10x9x8x7x 6! 10x9x8x7
=
=
=
= 210
(10 − 6)!6! 4! 6!
4x3x21x 6!
4x3x2x1
•
5
•
10
African Virtual University 45
DO THIS
Work out the following;
1).
2)
3)
4)
5)
6)
7)
8)
P
C
8 3
C
15 10
C
6 3
P
15 4
C
9 3
C
10 8
P
7 4
8 3
Answer key
1) 336
2) 56
3) 3003
4) 20
5) 32 760
6)84
7)90
8) 840
African Virtual University 46
Rules of Probability
Addition Rules
1) Rule 1: When two events A and B are mutually exclusive, then
P(A or B)=P(A)+P(B)
Example: When a is tossed, find the probability of getting a 3 or 5.
Solution: P(3) =1/6 and P(5) =1/6.
Therefore P( 3 or 5) = P(3) + P(5) = 1/6+1/6 =2/6=1/3.
2) Rule 2: If A and B are two events that are NOT mutually exclusive, then
P(A or B) = P(A) + P(B) - P(A and B), where A and B means the number of outcomes that event A and B have in common.
Example: When a card is drawn from a pack of 52 cards, find the probability that the card is a 10 or a heart.
Solution
P( 10) = 4/52 and P( heart)=13/52
P ( 10 that is Heart) = 1/52
P( A or B) = P(A) +P(B)-P( A and B) = 4/52 _ 13/52 – 1/52 = 16/52.
Multiplication Rules
1) Rule 1: For two independent events A and B, then P( A and B) = P(A) x P(B).
Example: Determine the probability of obtaining a 5 on a die and a tail on a coin
in one throw.
Solution: P( 5) =1/6 and P(T) =1/2.
P(5 and T)= P( 5) x P(T) = 1/6 x ½= 1/12.
2) Rule 2: When to events are dependent, the probability of both events occurring
is P(A and B)=P(A) x P(B|A), where P(B|A) is the probability that event B occurs
given that event A has already occurred.
African Virtual University 47
Example: Find the probability of obtaining two Aces from a pack of 52 cards without
replacement.
Solution: P( Ace) =2/52 and P( second Ace if NO replacement) = 3/51
Therefore P(Ace and Ace) = P(Ace) x P( Second Ace) = 4/52 x 3/51 = 1/221
Conditional Probability
The conditional probability of two events A and B is P(A|B) =
P (A and B)
,
P (B)
where P(A and B) means the probability of the outcomes that events A and B have
in common.
Example: When a die is rolled once, find the probability of getting a 4 given that
an even number occurred in an earlier throw.
Solution: P( 4 and an even number) = 1/6 ie. P(A and B) =1/6. P(even number) =3/6
=1/2.
P( A|B) =
P (A and B)
=
P (B)
1
1
6
2
=
1
3
Examples
1) A bag contains 3 orange, 3 yellow and 2 white marbles. Three marbles are selected without replacement. Find the probability of selecting two yellow and a
white marble.
Solution. P( 1st Y) =3/8, P( 2nd Y) = 2/7 and P( W)= 2/6
P(Y and Y and W)=P(Y) x P(Y) x P(W) = 3/8 x 2/7 x 2/6 = 1 / 28
2) In a class, there are 8 girls and 6 boys. If three students are selected at random
for debating, find the probability that all girls.
Solution: P( G) =8/14 and P(B) =6/14. P( 1st G)=8/14, P(2nd G) 7/13 and P(3rdG)=
6/12.
P( three girls) 8/14 x 7/13 x 6/12= 2/13
3) In how many ways can 3 drama officials be selected from 8 members?
Solution:
C3
8
= 56 ways.
African Virtual University 48
4) A box has 12 bulbs, of which 3 are defective. If 4 bulbs are sold, find the probability that exactly one will be defective.
Solution
P( defective bulb)= 3C1 and P( non-defective bulbs) = 9C3
3
C1 x 9C3 =
3!
9!
x
= 252
(3 − 1)!1! (9 − 3)!3!
P( 4 bulbs from 12) = 12C4 = 495.
P( 1 defective bulb and 3 okey bulbs) = 295/495=0.509.
DO THIS
1) In how many ways can 7 dresses be displayed in a row on a shelf?
2) In how many ways can 3 pens be selected from 12 pens?
3) From a pack of 52 cards, 3 cards are selected. What is the probability that
they will all be diamonds?
Answer Key
1) 5040
2) 220
3) 0.013
African Virtual University 49
READ:
An Introduction to Probability & Random
Processes By Kenneth B & Gian-Carlo R,
pages
1. 1.20 -1.22
• Exercise Chapter 1: Sets, Events &
Probability Pg 1.23-1.28 Nos. 1-12
& 14-20
2. 2.1-2.33
• Exercise Chapter 2: Finite Processes Pg 2.33 Nos. 1,2,3,13-20,
22-27
3. Introduction to Probability, By Charles
M. Grinstead pages139-141
Random Variables
Random Variables ( r.v)
Definition: A random variable is a function that assigns a real number to every possible result of a random experiment.
(Harry Frank & Steve C Althoen,CUP, 1994, pg 155)
A random variable is a variable in the sense that it can be used as a placeholder for
a number in equations and inequalities. Its randomness is completely described by
its cumulative distribution function which can be used to determine the probability
it takes on particular values.
Formally, a random variable is a measurable function from a probability space to the
real numbers. For example, a random variable can be used to describe the process
of rolling a fair die and the possible outcomes { 1, 2, 3, 4, 5, 6 }. The most obvious
representation is to take this set as the sample space, the probability measure to be
uniform measure, and the function to be the identity function.
Random variable
Some consider the expression random variable a misnomer, as a random variable is
not a variable but rather a function that maps outcomes (of an experiment) to numbers.
Let A be a σ-algebra and Ω the space of outcomes relevant to the experiment being
performed. In the die-rolling example, the space of outcomes is the set Ω = { 1, 2,
3, 4, 5, 6 }, and A would be the power set of Ω. In this case, an appropriate random
African Virtual University 50
variable might be the identity function X(ω) = ω, such that if the outcome is a ‘1’,
then the random variable is also equal to 1. An equally simple but less trivial example
is one in which we might toss a coin: a suitable space of possible outcomes is Ω = {
H, T } (for heads and tails), and A equal again to the power set of Ω. One among the
many possible random variables defined on this space is
Mathematically, a random variable is defined as a measurable function from a sample
space to some measurable space.
Convergence of Random Variables
In probability theory, there are several notions of convergence for random variables.
They are listed below in the order of strength, i.e., any subsequent notion convergence
in the list implies convergence according to all of the preceding notions.
Convergence in distribution: As the name implies, a sequence of random variables
converges to the random variable
in distribution if their respective cumulative distribution functions
distribution function of
, wherever
converge to the cumulative
is continuous.
Weak convergence: The sequence of random variables
is said to conver-
ge towards the random variable
weakly if
for every ε > 0. Weak convergence is also called convergence in probability.
Strong convergence: The sequence of random variables
is said to
converge towards the random variable
strongly if
Strong convergence is also known as almost sure convergence.
Intuitively, strong convergence is a stronger version of the weak convergence, and
in both cases the random variables
show an increasing correlation
with
. However, in case of convergence in distribution, the realized values of the
random variables do not need to converge, and any possible correlation among them
is immaterial.
African Virtual University 51
Law of Large Numbers
If a fair coin is tossed, we know that roughly half of the time it will turn up heads,
and the other half it will turn up tails. It also seems that the more we toss it, the more
likely it is that the ratio of heads:tails will approach 1:1. Modern probability allows
us to formally arrive at the same result, dubbed the law of large numbers. This
result is remarkable because it was nowhere assumed while building the theory and
is completely an offshoot of the theory. Linking theoretically-derived probabilities
to their actual frequency of occurrence in the real world, this result is considered as
a pillar in the history of statistical theory.
The strong law of large numbers (SLLN) states that if an event of probability p is
observed repeatedly during independent experiments, the ratio of the observed frequency of that event to the total number of repetitions converges towards p strongly
in probability.
In other words, if
are independent Bernoulli random variables taking
values 1 with probability p and 0 with probability 1-p, then the sequence of random
numbers
converges to p almost surely, i.e.
Central Limit Theorem
The central limit theorem is the reason for the ubiquitous occurrence of the normal
distribution in nature, for which it is one of the most celebrated theorems in probability and statistics.
The theorem states that the average of many independent and identically distributed
random variables tends towards a normal distribution irrespective of which distribution
the original random variables follow. Formally, let
random variables with means
sequence of random variables
, and variances
converges in distribution to a standard normal random variable.
be independent
Then the
African Virtual University 52
Functions of Random Variables
If we have a random variable X on Ω and a measurable function f: R → R, then Y
= f(X) will also be a random variable on Ω, since the composition of measurable
functions is also measurable. The same procedure that allowed one to go from a
probability space (Ω, P) to (R, dFX) can be used to obtain the distribution of Y. The
cumulative distribution function of Y is
Example
Let X be a real-valued, continuous random variable and let Y = X2. Then,
If y < 0, then P(X2 ≤ y) = 0, so
If y ≥ 0, then
So
Probability Distributions
Certain random variables occur very often in probability theory due to many natural
and physical processes. Their distributions therefore have gained special importance in
probability theory. Some fundamental discrete distributions are the discrete uniform,
Bernoulli, binomial, negative binomial, Poisson and geometric distributions. Important continuous distributions include the continuous uniform, normal, exponential,
gamma and beta distributions.
Distribution Functions
If a random variable
defined on the probability space (Ω,A,P) is given,
we can ask questions like “How likely is it that the value of X is bigger than 2?”.
This is the same as the probability of the event
is often written as P(X > 2) for short.
which
Recording all these probabilities of output ranges of a real-valued random variable X
African Virtual University 53
yields the probability distribution of X. The probability distribution “forgets” about
the particular probability space used to define X and only records the probabilities
of various values of X. Such a probability distribution can always be captured by its
cumulative distribution function
and sometimes also using a probability density function. In measure-theoretic terms,
we use the random variable X to “push-forward” the measure P on Ω to a measure
dF on R. The underlying probability space Ω is a technical device used to guarantee
the existence of random variables, and sometimes to construct them. In practice, one
often disposes of the space Ω altogether and just puts a measure on R that assigns
measure 1 to the whole real line, i.e., one works with probability distributions instead
of random variables.
