INDEX N.S.M.-2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY GTU Syllabus Tutorial No. Name Of Topic Statistics Unit 1 Basic Probability Unit 2 Probability Unit 3 Descriptive Statistics Unit 4 Correlation and Regression Unit 5 Curve Fitting Numerical Methods Unit 6 Finite Differences and Interpolation Unit 7 Numerical Integration Unit 8 Solution of a System of Linear Equations Unit 9 Roots of Algebraic and Transcendental Equations Unit 10 Numerical solution of Ordinary Differential Equations GTU Papers June-2015 December-2015 June-2016 December-2016
125
Embed
INDEX [] TECHNOLOGICAL UNIVERSITY CIVIL ENGINEERING (06) NUMERICAL AND STATISTICAL METHODS FOR CIVIL ENGINEERING SUBJECT CODE: 2140606 B.E. 4th Semester Type of course: Engineering
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
INDEX
N.S.M.-2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
GTU Syllabus
Tutorial No. Name Of Topic
Statistics
Unit 1 Basic Probability
Unit 2 Probability
Unit 3 Descriptive Statistics
Unit 4 Correlation and Regression
Unit 5 Curve Fitting
Numerical Methods
Unit 6 Finite Differences and Interpolation
Unit 7 Numerical Integration
Unit 8 Solution of a System of Linear Equations
Unit 9 Roots of Algebraic and Transcendental Equations
Unit 10 Numerical solution of Ordinary Differential Equations
GTU Papers
June-2015
December-2015
June-2016
December-2016
GUJARAT TECHNOLOGICAL UNIVERSITY
CIVIL ENGINEERING (06) NUMERICAL AND STATISTICAL METHODS FOR CIVIL ENGINEERING
SUBJECT CODE: 2140606
B.E. 4th Semester
Type of course: Engineering Mathematics
Prerequisites: The students are required to have a reasonable mastery over calculus,
Differential equations and Linear algebra and introductory knowledge of probability and statistics.
Rationale: Mathematics is a language of Science and Engineering.
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Introduction
The words PROBABLE and POSSIBLE CHANCES are quite familiar to us.
We use these words when we are sure of the result of certain events. These words convey the sense of uncertainty of occurrence of events.
PROBABILITY is the word we use to calculate the degree of the certainty of events.
There are two types of approaches in the theory of PROBABILITY.
(1) Classical Approach – By Blaise Pascal
(2) Axiomatic Approach – By A. Kolmogorov
An experiment, in which we know all the possible outcomes in advance but which of them will occur is known only after the experiment is performed, is called a Random Variable.
Examples:
Event Outcomes
Toss Coin Head or Tail
Roll Die 1, 2, 3, 4, 5, 6
Pick a Card Ace,2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King (Clubs, Diamonds, Hearts, Spades)
Throw a stone It comes back surely. ( Why do we think about possibility )
Random Experiments
An experiment is called random experiment if it satisfies conditions:
(1) It has more than one possible outcome.
(2) It is not possible to predict the outcome in advance.
Sample Space
The set of outcomes is called the sample space of the experiment.
It is denoted by “ U ”.
Each element of the sample space is called a SAMPLE POINT.
If a sample space is in one-one correspondence with a finite set, {𝑥 ∈ ℕ | 1 ≤ 𝑥 ≤ 𝑛 , 𝑛 ∈ ℕ},
then it is called a finite sample space. Otherwise it is knowing as an infinite sample space.
1. Basic Concept Of Probability PAGE | 2
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Examples:
(1) Finite Sample Space: An experiment of tossing a coin twice.
U = {H, T} × {H, T} = {HH, HT, TH, TT}
(2) Infinite Sample Space: An experiment of tossing a coin until a HEAD comes up for the first time.
U = {H, TH, TTH, TTTH, TTTTH, TTTTTH, … }
Event
A subset of sample space is known as Event. Each member is called Sample Point.