Discrete Probability Theory
Discrete probability theory deals with events which occur in countable sample
spaces.
Examples: Throwing dice, experiments with decks of cards, and random walk.
Classical definition: Initially the probability of an event to occur was defined as number of cases favorable for the event, over the number of total outcomes possible.
For example, if the event is “occurrence of an even number when a die is rolled”, the
probability is given by
, since 3 faces out of the 6 have even numbers.
Modern definition: The modern definition starts with a set called the sample
space which relates to the set of all possible outcomes in classical sense, denoted by
. It is then assumed that for each element
“probability” value
1.
2.
, an intrinsic
is attached, which satisfies the following properties:
An event is defined as any subset
event defined as
of the sample space
. The probability of the
So, the probability of the entire sample space is 1, and the probability of the null
event is 0.
African Virtual University 54
The function
mapping a point in the sample space to the “probability” value
is called a probability mass function abbreviated as pmf. The modern definition
does not try to answer how probability mass functions are obtained; instead it builds
a theory that assumes their existence.
Continuous Probability Theory
Continuous probability theory deals with events which occur in a continuous
sample space.
If the sample space is the real numbers, then a function called the cumulative distribution function or cdf
.
is assumed to exist, which gives
The cdf must satisfy the following properties.
1.
is a monotonically non-decreasing right-continuous function
2.
3.
If
is differentiable, then the random variable is said to have a probability density
function or pdf or simply density
For a set
.
, the probability of the random variable being in
is defined as
In case the density exists, then it can be written as
Whereas the pdf exists only for continuous random variables, the cdf exists for all
random variables (including discrete random variables) that take values on .
These concepts can be generalized for multidimensional cases on
.
African Virtual University 55
Probability Density Function
Discrete Distribution
If X is a variable that can assume a discrete set of values X1, X2, X3,…….., Xk wih
respet to probabilities p1, p2, p3,……., pk, where p1+ p2 + p3,……., + pk = 1, we say
that a discrete probability distribution for X has been defined. The function p(X),
which has the respective values p1, p2, p3,……., pk for X= X1, X2, X3,…….., Xk is
called the probability function, or frequency function, of X. Because X can assume
certain values with given probabilities, it is often called a discrete random variable. A
random variable is also known as a chance variable or stochastic variable. { Murray
R, 2006 pg 130}
Continuous Distribution
Suppose X is a continuous random variable. A continuous random variable X is specified by its probability density function which is written f(x) where f(x) ≥ 0 throughout
the range of values for which x is valid. This probability density function can be
represented by a curve, and the probabilities are given by the area under the curve.
The total area under the curve is equal to 1. The are under the curve between the
lines x=a and x=b ( shaded) gives the probability that X lies between a and b, which
can be denoted by P(a<X<b). p(X) is called a probability density function and the
variable X is often called a continuous random variable
Since the total area under the curve is equal to 1, it follows that the probability between
a range space a and b is given by
P (a ≤ X ≤ b) =
b
∫ f (x)dx ,
a
which is the shaded area.
African Virtual University 56
Note: when computing area from a to b, we need not distinguish
(≤ and ≥) and (< and >) inequalities. We assume the lines at a and b have no
thickness and its area is zero.
Solved Examples
1) The continuous random variable X is distributed with probability density function
f defined by
f(x) = kx(16-x2), for 0<x<4.
Evaluate
a). The value of constant k
b). The probability of range space P(1<X<2)
c). The probability P(x ≥ 3)
Solution
f(x)
x
b
a
For any function f(x) such tha
f(x) ≥ 0, for a ≤ x ≤ b,
and
b
∫a f (x)dx = 1
may be taken as the probability density function (p.d.f) of a continuous random variable in the range space a ≤ x ≤ b.
African Virtual University 57
Procedure
Step 1: In general, if X is a continuous random variable (r.v) with p.d.f f(x) valid
over the range a ≤ x ≤ b, then
∫ f (x)dx = 1 i.e.
all x
∫
b
a
f (x)dx = 1
Step 2
a). To determine k, we use the fact that in f(x) = kx(16-x2), for 0<x<4, then
∫
4
0
kx(16 − x 2 )dx = 1
4
⇒ k ∫ 16x − x 3 )dx = 1
0
⇒ k=
Step 3
b). 1
64
Find P(1<X<2)
Solution
2
P(1<X<2)= ∫ f (x)dx
1
=
1 2
81
(16x − x 3 )dx =
∫
1
64
256
Step 4
c). To find P(x ≥ 3) P (x ≥ 3) =
1 4
49
(16x − x 3 )dx =
∫
64 3
256
African Virtual University 58
Example 2
2). X is the continuous random variable ‘the mass of a substance, in kg, per minute in an industrial production process’, where
⎧1
⎪ x(6 − x)
f (x) = ⎨12
⎪⎩
0
(0 ≤ x ≤ 3)
otherwise
Find the probability that the mass is more than 2 kg.
Solution
X can take values from 0 to 3 only. We sketch f(x), and shade the area required.
f ( x) =
f(x)
1
x( 6 − x)
12
x
0
P (x > 2) =
=
2
∫
3
2
3
1
x(6 − x)dx
12
1 3
(6x − x 2 )dx
12 ∫2
3
1 ⎡
x3 ⎤
= ⎢ 3x 2 − ⎥
12 ⎣
3 ⎦2
= 0.722 (3 d.p)
The probability that the mass is more than 2 kg is 0.722.
African Virtual University 59
Worked example
3). A continuous random variable has p.d.f f(x) where
f (x) = kx 2 , 0 ≤ x ≤ 6.
a).
Find the value of k
b).
Find P (2 ≤ X ≤ 4)
Solution
a). Since X is a random variable the total probability is 1. i.e.
∫ f (x)dx = 1
all
⇒
∫
6
0
kx 2 dx = 1
6
⎡ kx 3 ⎤
⎢ 3 ⎥ = =1
⎣
⎦0
216k
=1
3
3
⇒k=
216
Therefore f(x)=
3 2
1 2
x =
x , 0≤x≤6
216
72
b).
f ( x) =
f(x)
x
0
2
4
6
1
x2
72
African Virtual University 60
P (2 ≤ x ≤ 4) =
1 3
x
216
= 0.259
=
]
∫
4
2
1 2
x dx
72
4
2
Therefore the probability P (2 ≤ X ≤ 4) = 0.259
Worked Example
4). The continuous random variable (r.v) has a probability density function(p.d.f)
where
⎧k
⎪
f (x) = ⎨ k(2x − 3)
⎪0
⎩
0≤x<2
2≤x≤5
otherwise
a). Find the value of the constant k
b). Sketch y=f(x)
c). Find P(X ≤ 1)
d). Find P(X>2.5)
Solution
a). Since X is a r.v, then
∫
f (x)dx = 1
all x
Therefore
∫
2
0
5
kdx + ∫ k(2x − 3)dx = 1
2
African Virtual University 61
kx
2
0
5
+ k ⎡⎣ x 2 − 3x ⎤⎦ 2
2k + 19k = 1
1
⇒k=
21
b). So the p.d.f of X is
⎧ 1
⎪ 21
⎪
⎪
⎪1
⎪ (2x − 3)
f (x) = ⎨ 21
⎪
⎪
⎪0
⎪
⎪
⎩
0≤x<2
2≤x≤5
otherwise
Sketch
1
3
1
21
0
1
2
2.5
3
4
5
African Virtual University 62
c). P(x ≤ 1) = area between zero and 1 = L x W= 1 x
1 1
=
= 0.048
21 21
d). Find P(X>2.5) = area of rectangle + area of trapezium.
=(
1
1
1
2
11
x 2 ) + ( {0.5}{ + } =
= 0.131
21
2
21 21 84
African Virtual University 63
q
Reflection : Teachers may find graph drawing software useful in
the teaching of statistics.
An example of Open Source software is Graph. See: http://www.
padowan.dk/graph/
If you have computer access, download graph and explore its statistical features.
Here is an example of different trendlines which can be drawn using
Graph.
African Virtual University 64
DO THIS
1). The continuous random variable X has p.d.f f(x) where f(x)= k, 0 ≤ x ≥ 3 .
a) Sketch y=f(x)
b). Find the value of the constant k
c). Find P(0.5 ≤ X ≤ 1
2). The continuous random variable has p.d.f f(x) where f(x)=kx2, 1 ≤ x ≤ 4 .
a). Find the value of the constant
b). Find P(x ≥ 2)
c). Find P(2.5 ≤ x ≤ 3.5
3). The continuous random variable has p.d.f f(x) where
⎧k
⎪
f (x) ⎨ k(2x − 1)
⎪0
⎩
0≤x<2
2≤x≤3
otherwise
a) find the value of the constant k.
b) Sketch y=f(x)
c) Find P(X ≤ 2 )
d) Find P(1 ≤ X ≤ 2.2)
African Virtual University 65
Expectation
Definition
If X is a continuous variable (r.v) with probability density function (p.d.f) f(x), then
the expectation of X is E(X) where
E (X ) =
∫
x f (x)dx
all x
NB: E(X) is often denoted by μ and referred to as the mean of X
Example
1
x2 ,
1). If X is a continuous variable ( r.v) with a p.d.f f (x) =
16
find E(X).
0 ≤ x ≤ 3, Solution
E (X ) =
x f (x)dx
∫
all x
⇒∫
3
0
1
{x} x 2 dx
16
3
1 ⎡ x4 ⎤
81
= ⎢ ⎥ =
= 1.265
16 ⎣ 4 ⎦ 0 64
2). If the continuous random variable X has p.d.f f (x) =
E (X ) =
2
(3 + x)(x − 1),
5
∫
all x
x f (x)dx
1 ≤ x ≤ 3 , find E(X).
African Virtual University 66
E (x) =
∫
3
1
2
{x} (3 + x)(x − 1)dx
5
3
2 ⎡ x 4 2x 3 3x 2 ⎤
= ⎢ +
−
5⎣ 4
3
2 ⎥⎦1
608
=
60
= 10.13
Generalisation
If g( x) is any function of the continuous random variable r.v X having p.d.f f(x),
then
E [ g(X )] =
∫
g(x) f (x)dx
all x
and in particular
E (X 2 ) =
∫
x 2 f ( x ) dx
all x
The following conclusions hold
1. E (a) = a
2. E (aX ) = aE (X )
3. E (aX + b) = aE (X ) + b
4. E [ ( f1 (X ) + f2 (X )] = E [ f2 (X )]
Example
1). The continuous random variable X has p.d.f f(x) where f(x)=
Find
a).
E(X)
b).
E(X2)
c).
E(2X +3)
1
x, 0 ≤ x ≤ 3.
2
African Virtual University 67
Solution
a) E (X ) =
x f (x)dx
∫
all x
=
∫
3
0
1 2
x dx
2
3
1 ⎡ x3 ⎤
= ⎢ ⎥
2 ⎣ 3 ⎦0
= 4.5
b)
E (X 2 ) =
=
∫
all x
x 2 f (x)dx
1 3 3
x dx
2 ∫0
3
1 ⎡ x4 ⎤
= ⎢ ⎥
2 ⎣ 4 ⎦0
=
81
= 10.125
8
c). E(2X +3) = E (2X) + 3
= 2E(X) +3
= 2(10.125)+5
= 25.25 ( from (b) above)
African Virtual University 68
DO THIS
1) The continuous random variable X has p.d.f f(x) where
⎧
⎪ kx
⎪⎪
f (x) = ⎨ k
⎪ k(4 − x)
⎪
⎪⎩0
0 ≤ x <1
1≤ x < 3
3≤ x ≤5
otherwise
a). Find k
b). Calculate E(X)
2) The continuous random variable has p.d.f f(x) where f(x) = 1 (x + 3), 0 ≤ x ≤ 5
10
a).
b).
c).
d).
Find E(X)
Find E(2X+4)
Find E(X2).
Find E( X2 + 2X – 1).
African Virtual University 69
Bernoulli Distribution
In probability theory and statistics, the Bernoulli distribution, named after Swiss
scientist Jakob Bernoulli, is a discrete probability distribution, which takes value 1
with success probability p and value 0 with failure probability q = 1 − p. So if X is a
random variable with this distribution, we have:
The probability mass function f of this distribution is
The expected value of a Bernoulli random variable X is
riance is
, and its va-
The kurtosis goes to infinity for high and low values of p, but for p = 1 / 2 the Bernoulli
distribution has a lower kurtosis than any other probability distribution, namely -2.
The Bernoulli distribution is a member of the exponential family.
Binomial Distribution
In probability theory and statistics, the binomial distribution is the discrete probability distribution of the number of successes in a sequence of n independent yes/no
experiments, each of which yields success with probability p. Such a success/failure
experiment is also called a Bernoulli experiment or Bernoulli trial. In fact, when n =
1, the binomial distribution is a Bernoulli distribution. The binomial distribution is
the basis for the popular binomial test of statistical significance.
Examples
An elementary example is this: roll a die ten times and count the number of 1s as
outcome. Then this random number follows a binomial distribution with n = 10 and
p = 1/6.
For example, assume 5% of the population is green-eyed. You pick 500 people
randomly. The number of green-eyed people you pick is a random variable X which
follows a binomial distribution with n = 500 and p = 0.05 (when picking the people
with replacement).
African Virtual University 70
Examples
1) A coin is tossed 3 times. Find the probability of getting 2 heads and a tail in any
given order.
Formula
We can use the formula
Cx. (p)x.(1-p)n-x
n
Where n = the total number of trials
x = the number of successes ( 1,2,…)
p= the probability of a success.
1st)
n
Cx
determines the number of ways a success can occur.
2nd)
(p)x is the probability of getting x successes and
3rd)
(1-p)n-x is the probability of getting n-x failures
Solution
Tossing 3 times means n=3
Two heads means x=2
P(H)=1/2;
P(T)=1/2
P( 2 heads) = 3C2.
1
2
( )2.(1-
1 3-1
) = 3(1/4)(1/2)= 3/8
2
African Virtual University 71
DO THIS
1) Find the probability of exactly one 5 when a die is rolled 3 times
2) Find the probability of getting 3 heads when 8 coins are tossed.
3) A bag contains 4 red and 2 green balls. A ball is drawn and replaced 4 times. What is the probability of getting exactly 3 red balls and 1 green ball.
Answer
1
6
5 2
) = 25/72 = 0.347 i.e n=3, x=1, p=1/6
6
1
2
1 5
) = 7/32 = 0.218. i.e n=8, x=3, p=1/2
2
= 3C1.
( )1.(
2). P ( 3 heads) = 8C3.
( )3.(
1). P( one 5)
3). P( 3 Red balls) = 4C3.
2
3
( )3.(
1 1
) = 32/81= 0.395 i.e. n=4, x=3, p=2/3
3
READ:
1. Lectures on Statistics, By Robert B. Ash, , page 1-4
• Exercise Nos.1, 2 and 3 on pg 4.
2. An Introduction to Probability & Random Processes By
Kenneth B & Gian-Carlo R, pages 3.1-3.63
• Exercise Chapter 3: Random Variables pg 3.64-3.82
Nos. 1-7, 11-17, 20-24, 34-36
3. An Introduction to Probability By Charles M. Grinstead
pages 96-107, & 184
• Exercise on pages 113-118
Nos. 1,2,3,4,5,8,9,10,19,20
Ref: http://en.wikipedia.org/wiki/measurable_space
Ref: http://en.wikipedia.org/wiki/Probability_theory
Ref: http://en.wikipedia.org/wiki/Bernoulli_distribution
African Virtual University 72
Poisson Distribution
In probability theory and statistics, the Poisson distribution is a discrete probability
distribution that expresses the probability of a number of events occurring in a fixed
period of time if these events occur with a known average rate, and are independent
of the time since the last event.
The distribution was discovered by Siméon-Denis Poisson (1781–1840)
The Poisson distribution is sometimes called a Poissonian, analagous to the term
Gaussian for a Gauss or normal distribution.
The Poisson distribution is used when the variable occurs over a period of time,
volume, area etc…it can be used for the arrival of airplanes at airports, the number
of phone calls per hour for a station, the number of white blood cells on a certain
area.
The probability of x successes is
e− λ λ x
x!
where e is a mathematical constant = 2.7183
λ is the mean or expected value of the variables.
q
Group Work
1. Study the probability computations
and attempt the given question.
Example
If there are 100 typographical errors randomly distributed. In 500 pages manuscripts
find the probability that any given page has exactly 4 errors.
Solution
Find the mean number of errors λ = 100/500 = 1 / 5 = 0,2
In other words there is an average of 0.2 errors per page. In this case λ = 4 so the
probability of selecting a page with exactly 4 errors
e .λ x ( 2.7183) ( 0.2 )
=
x!
41
λ
−0.2
= 0.00168
Amount 0.2%
4
African Virtual University 73
Worked Example
A hot line with a full free number receives an average of 4 calls per hour for any
given hour. Find the probability that it will receive exactly 5 calls.
e .λ x ( 2.7183)
=
x!
5!
= 0.1001
Which is 10%
λ
−3
( 3)
5
DO THIS
1) A telephone Marketing Company gets an average of 5 orders per 1000 calls. If
a company calls 500 people find the probability of getting 2 orders.
Solution
0.26
Which is 26%
READ:
1. An Introduction to Probability & Random Processes By
Kenneth B & Gian-Carlo R, pages187-192
2. Robert B. Ash, Lectures on Statistics, page 1 and Answer
problems 1,2,3 on pg 15.
Ref: http://en.wikipedia.org/wiki/Normal_distribution
African Virtual University 74
Geometric Distribution
In probability theory and statistics, the geometric distribution is either of two discrete probability distributions:
the probability distribution of the number X of Bernoulli trials needed to get
one success, supported on the set { 1, 2, 3, ...}, or
• the probability distribution of the number Y = X − 1 of failures before the first
success, supported on the set { 0, 1, 2, 3, ... }.
Which of these one calls “the” geometric distribution is a matter of convention and
convenience.
•
If the probability of success on each trial is p1, then the probability that k trials are
needed to get one success is
for k = 1, 2, 3, ....
Equivalently, if the probability of success on each trial is p0, then the probability that
there are k failures before the first success is
for k = 0, 1, 2, 3, ....
In either case, the sequence of probabilities is a geometric sequence.
For example, suppose an ordinary die is thrown repeatedly until the first time a “1”
appears. The probability distribution of the number of times it is thrown is supported
on the infinite set { 1, 2, 3, ... } and is a geometric distribution with p1 = 1/6.
Solutions Using The Geometric Distribution Formula
The formula for the probability that the first success occurs on the nth trial is
(1-p)n-1p or simply
, where p is the probability
of a success and n is the trial number of the first success.
Example
1) Find the probability that the first tail occurs on the third toss of a coin.
Solution
The outcome of a tail on the third throw implies HHT. From
and therefore P(HHT) = ( 1-
1 3-1 ( 1 ) 1
1 1
)
= ( ) .. ( ) ( ) =1/8
2
2
2
2 2
(1-p)n-1p , n=3, p=1/2
African Virtual University 75
Examples In Geometric Distribution
Flipping a coin several times we apply the geometric to distribution to get the answer
of flipping a coin several times.
Example
1) A coin is tossed find the probability that the first head occurs on the third toss
solution. Out come is TTH
n = 3 and p=1/2
Probability of getting 2 tails and then one head is
1 1 1 1
− − =
2 2 2 8
Or by the formula
1⎞
⎛
⎜⎝ 1 − ⎟⎠
2
3−1
1 ⎛ 1⎞ ⎛ 1⎞ 1
. =⎜ ⎟ ⎜ ⎟ = .
2 ⎝ 2⎠ ⎝ 2⎠ 8
2
2) A die is rolled; find the probability of getting the first 3 on the fourth roll.
Solution
n=4 p=1/6
1⎞
⎛
∴ ⎜1 − ⎟
⎝
6⎠
4 −1
⎛ 1 ⎞ ⎛ 5 ⎞ ⎛ 5 ⎞ ⎛ 1 ⎞ 125
= 0.96
⎜⎝ ⎟⎠ = ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ =
6
6
6
6
1296
3
3
Example
If cards are selected from a deck and replaced, how many trials would it take on
average to get two clubs?