Example:
Experiment U : Tossing a coin twice. U = {HH, HT, TH, TT}
Event A : getting TAIL both times. A = {TT}
Event B : getting TAIL exactly once. A = {HT, TH}
Definitions
1. The subset ∅ of a sample space is called “Impossible Events”. 2. The subset U(itself) of a sample space is called “Sure/Certain Events”. 3. If Subset contains only one element, it is called “Elementary/Sample Events”. 4. If Subset contains more than one element, it is called “Compound/Decomposable Events”. 5. The set contains all elements other than event A is known as “Complementary Event” of A.
It is known as A’. 6. A Union of Events A and B is Union of sets A and B (As per set theory). 7. An Intersection of Events A and B is Intersection of sets A and B (As per set theory). 8. If A ∩ B = ϕ, Events are called Mutually Exclusive Events (Disjoint set).
Set Notation: A ∩ B = { 𝑥 | 𝑥 ∈ A AND 𝑥 ∈ B }
9. If A ∪ B = U, Events are called Mutually Exhaustive Events.
Set Notation: A ∪ B = { 𝑥 | 𝑥 ∈ A OR 𝑥 ∈ B }
10. If A ∩ B = ϕ and A ∪ B = U, Events are called Mutually Exclusive & Exhaustive Events. 11. Suppose A1, A2, A3,…, An are Events.
If ,
1. Ai ∩ Aj = ϕ for i = 1,2, … , n ; j = 1,2, … , n ; i ≠ j.
2. ⋃ Ai
n
i=1
= U
Events A1, A2, A3,…, An are said to be Mutually Exclusive and Exhaustive Events. 12. Difference Events
A − B = A − (A ∩ B) = {x |x ∈ U, x ∈ A AND x ∉ B }
B − A = B − (A ∩ B) = {x |x ∈ U, x ∈ B AND x ∉ A }
13. Symmetric Difference Set A Δ B = (A − B) ∪ (B − A)
1. Basic Concept Of Probability PAGE | 3
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Exercise-1
C Que. 1 Describe the sample space for the indicated random experiments.
1. A coin is tossed 3 times. 2. A coin and die is tossed simultaneously.
H Que. 2 A balanced coin is tossed thrice. If three tails are obtained, a balance die is rolled. Otherwise the experiment is terminated. Write down the elements of the sample space.
Ans. 𝐒 = { 𝐇𝐇𝐇, 𝐇𝐇𝐓, 𝐇𝐓𝐇, 𝐇𝐓𝐓, 𝐓𝐇𝐇, 𝐓𝐇𝐓, 𝐓𝐓𝐇,
𝐓𝐓𝐓𝟏, 𝐓𝐓𝐓𝟐, 𝐓𝐓𝐓𝟑, 𝐓𝐓𝐓𝟒, 𝐓𝐓𝐓𝟓, 𝐓𝐓𝐓𝟔}
C Que. 3 Let a coin be tossed. If it shows head we draw a ball from a box containing 3 identical red and 4 identical green balls and if it shows a tail, we throw a die. What is the sample space of experiments?
H Que. 4 An experiment consists of rolling a die and then tossing a coin once if the number on the die is odd. If the number on the die is even the coin is tossed twice. Write sample space.
C Que. 5 Four card are labeled with A, B, C and D. We select and two cards at random without replacement. Describe the sample space for the experiments.
Ans. 𝐒 = {𝐀𝐁, 𝐀𝐂, 𝐀𝐃, 𝐁𝐂, 𝐁𝐃, 𝐂𝐃}
C Que. 6
A coin is tossed 3 times. Give the elements of the following events: 1. Event A: Getting at least two heads 2. Event B: Getting exactly two tails 3. Event C: Getting at most one tail 4. Event D: Getting at least one tail
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
H Que. 7
Describe the sample space associated with the experiment of selecting a child at random from three families each with a boy and a girl. Also write the elements of the following events:
1. There is at most one boy in the selection. 2. The selection consists of only girls. 3. The selection has exactly two girls.
An integer from 1 to 50 is selected at random. Write the elements of the following events. Event A: Randomly selected integer is a multiple of 2. Event B: Randomly selected integer is a multiple of 10. Event C: Randomly selected integer is a multiple of 4.
There are three identical balls, marked with a, b, c in a box. One ball is picked up from the box at random. The letter on it is noted and the ball is put back in the box, then another ball is picked up from the box and the letter on it is noted. Write the sample space of the experiment. Write down the elements of the following events: Event A: Ball marked “ a ” is selected exactly once. Event B: Balls selected have same letters marked. Event C: Ball with mark “ c ” is selected at least once.
Let U = {x1, x2, . . , xn} be a finite sample space. If P({x1}) = P({x2}) = P({x3}) = ⋯ = P({xn}), then the elementary events {x1}, {x2}, {x3}, … , {xn} are called Equiprobable Events.
Definition: Probability of an Event
If a finite sample space associated with a random experiments has "n" equally likely (Equiprobable) outcomes (elements) and of these "r“ (0 ≤ r ≤ n) outcomes are favorable for the occurrence of an event A, then probability of A is defined as follow.
𝐏(𝐀) =𝐟𝐚𝐯𝐨𝐫𝐚𝐛𝐥𝐞 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬
𝐭𝐨𝐭𝐚𝐥 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬=
𝐫
𝐧
1. Basic Concept Of Probability PAGE | 6
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Theorems
Thm. 1. For the Impossible Event , P(𝜙) = 0.
Thm. 2. Complementation Rule: For every Event A, P(A′) = 1 − P(A).
Thm. 3. If A ⊂ B, than P(B − A) = P(B) − P(A) and P(A) ≤ P(B)
Corollary 1. For every event A, 0 ≤ P(A) ≤ 1.
Corollary 2. For any events A and B,
P(A ∩ B′) = P(A) − P(A ∩ B)
P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) (De Morgan’s Rule)
P(A′ ∪ B′) = P(A ∩ B)′ = 1 − P(A ∩ B) (De Morgan’s Rule)
Thm. 4. Addition Rule for Arbitrary Events
Let S be sample space and A, B and C be any events in S, then
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(C ∩ A) + P(A ∩ B ∩ C)
Exercise - 2
C Que. 1 If A and B are two mutually exclusive events with P(A) = 0.30 , P(B) = 0.45. Find the probability of A′ , A ∩ B , A ∪ B , A′ ∩ B′ .
[𝟎. 𝟕 , 𝟎 , 𝟎. 𝟕𝟓, 𝟎. 𝟐𝟓]
C Que. 2 If probability of an event A id
9
10, what is the probability of the
event “ not A ”. [𝟎. 𝟏]
H Que.3
A fair coin is tossed twice. Find the probability of (1) getting H exactly once. (2) getting T at least once.
[𝟏
𝟐,𝟑
𝟒]
T Que. 4
One card is drawn at random from a well shuffled pack of 52 cards. Calculate the probability that the card will be
(1) An Ace (2) A card of black colour (2) A diamond (4) Not an ace
[𝟏
𝟏𝟑,𝟏
𝟐 ,
𝟏
𝟒 ,
𝟏𝟐
𝟏𝟑]
C Que. 5
Four cards are drawn from the pack of cards. Find the probability that (i) all are diamonds (ii) there is one card of each suit (iii) there are two spades and two hearts.
[𝟎. 𝟎𝟎𝟐𝟔, 𝟎. 𝟏𝟎𝟓𝟓, 𝟎. 𝟎𝟐𝟐𝟓]
Nov-16
C Que. 6
A box contains 5 red, 6 white and 2 black balls. The balls are identical in all respect other than colour.
1. One ball is drawn at random from the box. Find the probability that the selected ball is black.
2. Two balls are drawn at random from the box. Find the probability that one ball is white and one is red.
[𝟐
𝟏𝟑 ,
𝟓
𝟏𝟑]
1. Basic Concept Of Probability PAGE | 7
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Conditional Probability
Let S be a sample space and A and B be any two events in S. Then the probability of the
occurrence of event A when it is given that B has already occurred is expressed by the
symbol P(A B⁄ ) and is defined as
P(A B⁄ ) =P(A ∩ B)
P(B); P(B) > 0,
Which is known as conditional probability of the event A relative to eventB.
Similarly, the conditional probability of the event B relative to eventA is
P(B A⁄ ) =P(B ∩ A)
P(A); P(A) > 0.