P (Club) = 13/52=1/4
Expected no. of trials for selecting 2 clubs would be
2
4
= 2x = 8
1
1
4
African Virtual University 76
DO THIS
1. A card from an ordinary deck of cards is selected and then replaced with another
card selected etc… find the probability that the first club will occur on he fourth
draw. 2. A die is tossed until 5 or 6 is obtained. Find the expected number of tosses.
Answer Key
1) Fourth
2) 3
Hypergeometric Distribution
In probability theory and statistics, the hypergeometric distribution is a discrete
probability distribution that describes the number of successes in a sequence of n
draws from a finite population without replacement.
A typical example is illustrated by the contingency table above: there is a shipment
of N objects in which D are defective. The hypergeometric distribution describes the
probability that in a sample of n distinctive objects drawn from the shipment exactly
k objects are defective.
In general, if a random variable X follows the hypergeometric distribution with parameters N, D and n, then the probability of getting exactly k successes is given by
The probability is positive when k is between max{ 0, D + n − N } and min{ n, D }.
The formula can be understood as follows: There are
replacement). There are
possible samples (without
ways to obtain k defective objects and
African Virtual University 77
there are
objects.
ways to fill out the rest of the sample with non-defective
When the population size is large compared to the sample size (i.e., N is much larger
than n) the hypergeometric distribution is approximated reasonably well by a binomial distribution with parameters n (number of trials) and p = D / N (probability of
success in a single trial).
Hypergeometric Formula
When there are two groups of items such that there are ‘a’ items in the first group
and ‘b’ items in the second group, so that the total number of items is (a + b), the
probability of selecting x items from the first group and (n-x) items from the second
group is
C
C
a x . b n− x
, where n is the total of items selected without replacement.
C
a+b n
Examples
1. A bag contains 3 blue chips and 3 green chips. If two chips are selected at random, find the probability that both are blue.
Solution
C
C
a x . b n− x
From the formula
; a = 3, b= 3, x=2, n=2, n-x=2-2=0
C
a+b n
The probability of both blue =
C
C
3 2 . 3 2− 2 3 x1 1
=
= = 0.2
C
15
5
3+ 3 2
2. A committee of 3 people is selected at random without replacement from a group
of 6 men and 3 women. Find the probability that the committee consists of 2
men and 2 women.
Solution
So into
a=6 b=3
n = 6+3=9
African Virtual University 78
since the committee consists 2 men and 2 women
x=2 n-.x= 3-2=1
Pr = 6C 2 3C 1 15x3 15
=
=
= 0.536
9C 3
84
28
3. A group of 10 tanks contains 3 defective tanks. If 4 tanks are randomly selected
and tested find the probability that exactly one will be defective solution.
3 are defective a=3
7 are good
b=7
Pr (one to be defective)
n = 4 x=1
n-x=4-1=3
Pr (one to be exactly defective)
3C 1 .7C 3 105
=
= 0.5
10C 4
210
DO THIS
1. In a box of 10 shirts there are five (5) defective ones. If 5 shirts are sold at random
find the probability that exactly two are defective.
Answer
2. In a shipment of 12 lawn chairs 86 are brown and 4 are blue. If 3 chairs
are sold at random find the probability that that all are brown.
Answer Key
1). 0.397
2) 0.255
q
Group Work
1. Revise the following probability
questions and answers
2. Discuss any problems encountered in
the computations of the probabilities.
African Virtual University 79
1) Find the probability of choosing 5 women from a committee of 15 women
P(choosing 5) =
15
1
1
=
C 5 3003
2) What is the probability of drawing an ace or a spade from deck of playing cards?
4
∴ P ( AUB ) = P ( A ) + P ( B ) − P ( AUB )
52
P (Ace) =
P ( spade) =
13
52 3)
=
4 13 1
+
−
52 52 52
=
16 4
=
52 13
1
There are problems pregnant for women. The probability of dying is
what
5
is the probability that at least one will die in every 5 women
P ( A) =
1
51
P ( A )1 −
⎛ 50 ⎞
⎟ = use calculator
51 ⎠
5
P(At least one will die) ⎜
⎝
1 50
=
51 51
Application and Example
The classical application of the hypergeometric distribution is sampling without replacement. Think of an urn with two types of marbles, black ones
and white ones. Define drawing a white marble as a success and drawing a
black marble as a failure (analogous to the binomial distribution). If the variable N describes the number of all marbles in the urn (see contingency table
above) and D describes the number of white marbles (called defective in the
example above), then N − D corresponds to the number of black marbles.
Now, assume that there are 5 white and 45 black marbles in the urn. Standing next
to the urn, you close your eyes and draw 10 marbles without replacement. What’s
the probability p (k=4) that you draw exactly 4 white marbles (and - of course - 6
black marbles) ?
African Virtual University 80
This problem is summarized by the following contingency table:
white
marbles
black
marbles
total
drawn
not drawn
total
4 (k)
1 = 5 − 4 (D − k)
5 (D)
6 = 10 − 4 (n − k)
39 = 50 + 4 − 10 − 5 (N + k− n − D)
45 (N− D)
10 (n)
40 (N − n)
50 (N)
The probability Pr (k = x) of drawing exactly x white marbles (= number of successes)
can be calculated by the formula
Hence, in this example x = 4, calculate
So, the probability of drawing exactly 4 white marbles is quite low (approximately
0.004) and the event is very unlikely. It means, if you repeated your random experiment (drawing 10 marbles from the urn of 50 marbles without replacement) 1000
times you just would expect to obtain such a result 4 times.
But what about the probability of drawing even (all) 5 white marbles? You will intuitively agree upon that this is even more unlikely than drawing 4 white marbles.
Let us calculate the probability for such an extreme event.
The contingency table is as follows:
drawn
not drawn
total
white marbles
5 (k)
0 = 5 − 5 (D − k)
5 (D)
black marbles
5 = 10 − 5 (n − k)
40 = 50 + 5 − 10 − 5 (N + k − n − D)
45 (N − D)
total
10 (n)
40 (N − n)
50 (N)
And we can calculate the probability as follows (notice that the denominator always
stays the same):
African Virtual University 81
As expected, the probability of drawing 5 white marbles is even much lower than
drawing 4 white marbles.
Conclusion
Consequently, one could expand the initial question as follows: If you draw 10 marbles from an urn (containing 5 white and 45 black marbles), what’s the probability
of drawing at least 4 white marbles? Or, what’s the probability of drawing 4 white
marbles and more extreme outcomes such as drawing 5)? This corresponds to calculating the cumulative probability p(k>=4) and can be calculated by the cumulative
distribution function (cdf). Since the hypergeometric distribution is a discrete
probability distribution the cumulative probability can be calculated easily by
adding all corresponding single probability values.
In our example you just have to sum-up Pr (k = 4) and Pr (k = 5):
Pr (k ≥ 4) = 0.003964583 + 0.0001189375 = 0.004083520
READ:
1. An Introduction to Probability & Random Processes By
Kenneth B & Gian-Carlo R, pages 184-195
African Virtual University 82
Bivariate Frequency Distributions
The bivariate normal distribution is the statistical distribution with probability function
where
and
is the correlation of
and
(Kenney and Keeping 1951, pp. 92 and 202-205; Whit-
taker and Robinson 1967, p. 32)
are commonly used in place of
and
The marginal probabilities are then
=
=
.
African Virtual University 83
And
=
=
Joint Probability Tables
This table is a correctly formatted joint probability table.
Days Listed Until Sold
Initial Asking Price
Totals
Under 30
31-90
Over 90
Under $50,000
0.06
0.05
0.01
0.13
$50,000-99,999
0.03
0.19
0.10
0.31
$100,000-150,000
0.03
0.35
0.13
0.50
Over $150,000
0.01
0.04
0.01
0.06
Totals
0.13
0.63
0.25
1.00
Marginal Probabilities
Let be partitioned into
disjoint sets
and
denoted
. Then the marginal probability of
where the general subset is
is
READ:
1. An Introduction to Probability & Random Processes
By Kenneth B & Gian-Carlo R, pages 142-150
2. Exercise pg 150 Nos. 1,23,4,5,6,7,8,9,14,15,16,17,26.
African Virtual University 84
q
REFLECTION: ICT resources are difficult
to access!! The link opens up avenue
for Mathematics teachers to access ICT
resources.
http://www.tsm-resources.com/suppl.html
African Virtual University 85
Unit 2 ( 40 Hours)
Random Variables And Test Distributions
Moments
The probability distribution of a random variable is often characterised by a small
number of parameters, which also have a practical interpretation. For example, it is
often enough to know what its “average value” is. This is captured by the mathematical
concept of expected value of a random variable, denoted E[X]. Note that in general,
E[f(X)] is not the same as f(E[X]). Once the “average value” is known, one could then
ask how far from this average value the values of X typically are, a question that is
answered by the variance and standard deviation of a random variable.
Mathematically, this is known as the (generalised) problem of moments: for a given
class of random variables X, find a collection {fi} of functions such that the expectation
values E[fi(X)] fully characterize the distribution of the random variable X.
Equivalence of random variables
There are several different senses in which random variables can be considered to be
equivalent. Two random variables can be equal, equal almost surely, equal in mean,
or equal in distribution.
In increasing order of strength, the precise definition of these notions of equivalence
is given below.
Equality in distribution
Two random variables X and Y are equal in distribution if they have the same distribution functions:
Two random variables having equal moment generating functions have the same
distribution.
Equality in mean
Two random variables X and Y are equal in p-th mean if the pth moment of |X − Y|
is zero, that is,
African Virtual University 86
Equality in pth mean implies equality in qth mean for all q<p. As in the previous case,
there is a related distance between the random variables, namely
Equality
Finally, the two random variables X and Y are equal if they are equal as functions on
their probability space, that is,
Moment-generating Function
In probability theory and statistics, the moment-generating function of a random
variable X is
wherever this expectation exists. The moment-generating function generates the
moments of the probability distribution.
For vector-valued random variables X with real components, the moment-generating
function is given by
where t is a vector and
is the dot product.
Provided the moment-generating function exists in an interval around t = 0, the nth
moment is given by
If X has a continuous probability density function f(x) then the moment generating
function is given by
African Virtual University 87
where mi is the ith moment. MX( − t) is just the two-sided Laplace transform of f(x).
Regardless of whether the probability distribution is continuous or not, the momentgenerating function is given by the Riemann-Stieltjes integral
where F is the cumulative distribution function.
If X1, X2, ..., Xn is a sequence of independent (and not necessarily identically distributed) random variables, and
where the ai are constants, then the probability density function for Sn is the convolution of the probability density functions of each of the Xi and the moment-generating
function for Sn is given by
Related to the moment-generating function are a number of other transforms that are
common in probability theory, including the characteristic function and the probability-generating function.
Markov’s Inequality
ε
f(x)
{X εX
| f ( x) ≥ ε
}
Markov’s inequality gives an upper bound for the probability that X lies within
{ Xε X | f (x) ≥ ε }
African Virtual University 88
In probability theory, Markov’s inequality gives an upper bound for the probability
that a non-negative function of a random variable is greater than or equal to some positive constant. It is named after the Russian mathematician Andrey Markov, although
it appeared earlier in the work of Pafnuty Chebyshev (Markov’s teacher).
Markov’s inequality (and other similar inequalities) relate probabilities to expectations,
and provide (frequently) loose but still useful bounds for the cumulative distribution
function of a random variable.
Special case: probability theory
For any event E, let IE be the indicator random variable of E, that is, IE = 1 if E occurs
and = 0 otherwise. Thus I(|X| ≥ a) = 1 if the event |X| ≥ a occurs, and I(|X| ≥ a) = 0 if |X| < a.
Then, given a>0,
Therefore
Now observe that the left side of this inequality is the same as
Thus we have
and since a > 0, we can divide both sides by a.
READ:
1. Robert B. Ash, Lectures on Statistics, page 9-13
2. An Introduction to Probability & Random Processes
By Kenneth B & Gian-Carlo R, pages 366 -374 & 404 - 407
• Exercise on pg 376 -376 Nos. 1,3,7,8
• Exercise on pg 442 Nos. 1,2,3,4,5
Ref:
• http://en.wikipedia.org/wiki/Moment-generating_
function
• http://en.wikipedia.org/wiki/characteristic_function_
%28probability_theory%29.
• http://en.wikipedia.org/wiki/Integral_transform
African Virtual University 89
Chebyshev’s Inequality
In probability theory, Chebyshev’s inequality (also known as Tchebysheff’s inequality, Chebyshev’s theorem, or the Bienaymé-Chebyshev inequality), named
after Pafnuty Chebyshev, who first proved it, states that in any data sample or probability distribution, nearly all the values are close to the mean value, and provides
a quantitative description of “nearly all” and “close to”. For example, no more than
1/4 of the values are more than 2 standard deviations away from the mean, no more
than 1/9 are more than 3 standard deviations away, no more than 1/25 are more than
5 standard deviations away, and so on.
Probabilistic statement
Let X be a random variable with expected value μ and finite variance σ2. Then for
any real number k > 0,
Only the cases k > 1 provide useful information.
As an example, using k=√2 shows that at least half of the values lie in the interval
(μ − √2 σ, μ + √2 σ).
Typically, the theorem will provide rather loose bounds. However, the bounds provided by Chebyshev’s inequality cannot, in general (remaining sound for variables of
arbitrary distribution), be improved upon. For example, for any k > 1, the following
example (where σ = 1/k) meets the bounds exactly.
The theorem can be useful despite loose bounds because it applies to random variables
of any distribution, and because these bounds can be calculated knowing no more
about the distribution than the mean and variance.
Chebyshev’s inequality is used for proving the weak law of large numbers.
Example application
For illustration, assume we have a large body of text, for example articles from a
publication. Assume we know that the articles are on average 1000 characters long
with a standard deviation of 200 characters. From Chebyshev’s inequality we can
then deduce that at least 75% of the articles have a length between 600 and 1400
characters (k = 2).
African Virtual University 90
Probabilistic proof
Markov’s inequality states that for any real-valued random variable Y and any positive
number a, we have Pr(|Y| > a) ≤ E(|Y|)/a. One way to prove Chebyshev’s inequality
is to apply Markov’s inequality to the random variable Y = (X − μ)2 with a = (σk)2.
It can also be proved directly. For any event A, let IA be the indicator random variable
of A, i.e. IA equals 1 if A occurs and 0 otherwise. Then
The direct proof shows why the bounds are quite loose in typical cases: the number 1
to the left of “≥” is replaced by [(X − μ)/(kσ)]2 to the right of “≥” whenever the latter
exceeds 1. In some cases it exceeds 1 by a very wide margin.
READ:
1. An Introduction to Probability & Random Processes
By Kenneth B & Gian-Carlo R, pages 305-318
•
•
Exercise on pg 309 nos. 1,2,3,4,5
Exercise on pg 320-324 Nos. 1,3,10,12
African Virtual University 91
Correlation Types
Correlation is a measure of association between two variables. The variables are not
designated as dependent or independent. The two most popular correlation coefficients are: Spearman’s correlation coefficient rho and Pearson’s product-moment
correlation coefficient.
When calculating a correlation coefficient for ordinal data, select Spearman’s technique. For interval or ratio-type data, use Pearson’s technique.
The value of a correlation coefficient can vary from minus one to plus one. A minus
one indicates a perfect negative correlation, while a plus one indicates a perfect positive correlation. A correlation of zero means there is no relationship between the
two variables. When there is a negative correlation between two variables, as the
value of one variable increases, the value of the other variable decreases, and vise
versa. In other words, for a negative correlation, the variables work opposite each
other. When there is a positive correlation between two variables, as the value of
one variable increases, the value of the other variable also increases. The variables
move together.
The standard error of a correlation coefficient is used to determine the confidence
intervals around a true correlation of zero. If your correlation coefficient falls outside
of this range, then it is significantly different than zero. The standard error can be
calculated for interval or ratio-type data (i.e., only for Pearson’s product-moment
correlation).
The significance (probability) of the correlation coefficient is determined from the
t-statistic. The probability of the t-statistic indicates whether the observed correlation
coefficient occurred by chance if the true correlation is zero. In other words, it asks
if the correlation is significantly different than zero. When the t-statistic is calculated for Spearman’s rank-difference correlation coefficient, there must be at least 30
cases before the t-distribution can be used to determine the probability. If there are
fewer than 30 cases, you must refer to a special table to find the probability of the
correlation coefficient.
African Virtual University 92
Example
A company wanted to know if there is a significant relationship between the total number of salespeople and the total number of sales. They collect data for five months.
Variable 1
207
180
220
205
190
Variable 2
6907
5991
6810
6553
6190
Correlation coefficient = .921
Standard error of the coefficient = ..068
t-test for the significance of the coefficient = 4.100
Degrees of freedom = 3
Two-tailed probability = .0263
Another Example
Respondents to a survey were asked to judge the quality of a product on a four-point
Likert scale (excellent, good, fair, poor). They were also asked to judge the reputation
of the company that made the product on a three-point scale (good, fair, poor). Is
there a significant relationship between respondents perceptions of the company and
their perceptions of quality of the product?
Since both variables are ordinal, Spearman’s method is chosen. The first variable is
the rating for the quality the product. Responses are coded as 4=excellent, 3=good,
2=fair, and 1=poor. The second variable is the perceived reputation of the company
and is coded 3=good, 2=fair, and 1=poor.
Variable 1
4
2
1
3
4
1
2
Variable 2
3
2
2
3
3
1
1
Correlation coefficient rho = .830
t-test for the significance of the coefficient = 3.332
Number of data pairs = 7
Probability must be determined from a table because of the small sample size.
African Virtual University 93
Regression
Simple regression is used to examine the relationship between one dependent and
one independent variable. After performing an analysis, the regression statistics can
be used to predict the dependent variable when the independent variable is known.
Regression goes beyond correlation by adding prediction capabilities.
People use regression on an intuitive level every day. In business, a well-dressed
man is thought to be financially successful. A mother knows that more sugar in her
children’s diet results in higher energy levels. The ease of waking up in the morning
often depends on how late you went to bed the night before. Quantitative regression
adds precision by developing a mathematical formula that can be used for predictive
purposes.
For example, a medical researcher might want to use body weight (independent
variable) to predict the most appropriate dose for a new drug (dependent variable).
The purpose of running the regression is to find a formula that fits the relationship
between the two variables. Then you can use that formula to predict values for the
dependent variable when only the independent variable is known. A doctor could
prescribe the proper dose based on a person’s body weight.
The regression line (known as the least squares line) is a plot of the expected value
of the dependent variable for all values of the independent variable. Technically, it
is the line that “minimizes the squared residuals”. The regression line is the one that
best fits the data on a scatterplot.
Using the regression equation, the dependent variable may be predicted from the independent variable. The slope of the regression line (b) is defined as the rise divided by
the run. The y intercept (a) is the point on the y axis where the regression line would
intercept the y axis. The slope and y intercept are incorporated into the regression
equation. The intercept is usually called the constant, and the slope is referred to as
the coefficient. Since the regression model is usually not a perfect predictor, there is
also an error term in the equation.
In the regression equation, y is always the dependent variable and x is always the
independent variable. Here are three equivalent ways to mathematically describe a
linear regression model.
y = intercept + (slope x) + error
y = constant + (coefficient x) + error
y = a + bx + e
The significance of the slope of the regression line is determined from the t-statistic.
It is the probability that the observed correlation coefficient occurred by chance if
the true correlation is zero. Some researchers prefer to report the F-ratio instead of
the t-statistic. The F-ratio is equal to the t-statistic squared.
African Virtual University 94
The t-statistic for the significance of the slope is essentially a test to determine if the
regression model (equation) is usable. If the slope is significantly different than zero,
then we can use the regression model to predict the dependent variable for any value
of the independent variable.
On the other hand, take an example where the slope is zero. It has no prediction ability
because for every value of the independent variable, the prediction for the dependent
variable would be the same. Knowing the value of the independent variable would
not improve our ability to predict the dependent variable. Thus, if the slope is not
significantly different than zero, don’t use the model to make predictions.
The coefficient of determination (r-squared) is the square of the correlation coefficient. Its value may vary from zero to one. It has the advantage over the correlation
coefficient in that it may be interpreted directly as the proportion of variance in the
dependent variable that can be accounted for by the regression equation. For example,
an r-squared value of .49 means that 49% of the variance in the dependent variable
can be explained by the regression equation. The other 51% is unexplained.
The standard error of the estimate for regression measures the amount of variability
in the points around the regression line. It is the standard deviation of the data points
as they are distributed around the regression line. The standard error of the estimate
can be used to develop confidence intervals around a prediction.