Properties of Conditional Probability
Let A1, A2 and B be any three events of a sample space S,then
Let A and B be any two events of a sample space S, then A and B are called independent events if P(A ∩ B) = P(A) ⋅ P(B) . It also mean that, P(A B⁄ ) = P(A) and P(B A⁄ ) = P(B). This means that the probability of A does not depend on the occurrence or nonoccurrence of B, and conversely.
Remarks:
Let A , B and C are said to be Mutually independent, if
P(A ∩ B) = P(A) ⋅ P(B) , P(B ∩ C) = P(B) ⋅ P(C) P(C ∩ A) = P(C) ⋅ P(A) , P(A ∩ B ∩ C) = P(A) ⋅ P(B) ⋅ P(C)
Let A , B and C are said to be Mutually independent, if
P(A ∩ B) = P(A) ⋅ P(B) P(B ∩ C) = P(B) ⋅ P(C) P(C ∩ A) = P(C) ⋅ P(A)
Exercise – 3
C Que. 1
For two independent events A and B if P(A) = 0.3 and P(A ∪ B) = 0.6, find P(B).
[ 𝟑
𝟕 ]
H Que. 2
If A and B are independent events, where P(A) =1
4, P(B) =
2
3.
Find P(A ∪ B).
[ 𝟑
𝟒 ]
H Que. 3 If P(A) =
1
3, P(B) =
3
4 and P(A ∪ B) =
11
12 find P(A B⁄ ).
[ 𝟐
𝟗 ]
H Que. 4
If A and B are independent events, with P(A) =3
8, P(B) =
7
8.
Find P(A ∪ B) Find P(A B⁄ )and P(B A⁄ ).
[𝟓𝟗
𝟔𝟒,𝟑
𝟖 ,
𝟕
𝟖 ]
C Que. 5
If P(A) =1
3, P(B′) =
1
4 and P(A ∩ B) =
1
6 , then find P(A ∪ B),
P(A′ ∩ B′) and P(A′ B′⁄ ).
[𝟏𝟏
𝟏𝟐,
𝟏
𝟏𝟐,𝟏
𝟑]
1. Basic Concept Of Probability PAGE | 9
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
C Que. 6
In a group of 200 students 40 are taking English,50 are taking mathematics,and 12 are taking both (a) If a student is selected at random, what is the probability that the student is taking English? (b) A student is selected at random from those taking mathematics. What is the probability that the student is taking English? (c) A student is selected at random from those taking English, what is the probability that the student is taking mathematics?
[𝟎. 𝟐𝟎 , 𝟎. 𝟐𝟒 , 𝟎. 𝟑]
H Que. 7
In a box, 100 bulbs are supplied out of which 10 bulbs have defects of type A, 5 bulbs have defects of type B and 2 bulbs have defects of both the type. Find the probability that
(a) a bulb to be drawn at random has a B type defect under the condition that it has an A type defect, and
(b) a bulb to be drawn at random has no B type defect under the condition that it has no A type defect.
[𝟎. 𝟐] , [𝟎. 𝟗𝟔𝟔𝟕]
C Que. 8
A person is known to hit the target in 3 out of 4 shots, whereas another person is known to hit the target in 2 out of 3 shots. Find the probability of the target being hit at all when they both try. What is the probability that the target will be hit?
[𝟔
𝟏𝟐,𝟏𝟏
𝟏𝟐]
May -15
Dec-15
C Que. 9
An urn contains 10 white and 3 black balls, while another urn contains 3 white and 5 black balls. Two balls are drawn from the first urn and put into the second urn and then a ball is drawn from the later. What is the probability that it is a white ball?
[𝟓𝟗
𝟏𝟑𝟎]
Nov-16
H Que. 10
There are two boxes A and B containing 4 white, 3 red and 3 white,7 red balls respectively. A box is chosen at random and a ball is drawn from it, if the ball is white, find the probability that it is from box A.