Example
A company wants to know if there is a significant relationship between its advertising
expenditures and its sales volume. The independent variable is advertising budget and
the dependent variable is sales volume. A lag time of one month will be used because
sales are expected to lag behind actual advertising expenditures. Data was collected
for a six month period. All figures are in thousands of dollars. Is there a significant
relationship between advertising budget and sales volume?
Independent
Variable
4.2
6.1
3.9
5.7
7.3
5.9
Dependent
Variable
27.1
30.4
25.0
29.7
40.1
28.8
Model: y = 10.079 + (3.700 x) + error
Standard error of the estimate = 2.568
t-test for the significance of the slope = 4.095
Degrees of freedom = 4
Two-tailed probability = .0149
r-squared = .807
African Virtual University 95
You might make a statement in a report like this: A simple linear regression was
performed on six months of data to determine if there was a significant relationship
between advertising expenditures and sales volume. The t-statistic for the slope was
significant at the .05 critical alpha level, t(4)=4.10, p=.015. Thus, we reject the null
hypothesis and conclude that there was a positive significant relationship between
advertising expenditures and sales volume. Furthermore, 80.7% of the variability in
sales volume could be explained
READ:
1) An Introduction to Probability & Random Processes By
Kenneth B & Gian-Carlo R, pages 18-30, 212-215, 300303
2) Robert B. Ash, Lectures on Statistics, page 28-29.
Ref: http://en.wikipedia.org/wiki/Correlation
Ref: http://en.wikipedia.org/wiki/Regression
African Virtual University 96
Chi-square Test
A chi-square test is any statistical hypothesis test in which the test statistic has a chisquare distribution when the null hypothesis is true, or any in which the probability
distribution of the test statistic (assuming the null hypothesis is true) can be made
to approximate a chi-square distribution as closely as desired by making the sample
size large enough.
Specifically, a chi-square test for independence evaluates statistically significant
differences between proportions for two or more groups in a data set.
Pearson's chi-square test, also known as the Chi-square goodness-of-fit test
• Yates' chi-square test also known as Yates' correction for continuity
• Mantel-Haenszel chi-square test
• Linear-by-linear association chi-square test
•
In probability theory and statistics, the chi-square distribution (also chi-squared or
χ2 distribution) is one of the most widely used theoretical probability distributions
in inferential statistics, i.e. in statistical significance tests. It is useful because, under
reasonable assumptions, easily calculated quantities can be proven to have distributions that approximate to the chi-square distribution if the null hypothesis is true.
If Xi are k independent, normally distributed random variables with mean 0 and variance 1, then the random variable
is distributed according to the chi-square distribution. This is usually written
The chi-square distribution has one parameter: k - a positive integer that specifies the
number of degrees of freedom (i.e. the number of Xi)
The chi-square distribution is a special case of the gamma distribution.
The best-known situations in which the chi-square distribution is used are the common
chi-square tests for goodness of fit of an observed distribution to a theoretical one,
and of the independence of two criteria of classification of qualitative data. However,
many other statistical tests lead to a use of this distribution.
African Virtual University 97
Characteristic Function
The characteristic function of the Chi-square distribution is
Properties
The chi-square distribution has numerous applications in inferential statistics, for
instance in chi-square tests and in estimating variances. It enters the problem of estimating the mean of a normally distributed population and the problem of estimating
the slope of a regression line via its role in Student’s t-distribution. It enters all analysis
of variance problems via its role in the F-distribution, which is the distribution of the
ratio of two independent chi-squared random variables divided by their respective
degrees of freedom.
Various chi and chi-square distributions
Name
Statistic
chi-square distribution
noncentral chi-square
distribution
chi distribution
noncentral chi distribution
READ:
Ref: http://en.wikipedia.org/wiki/pearson%chi-square_test
Ref: http://en.wikipedia.org/wiki/Chi-Square_test
African Virtual University 98
Student’s T-test
A t test is any statistical hypothesis test for two groups in which the test statistic has
a Student’s t distribution if the null hypothesis is true.
History
The t statistic was introduced by William Sealy Gosset for cheaply monitoring the
quality of beer brews. “Student” was his pen name. Gosset was a statistician for the
Guinness brewery in Dublin, Ireland, and was hired due to Claude Guinness’s innovative policy of recruiting the best graduates from Oxford and Cambridge to apply
biochemistry and statistics to Guinness’ industrial processes. Gosset published the t test
in Biometrika in 1908, but was forced to use a pen name by his employer who regarded
the fact that they were using statistics as a trade secret. In fact, Gosset’s identity was
unknown not only to fellow statisticians but to his employer—the company insisted
on the pseudonym so that it could turn a blind eye to the breach of its rules.
Today, it is more generally applied to the confidence that can be placed in judgments
made from small samples.
Use
Among the most frequently used t tests are:
•
A test of the null hypothesis that the means of two normally distributed populations are equal. Given two data sets, each characterized by its mean, standard
deviation and number of data points, we can use some kind of t test to determine
whether the means are distinct, provided that the underlying distributions can
be assumed to be normal. All such tests are usually called Student’s t tests,
though strictly speaking that name should only be used if the variances of the
two populations are also assumed to be equal; the form of the test used when
this assumption is dropped is sometimes called Welch’s t test. There are different
versions of the t test depending on whether the two samples are
o independent of each other (e.g., individuals randomly assigned into two
groups), or
o paired, so that each member of one sample has a unique relationship with a
particular member of the other sample (e.g., the same people measured before
and after an intervention, or IQ test scores of a husband and wife).
If the t value that is calculated is above the threshold chosen for statistical significance
(usually the 0.05 level), then the null hypothesis that the two groups do not differ is
rejected in favor of an alternative hypothesis, which typically states that the groups
do differ.
•
A test of whether the mean of a normally distributed population has a value
specified in a null hypothesis.
•
A test of whether the slope of a regression line differs significantly from 0.
Once a t value is determined, a P value can be found using a table of values from
Student’s t-distribution.
African Virtual University 99
Confidence intervals using a small sample size
Consider a normally distributed population. To estimate the population’s variance
take a sample of size n and calculate the sample’s variance, s. An unbiased estimator
of the population’s variance is
Clearly for small values of n this estimation is inaccurate. Hence for samples of
small size instead of calculating the z value for the number of standard deviations
from the mean
and using probabilities based on the normal distribution, calculate the t value
The probability that the t value is within a particular interval may be found using the
t distribution. The sample’s degrees of freedom are the number of data that need to
be known before the rest of the data can be calculated.
e.g.
A random sample of things have weights
30.02, 29.99, 30.11, 29.97, 30.01, 29.99
Calculate a 95% confidence interval for the population’s mean weight.
Assume the population ~ N(μ,σ2)
The samples’ mean weight is 30.015 with standard deviation of 0.045. With the mean
and the first five weights it is possible to calculate the sixth weight. Consequently
there are five degrees of freedom.
The t distribution tells us that, for five degrees of freedom, the probability that t >
2.571 is 0.025. Also, the probability that t < −2.571 is 0.025. Using the formula for
t with t = ± 2.571 a 95% confidence interval for the populations mean may be found
by making μ the subject of the equation.
i.e.
(29.97 < μ < 30.06)
African Virtual University 100
READ:
1. Introduction to Probability By Charles M. Grinstead, pages
18-30, 212-215, 300-303
2. Robert B. Ash, Lectures on Statistics, page 23-29.
• Answer problems 1- 6 on pg 23.
Ref: http://en.wikipedia.org/wiki/Statistical_Hypothesis_testing
Ref: http://en.wikipedia.org/wiki/Null_hypothesis
African Virtual University 101
q
Reflection
The study of Correlation, Regression Hypothesis
testing and other Mathematical modelling maybe
simplified through ICT. The following link enables
trainees to learn modelling with ease
http://www.ncaction.org.uk/subjects/maths/ict-lrn.htm
African Virtual University 102
Unit 3 Probability Theory ( 40 Hours)
Indicator Function
In mathematics, an indicator function or a characteristic function is a function
defined on a set X that indicates membership of an element in a subset A of X.
The indicator function of a subset A of a set X is a function
defined as
The indicator function of A is sometimes denoted
χA(x) or
or even A(x).
Bonferoni Inequality
Let
be the probability that is true, and
be the probability that at
least one of , , ..., is true. Then “the” Bonferroni inequality, also known as
Boole’s inequality, states that
where denotes the union. If and are disjoint sets for all and , then the inequality becomes an equality. A beautiful theorem that expresses the exact relationship
between the probability of unions and probabilities of individual events is known as
the inclusion-exclusion principle.
A slightly wider class of inequalities are also known as “Bonferroni inequalities.”
African Virtual University 103
Generating Function
In mathematics a generating function is a formal power series whose coefficients
encode information about a sequence an that is indexed by the natural numbers.
There are various types of generating functions, including ordinary generating
functions, exponential generating functions, Lambert series, Bell series, and
Dirichlet series; definitions and examples are given below. Every sequence has a
generating function of each type. The particular generating function that is most
useful in a given context will depend upon the nature of the sequence and the details
of the problem being addressed.
Generating functions are often expressed in closed form as functions of a formal
argument x. Sometimes a generating function is evaluated at a specific value of x.
However, it must be remembered that generating functions are formal power series,
and they will not necessarily converge for all values of x.
If an is the probability mass function of a discrete random variable, then its ordinary
generating function is called a probability-generating function.
The ordinary generating function can be generalised to sequences with multiple
indexes. For example, the ordinary generating function of a sequence am,n (where n
and m are natural numbers) is
Characteristic Function (Probability Theory)
In probability theory, the characteristic function of any random variable completely
defines its probability distribution. On the real line it is given by the following formula,
where X is any random variable with the distribution in question:
where t is a real number, i is the imaginary unit, and E denotes the expected value.
If FX is the cumulative distribution function, then the characteristic function is given
by the Riemann-Stieltjes integral
African Virtual University 104
In cases in which there is a probability density function, fX, this becomes
If X is a vector-valued random variable, one takes the argument t to be a vector and
tX to be a dot product.
Every probability distribution on R or on Rn has a characteristic function, because
one is integrating a bounded function over a space whose measure is finite.
The continuity theorem
If the sequence of characteristic functions of distributions Fn converges to the characteristic function of a distribution F, then Fn(x) converges to F(x) at every value of
x at which F is continuous.