[𝟒𝟎
𝟔𝟏]
C Que. 11
State Bayes’ theorem. A microchip company has two machines that produce the chips. Machine-I produces 65% of the chips, but 5% of its chips are defective. Machine-II produces 35% of the chips, but 15% of its chips are defective. A chip is selected at random and found to be defective. What is the probability that it came from Machine-I? [𝟎. 𝟑𝟖𝟐𝟒]
Nov-16
H Que. 12
State Bayes’ theorem. In a bolt factory, three machines A, B and C manufacture 25%, 35% and 40% of the total product respectively. Out Of these outputs 5%, 4% and 2% respectively, are defective bolts. A bolt is picked up at random and found to be defective. What are the Probabilities that it was manufactured by machine A, B and C? [𝟎. 𝟑𝟔𝟐𝟑, 𝟎. 𝟒𝟎𝟓𝟕, 𝟎. 𝟐𝟑𝟏𝟖]
Dec-15
1. Basic Concept Of Probability PAGE | 10
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
H Que. 13
A company has two plants to manufacture hydraulic machine. Plat I manufacture 70% of the hydraulic machines and plant II manufactures 30%. At plant I, 80% of hydraulic machines are rated standard quality and at plant II, 90% of hydraulic machine are rated standard quality. A machine is picked up at random and is found to be of standard quality. What is the chance that it has come from plant I? [𝟎. 𝟔𝟕𝟒𝟕]
May-15
C Que. 14
In a certain assembly plant, three machines, B1, B2 and B3, make 30%, 45% and 25%, respectively, of the products. It is known form the past experience that 2%, 3% and 2% of the products made by each machine respectively, are defective. Now, suppose that a finished product is randomly selected. What is the probability that it is defective? [𝟎. 𝟐𝟒𝟓]
May-16
C Que. 15
If proposed medical screening on a population, the probability that the test correctly identifies someone with illness as positive is 0.99 and the probability that test correctly identifies someone without illness as negative is 0.95. The incident of illness in general population is 0.0001. You take the test, the result is positive then what is the probability that you have the illness ?
[𝟎. 𝟎𝟎𝟏𝟗]
2. Probability Distribution PAGE | 11
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Random Variable
An experiment, in which we know all the possible outcomes in advance but which of
them will occur is known only after the experiment is performed, is called a Random
Variable.
Types of Random Variables 1. Discrete random variable
2. Continuous random variable
Discrete Random Variable
A random variable, which can take only finite, countable, or isolated values in a
given interval, is called discrete random variable.
i.e. A random variable is one which can assume any of a set of possible values which
can be counted or listed.
For example, the numbers of heads in tossing coins, the number of auto passengers
can take on only the values 1 , 2 , 3 and so on.
Note: Discrete random variables can be measured exactly.
Continuous Random Variable
A random variable, which can take all possible values that are infinite in a given
interval is called Continuous random variable.
i.e. a continuous random variable is one which can assume any of infinite spectrum
of different values across an interval which can not be counted or listed
For example, measuring the height of a student selected at random, finding the
average life of a brand X tyre etc.
Note: Continuous random variables can not be measured exactly.
Probability Function
If for random variable X, the real valued function f(x) is such that P(X = 𝑥) = 𝑓(𝑥),
then 𝑓(𝑥) is called Probability function of random variable X.
Probability function 𝑓(𝑥) gives the measures of probability for different values of X
say 𝑥1, 𝑥2, … . , 𝑥𝑛 .
Sometimes Probability density function is denoted by p(𝑥).
2. Probability Distribution PAGE | 12
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Probability Density Function
If X is a continuous random variable then its probability function 𝑓(𝑥) is called continuous probability function OR probability density function. It is defined as below
P(𝑎 < 𝑥 < 𝑏) = ∫ 𝑓(𝑥)𝑑𝑥𝑏
𝑎
Note:
1. 𝑓(𝑥) ≥ 0
2. ∫ 𝑓(𝑥)𝑑𝑥∞
−∞= 1
Probability Mass Function If X is a discrete random variable then its probability function 𝑓(𝑥) is discrete probability function. It is also called probability mass function, then P(X = 𝑥𝑖) = f(𝑥𝑖).
Note:
1. 𝑓(𝑥) ≥ 0 2. ∑ 𝑓(𝑥𝑖)
𝑛𝑖=1 = 1
Properties of Probability function If 𝑓(𝑥) is a probability function of random variable X then it possess the following Properties:
1. (𝑥) is positive for any value of x i.e. 𝑓(𝑥) ≥ 0 for all 𝑥. 2. Total of all values of 𝑓(𝑥) for different x is always 1. i.e. ∑ 𝑓(𝑥) = 1,
summation is taken for all values of 𝑥. 3. 𝑓(𝑥) measures the probability for any given value of 𝑥. 4. 𝑓(𝑥) cannot be negative for any value of 𝑥.
Mathematical Expectation If X is a discrete random variable having various possible values 𝑥1 , 𝑥2 , … . , 𝑥𝑛 and if 𝑓(𝑥) is the probability function, the mathematical Expectation or simply expectation of X is defined and denoted by E(X).
E(X) = ∑ 𝑥𝑖 ⋅ 𝑓(𝑥𝑖)
n
i=1
= ∑ 𝑥𝑖 ⋅ p(𝑥𝑖)
n
i=1
= ∑ 𝑥𝑖 ⋅ pi
n
i=1
Where, ∑ pi
n
i=1
= p1 + p2 + ⋯ pn = 1
If X is a continuous random variable having probability density function 𝑓(𝑥), expectation of X
is defined as E(X) = ∫ 𝑥 𝑓(𝑥) 𝑑𝑥∞
−∞
Note:
E(X) is also called the mean value of the probability distribution of x and is denoted by μ.
2. Probability Distribution PAGE | 13
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Properties of Mathematical Expectation 1. Expected value of constant term is constant. i.e. E(c) = c
2. If c is constant, then E(cX) = c ∙ E(X)
3. If a and b are constants, then E(aX ± b) = aE(X) ± b
4. If a , b and c are constants, then E (aX+b
c) =
1
c[aE(X) + b]
5. If X and Y are two random variables , then E(X + Y) = E(X) + E(Y)
6. If X and Y are two independent random variable, then E(X ∙ Y) = E(X) ∙ E(Y)
7. If 𝑔(𝑥) is any function of random variable X and 𝑓(𝑥) is probability density
function , then E{𝑔(𝑥)} = ∑ 𝑔(𝑥) ∙ 𝑓(𝑥)
Variance of a Random Variable
Variance is a characteristic of random variable X and it is used to measure dispersion
(or variation) of X.
If X is a discrete random variable with probability density function 𝑓(𝑥), then expected
value of [X − E(X)]2 is called the variance of X and it is denoted by V(X).
i.e. V(X) = E[X − E(X)]2 = E(X2) − [E(X)]2
Properties of Variance 1. V(c) = 0, Where c is a constant.
2. V(cX) = c2 V(X) , where c is a constant.
3. V(X + c) = V(X), Where c is a constant.
4. If a and b are constants , then V(aX + b) = a2V(X)
5. If X and Y are the independent random variables, then
V(X + Y) = V(X) + V(Y)
6. V(X) = E(X2) − μ2
Standard Deviation of Random Variable The positive square root of V(X) (Variance of X) is called standard deviation of random
variable X and is denoted by σ.
i.e. S.D. σ = √V(X).
Note : σ2 is called variance of V(X).
2. Probability Distribution PAGE | 14
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Exercise - 1
C Que.1
Which of the following functions are probability function?
(i) 𝑓(𝑥) = (1
2)
𝑥
(1
2)
1−𝑥
; 𝑥 = 0,1
(ii) 𝑓(𝑥) = (−1
2)
𝑥
, 𝑥 = 0,1,2
(iii) X −1 0 1
𝑓(𝑥) 0.5 0.8 −3 (iv)
X 1 2 3
𝑓(𝑥) 0.3 0.5 0.2 [YES, NO, NO , YES]
T Que.2
The probability function of a random variable X is
𝑝(𝑥) =2𝑥+1
48, 𝑥 = 1,2,3,4,5,6.
(a)Verify whether 𝑝(𝑥) is probability function ? (b)Also find E(X).
[𝐘𝐞𝐬, 𝟒. 𝟐𝟑]
H Que.3
Find the expected value of a random variable X having the following probability distribution.
X −5 −1 0 1 5 8
P(X = 𝑥) 0.12 0.16 0.28 0.22 0.12 0.1
[𝟎. 𝟖𝟔𝟎𝟎]
C Que.4
A random variable X has the following function.