Uses Of Characteristic Functions
Characteristic functions are particularly useful for dealing with functions of independent random variables. For example, if X1, X2, ..., Xn is a sequence of independent
(and not necessarily identically distributed) random variables, and
where the ai are constants, then the characteristic function for Sn is given by
In particular,
of characteristic function:
. To see this, write out the definition
Observe that the independence of X and Y is required to establish the equality of the
third and fourth expressions.
Because of the continuity theorem, characteristic functions are used in the most
frequently seen proof of the central limit theorem.
Characteristic functions can also be used to find moments of random variable. Provided
that nth moment exists, characteristic function can be differentiated n times and
African Virtual University 105
READ:
1. Robert B. Ash, Lectures on Statistics, page 32 of 45:
Ref : http://en.wikipedia.org/wiki/Characteristic_function_
%28probability_theory%29
Statistical Independence
In probability theory, to say that two events are independent intuitively means that
the occurrence of one event makes it neither more nor less probable that the other
occurs. For example:
•
The event of getting a "6" the first time a die is rolled and the event of getting a
"6" the second time are independent.
•
By contrast, the event of getting a "6" the first time a die is rolled and the event
that the sum of the numbers seen on the first and second trials is "8" are dependent.
•
If two cards are drawn with replacement from a deck of cards, the event of
drawing a red card on the first trial and that of drawing a red card on the second
trial are independent.
•
By contrast, if two cards are drawn without replacement from a deck of cards,
the event of drawing a red card on the first trial and that of drawing a red card
on the second trial are dependent.
Similarly, two random variables are independent if the conditional probability distribution of either given the observed value of the other is the same as if the other’s
value had not been observed.
Independent Events
The standard definition says:
Two events A and B are independent if and only if Pr(A ∩ B) = Pr(A)Pr(B).
Here A ∩ B is the intersection of A and B, that is, it is the event that both events A
and B occur.
More generally, any collection of events -- possibly more than just two of them -- are
mutually independent if and only if for any finite subset A1, ..., An of the collection
we have
African Virtual University 106
This is called the multiplication rule for independent events.
If two events A and B are independent, then the conditional probability of A given B
is the same as the “unconditional” (or “marginal”) probability of A, that is,
There are at least two reasons why this statement is not taken to be the definition
of independence: (1) the two events A and B do not play symmetrical roles in this
statement, and (2) problems arise with this statement when events of probability 0
are involved.
When one recalls that the conditional probability Pr(A | B) is given by
(so long as Pr(B) ≠ 0 )
one sees that the statement above is equivalent to
which is the standard definition given above.
Random Sample
A sample is a subset chosen from a population for investigation. A random sample is
one chosen by a method involving an unpredictable component. Random sampling can
also refer to taking a number of independent observations from the same probability
distribution, without involving any real population. A probability sample is one in
which each item has a known probability of being in the sample.
The sample will usually not be completely representative of the population from
which it was drawn— this random variation in the results is known as sampling
error. In the case of random samples, mathematical theory is available to assess the
sampling error. Thus, estimates obtained from random samples can be accompanied
by measures of the uncertainty associated with the estimate. This can take the form
of a standard error, or if the sample is large enough for the central limit theorem to
take effect, confience intervals may be calculated.
African Virtual University 107
Types of random sample
•
A simple random sample is selected so that every possible sample has an equal
chance of being selected.
•
A self-weighting sample, also known as an epsem sample, is one in which every
individual, or object, in the population of interest has an equal opportunity of
being selected for the sample. Simple random samples are self-weighting.
•
Stratified sampling involves selecting independent samples from a number of
subpopulations (or strata) within the population. Great gains in efficiency are
sometimes possible from judicious stratification.
•
Cluster sampling involves selecting the sample units in groups. For example, a
sample of telephone calls may be collected by first taking a collection of telephone
lines and collecting all the calls on the sampled lines. The analysis of cluster
samples must take into account the intra-cluster correlation which reflects the
fact that units in the same cluster are likely to be more similar than two units
picked at random.
Multinomial Distribution
In probability theory, the multinomial distribution is a generalization of the binomial distribution.
The binomial distribution is the probability distribution of the number of “successes”
in n independent Bernoulli trials, with the same probability of “success” on each trial.
In a multinomial distribution, each trial results in exactly one of some fixed finite
number k of possible outcomes, with probabilities p1, ..., pk (so that pi ≥ 0 for
i = 1, ..., k and
), and
there are n independent trials. Then let the random variables Xi indicate the number
of times outcome number i was observed over the n trials.
follows a multinomial distribution with parameters n and p.
Solutions from Multinomial Distribution Formula
A short version of the multinomial formula for three consecutive outcomes
is given below.
If X consists of events E1, E2, E3, which have the corresponding probabilities
of p1, p2, and p3 of occurring, where x1 is the number of times E1 will occur,
x2 is the number of times E2 will occur, and x3 is the number of times E3 will
occur, then the probability of X is
African Virtual University 108
n!
.
x1 ! x2 ! x3 !
p .p .p
x1
x2
x3
1
2
3
where x1 + x2 + x3 = n and p1 + p2 + p3 = 1
Example
1) In a large city, 60% of the workers drive to work, 30% take the bus, and 10%
take the train. If 5 workers are selected at random, find the probability that 2 will
drive, 2 will take the us, and 1 will take the train.
Solution
n= 5, x1=2, x2 = 2, x3= 1 and p1=0.6, p2= 0.3, and p3 = 0.1
Hence, the probability that 2 workers will take the bus, and one will take the train
is
2
2
1
5!
. (0.6) (0.3) (0.1) = 0.0972
2 ! 2 !1!
2) A box contains 5 red balls, 3 blue balls, and 2 white balls. If 4 balls are selected
with replacement, find the probability of getting 2 red balls, one blue ball, and
one white ball.
Solution
n=4, x1=2, x2=1, x3=1, and p1=
5
3
2
, p2=
, and p3=
.
10
10
10
Hence, the probability of getting 2 red balls, one blue ball, and one white ball is
4! ⎛ 5 ⎞ ⎛ 3 ⎞ ⎛ 2 ⎞
⎛ 3 ⎞ 9
= 0.18
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = 12 ⎜⎝
⎟=
2!1!1! 10
10
10
200 ⎠ 50
2
1
{ Allan G, 2005, pg 132}
1
African Virtual University 109
Order Statistic
Probability distributions for the n = 5 order statistics of an exponential distribution
with θ = 3.
In statistics, the kth order statistic of a statistical sample is equal its kth-smallest
value. Together with rank statistics, order statistics are among the most fundamental
tools in non-parametric statistics and inference.
Important special cases of the order statistics are the minimum and maximum value
of a sample, and (with some qualifications discussed below) the sample median and
other sample quartiles.
When using probability theory to analyse order statistics of random samples from
a continuous distribution, the cumulative distribution function is used to reduce the
analysis to the case of order statistics of the uniform distribution.
READ:
1. Robert B. Ash, Lectures on Statistics, page 25 -26 and
Answer problems 1-4 on pg 26/27.
Ref: http://en.wikipedia.org/wiki/probability _distribution
Ref: http://en.wikipedia.org/wiki/Ranking
Ref: http://en.wikipedia.org/wiki/non-parametric_Statistics
Notation and examples
For example, suppose that four numbers are observed or recorded, resulting in a
sample of size n = 4. If the sample values are
6, 9, 3, 8,
they will usually be denoted
where the subscript i in xi indicates simply the order in which the observations were
recorded and is usually assumed not to be significant. A case when the order is significant is when the observations are part of a time series.
The order statistics would be denoted
where the subscript (i) enclosed in parentheses indicates the ith order statistic of the
sample.
African Virtual University 110
The first order statistic (or smallest order statistic) is always the minimum of the
sample, that is,
where, following a common convention, we use upper-case letters to refer to random
variables, and lower-case letters (as above) to refer to their actual observed values.
Similarly, for a sample of size n, the nth order statistic (or largest order statistic)
is the maximum, that is,
The sample range is the difference between the maximum and minimum. It is clearly
a function of the order statistics:
A similar important statistic in exploratory data analysis that is simply related to the
order statistics is the sample interquartile range.
The sample median may or may not be an order statistic, since there is a single middle value only when the number n of observations is odd. More precisely, if n = 2m
+ 1 for some m, then the sample median is X(m + 1) and so is an order statistic. On the
other hand, when n is even, n = 2m and there are two middle values, X(m) and X(m + 1),
and the sample median is some function of the two (usually the average) and hence
not an order statistic. Similar remarks apply to all sample quantiles.
Multivariate Normal Distribution
In probability theory and statistics, a multivariate normal distribution, also sometimes called a multivariate Gaussian distribution, is a specific probability
distribution, which can be thought of as a generalization to higher dimensions of the
one-dimensional normal distribution (also called a Gaussian distribution).
Higher moments
The kth-order moments of X are defined by
where
The central k-order moments are given as follows
(a) If k is odd,
(b) If k is even with k = 2λ, then
.
African Virtual University 111
where the sum is taken over all allocations of the set
into λ (unordeλ−1
red) pairs, giving (2λ − 1)! / (2 (λ − 1)!) terms in the sum, each being the product
of λ covariances. The covariances are determined by replacing the terms of the list
by the corresponding terms of the list consisting of r1 ones, then r2
twos, etc, after each of the possible allocations of the former list into pairs.
In particular, the 4-order moments are
For fourth order moments (four variables) there are three terms. For sixth-order moments there are 3 × 5 = 15 terms, and for eighth-order moments there are 3 × 5 × 7
= 105 terms.
African Virtual University 112
XV.Synthesis of the Module
At the end of this module learners are expected to compute various measures of dispersions and apply the laws of probability to various probability distributions. The
learners should be able to solve various coefficients of correlation and regression.
Unit one of Probability and Statistics covers Frequency distributions relative and cumulative distributions, various frequency curves, mean, Mode Median. Quartiles and
Percentiles, Standard deviation, symmetrical and skewed distributions. The learner
is introduced to various statistical measures and guided examples.
The examples are well illustrated and learners can follow without difficulty. It is
recommended that learners attempt the formative evaluations given to assess their
progress in the understanding of the content. Learners should take time to check the
recommended reference material in CD’S, attached open source materials and the
recommended websites. Most importantly, learners are encouraged to read the content
widely and attempt the questions after each topic. Unit two of the module takes learners
through Moment and moment generating function, Markov and Chebychev inequalities, special Univariate distributions, Bivariate probability distribution; Joint Marginal
and conditional distributions; Independence; Bivariate expectation Regression and
Correlation; Calculation of regression and correlation coefficient for bivariate data.
Distribution function of random variables, Bivariate normal distribution. Derived
distributions such as Chi-Square, t and F.
Unit two has various learning activities to aid learning and learners are advised
to master the content of the various sub-topics and assess themselves through the
formative evaluations. Failure to answer the formative assessments should be a
positive indicator that learners should revise the sub-topics before progressing to
other sub-topics. The tasks given under the different learning activities demands that
you demonstrate a high level of ICT skills competency. The learning objectives are
well stated in the beginning of the module and should guide learners in the level of
expectations of the module.
Unit three focuses on probability theory and concentrates on the various probability
distributions.
The summative evaluation will be used to judge the learners mastery of the module.
It is recommended that learners revise the module before sitting for the final summative evaluation.
African Virtual University 113
XVI. Summative Evaluation
Answer any four questions. Each question carries 15 marks.
Question 1: General Statistics
1) In the following table, the weights of 40 cows are recorded to the nearest kilogram.
128
157
144
135
165
161
138
146
146
168
135
150
140
142
138
142
147
176
142
147
145
140
154
149
152
156
125
148
119
153
150
144
163
134
136
145
173
164
158
126
Find;
a). the highest weight
b). the least weight
c). the range
d). construct a frequency distribution table starting with a class of 118-126
e). calculate the mean of the data
f). calculate the standard deviation
Question 2: General Probability
2) A). A coin and a die are thrown together. Draw a possibility space diagram and
find the probability of obtaining:
a).
b).
c).
d).
a head
a number greater than 4
a head and a number greater than 4
a head or a number greater than 4
B). Events M and N are such that P(M) =
P(M I N).
19
2
4
, P(N) = and P(M U N)= . Find
30
5
5
African Virtual University 114
Question 3: Poisson Distribution
3) A book contains 500 pages and has 750 misprints.
a). What is the average number of misprints per page?
b) Find the probability that page 427 contains
i). no misprints
ii). exactly 4 misprints
c). find the probability that pages 427 and 428 will contain no misprints
Question 4: Continuous Random Variable
4) A continuous random variable (r.v) X has a probability density function (p.d.f)
f(x) where
⎧
⎪
⎪
f (x) = ⎨
⎪
⎪⎩
k(x + 2)2
4k
0
−2≤xp0
0 ≤ x ≤ 113
otherwise
a) Find the value of the constant k
b) Sketch y=f(x)
c) Find P( - 1 ≤ X ≤ 1)
d) Find P(x>1)
Probability of an event
5). Given that P(AUB) =7/8, P(A I B)=1/4 and P(A’)=5/8, find the values of
a) P(A)
b) P(B)
c) P(A I B’)
d) P(A’U B’)
e) The probability that only one of A, B will occur.
African Virtual University 115
Expected Value
6). The continuous random variable r.v has the p.d.f
f (x) = x +
1
2
0 ≤ x ≤1
Find:
a).E(X)
b).E(24X +6)
c).E( 1-X)
1
2
7). The masses, to the nearest kg, of 50 boys are recorded below.
Mass (kg)
Frequency (f)
60-64
2
65-69
6
70-74
12
a).Construct a cumulative frequency curve
b).Use the curve to estimate the ;
i) Median
ii). Interquartile range
iii). 7th decile
iii). 60th percentile.
75-79
14
80-84
10
85-89
6
African Virtual University 116
Marking Scheme Of Summative Evaluation
1). a)176
b) 119
c) 176-119=57
d) Using 7 classes gives us a class interval of 9
Weight(kg)
118-126
127-135
136-144
145-153
154-162
163-171
172-180
Tally
///
////
//// ////
//// //// //
////
////
//
Frequency
3
5
9
12
5
4
2
Total 40
e) Accept any method of calculating the mean
f). Accept any method of calculating the standard deviation
2) A). A coin has either Head(H) or Tail(T) while a die has sides 1,2,3,4,5&6.
Coin / Die
Coin H
Coin T
1
2
3
4
5
6
H1 H2 H3 H4 H5 H6
T1 T2 T3 T4 T5 T6
Sample space=12.
a). 6/12=1/2
b). 4/12=1/3
c). 2/12=1/6
d). 8/12=2/3
African Virtual University 117
B) . P(M U N)= P(M)+P(N)-P(M I N).
4 19
2
=
+
- P(M I N).
5 30 5
⇒
⇒ P(M I N) =
19 12
+
30 30
24
30
−
7
30
=
3). a) average number per page = 750/500=1.5
b) Let X be ‘the number of misprints per page’. Then, assuming that misprints occur at random, X ~ P0(1.5)
i). P(X= 0) = e-1.5
P(there will be no misprints on page 427) = 0.223 ( 3d.p).
ii).
P( there will be 4 misprints on page 427) = 0.047 ( 3d.p)
= 0.2231…
P(X=4)= e-1.5
(1.5)4
= 0.0470…
4!
c). We expect 1.5 misprints on each page and so on two pages 427 & 428 we expect 1.5 + 1.5 = 3 misprints.
Let Y be the ‘’number of misprints on two pages’’
Y ~ P(3), so P0(Y=0)= e-3
= 0.4421
4). a). Since X is a random variable, then
all
Therefore
0
2
∫ k(x + 2) dx +
−2
k⎡
3 ⎤0
(x
+
2)
⎦⎥−2
3 ⎣⎢
∫
x
f (x)dx = 1
1
3
∫ 4kdx = 1
0
1
11
+ 4k [ x ]0 3
= 1
African Virtual University 118
k
⎛ 4⎞
(8) + 4k ⎜ ⎟
⎝ 3⎠
3
k=
= 1
8k=1
1
8
a) The p.d.f of X is
y
1
2
y
x
-2
0
11
3
c)
P(- 1 ≤
x≤ 0) =
and
0
1
Therefore
P (−1 ≤ X ≤ 1) =
2
−1
P (0 ≤ x ≤ 1) = area of
7
∫ 8 (x + 2) dx = 24
rec tan gle =
7 1 19
+ =
24 2 24
1
2
African Virtual University 119
d).
P(0 ≤ X ≤ 1) = area of rectangle=
Therefore P(x>1) =
1 1 1
× = .
3 2 6
1
6
5) a) P(A)=1-P(A’)=1- 5/8=3/8
b) P(AUB)=P(A) – P(B) – P(A I B)
7/8=3/8+P(B) – ¼
P(B)=3/4
c) P(A I B’)=P(A) – P(A I B)
= 3/8-1/4
=1/8
d) A’ U B’ = (A I B)’ and P(A’U B’) = 1 – P(A I B) = 3/4
e) Only one of A, B occurs = (A I B’)U((A’ I B).
P(only one of A,Boccurs) = P(A I B’)+P(A’ I B)
= { P(A)-P(A I B)} + { P(B)-P(A I B)}
= 1/8 + ½ =5/8
6). a).E(X)=7/8
b).E(24X+6)=20
c).E( 1-X)
1
2
=
1
1
3
2
(1
−
x)
(x + )dx =
∫0
2
5
1
African Virtual University 120
7). a). Medium= 76.3 kg.
b). Interquartile range = 9 kg
c). Estimate of
d). Estimate of
7
× 50 = 35 th
10
60
× 50 = 30 th
100
decile from curve .
percentile from curve
African Virtual University 121
XVII. References
http://en.wikipedia.org/wiki/Statistics
A concise Course in A-Level Statistics By J. Crawshaw and J.Chambers, Stanley
Thornes Publishers, 1994
http://en.wikipedia.org/wiki/Probability
Business Calculation and Statistics Simplified, By N.A. Saleemi, 2000
http://microblog.routed.net/wp-content/uploads/2007/01/onlinebooks.html
Statistics: concepts and applications, By Harry Frank and Steven C Althoen, Cambridge University Press, 2004
http://mathworld.wolfram.com/Statistics
http://mathworld.wolfram.com/Probability
Probability Demystified, By Allan G. Bluman, McGraw Hill, 2005.
http://directory.fsf.org/math/
http://microblog.routed.net/wp-content/uploads/2007/01/onlinebooks.html
Lectures on Statistics, By Robert B. Ash, 2005.
Introduction to Probability, By Charles M. Grinstead and J. Laurie Snell, Swarthmore
College.
http://directory.fsf.org/math/
Simple Statistics, By Frances Clegg, Cambridge University Press 1982.
Statistics for Advanced Level Mathematics, By I. Gwyn Evans University College
of Wales, 1984.
African Virtual University 122
XVIII. Student Records
Name of the EXCEL file
Mathematics: Probability and Statistics Student Records
African Virtual University 123
XIX. Main Author of the Module
Mr. Paul Chege (B.Ed(Sc), M.Ed)
paulamoud@yahoo.com
The module author is a teacher trainer at Amoud University, Borama, Republic of
Somaliland.
He has been a teacher trainer in Kenya, Republic of Seychelles, and Somalia. He has
been involved in strengthening Mathematics and Sciences at secondary and tertiary
levels under the Japan International Corporation Agency (JICA) programme in fifteen
African countries.
He is married with three children.
XX. File Structure
Module Developer Writing Tip. The file naming and structure must follow the AVU/
PI Consortium template as defined and explained by the AVU. Module Developers
still need to provide the name of all the files (module and other files accompanying
the module).
Daily, each module will be loaded in the personal eportfolio created for each consultant. For this, training will be provided by professor Thierry Karsenti and his team
(Salomon Tchaméni Ngamo and Toby Harper).
Name of the module (WORD) file : Mathematics: Probability and Statistics ( Word)
Name of all other files (WORD, PDF, PPT, etc.) for the module.
1. Mathematics: Probability and Statistics Student Records ( Excel)
2. Probability and Statistics: Marking Scheme for Summative Evaluation (
Word)
3. An Introduction to Probability and Random Processes, Textbook by Kenneth
Baclawski and Gian-Carlo Rota ( 1979) ( PDF)
4. Introduction To Probability, Textbook by Charles M. Grinstead and J. Laurie
Snell (PDF)
5. Lectures on Statistics, Textbook by Robert B. Ash (PDF)