X 0 1 2 3 4 5 6 7
P(X = 𝑥) 0 2k 3k k 2k k2 7k2 2k2 + k
Find the value of k and then evaluate (i)P(X < 6)(ii)P(X ≥ 6)(iii)P(0 < x < 5). [Hint: (i)P(X < 6) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) (ii)P(X ≥ 6) = 1 − P(X < 6) (iii) P(0 < X < 5) = P(1) + P(2) + P(3) + P(4) ]
[(𝐢)𝟎. 𝟖𝟏𝟎𝟎, (𝐢𝐢)𝟎. 𝟏𝟗𝟎𝟎, (𝐢𝐢𝐢)𝟎. 𝟖𝟎𝟎𝟎]
C Que.5
(a)A contestant tosses a coin and receives $5 if heads appears and $1 if tail appears. What is the expected value of a trial? (b)A contestant receives $4.00 if a coin turns up heads and pays $3.00 if
it turns tails. What is the expected value? [ Hint:P(H) = P(T) =1
2],
[(𝐚)$𝟑. 𝟎𝟎 , (𝐛)$𝟎. 𝟓𝟎]
C Que.6
A machine produces on average of 500 items during the first week of the month and average of 400 items during the last week of the month. The probability for these being 0.68 and 0.32. Determine the expected value of the production.
[𝟒𝟔𝟖]
May-15
2. Probability Distribution PAGE | 15
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
H Que.7
In a business, the probability that a trader can get profit of Rs. 5000 is 0.4 and probability for loss of Rs. 2000 is 0.6. Find his expected gain or loss.
[𝟖𝟎𝟎]
C Que.8 Three balanced coins are tossed, then find the mathematical expectation of tails.
[𝟏. 𝟓]
C Que.9 There are 8 apples in a box, of which 2 are rotten. A person selects 3 Apples at random from it. Find the expected value of the rotten Apples.
[𝟎. 𝟕𝟓𝟎𝟎]
C Que.10
There are 3 red and 2 white balls in a box and 2 balls are taken at random from it. A person gets Rs. 20 for each red ball and Rs. 10 for each white ball. Find his expected gain.
[𝐄(𝐱) = 𝟑𝟐]
H Que.11
There are 10 bulbs in a box, out of which 4 are defectives. If 3 bulbs are taken at random, find the expected number of defective bulbs.
[𝟏. 𝟐𝟎]
C Que.12
The probability distribution of a random variable x is given below.
C Que 2. 20% of the bulbs produced are defective .Find probability that at most 2 bulbs out of 4 bulbs are defective.
[𝟎. 𝟗𝟕𝟐𝟖]
C Que 3.
12% of the tablets produced by a tablet machine are defective. What is the probability that out of a random sample of 20 tablets produced by the machine, 5 are defective?
[𝟎. 𝟎𝟓𝟔𝟕]
C Que 4.
The probability that India wins a cricket test match against Australia is
given to be 1
3. If India and Australia play 3 tests matches, what is the
probability that (I) India will lose all the three test matches? (II) India will win at least one test match?
[(𝐢) 𝟎. 𝟐𝟗𝟔𝟑, (𝐢𝐢) 𝟎. 𝟕𝟎𝟑𝟕]
H Que 5.
What are the properties of Binomial Distribution ? The average percentage of failure in a certain examination is 40. What is the probability that out of a group of 6 candidates, at least 4 passed in examination ? [𝟎. 𝟓𝟒𝟒𝟑]
Dec-15
H Que 6.
The probability that in a university, a student will be a post-graduate is 0.6. Determine the probability that out of 8 students
(i) None (ii) Two (iii) At least two will be post-graduate
[(𝐢)𝟎. 𝟎𝟎𝟎𝟕, (𝐢𝐢)𝟎. 𝟎𝟒𝟏𝟑, (𝐢𝐢𝐢)𝟎. 𝟗𝟗𝟏𝟒]
T Que 7.
The probability that an infection is cured by a particular antibiotic drug within 5 days is 0.75. Suppose 4 patients are treated by this antibiotic drug. What is the probability that (a) no patient is cured (b) exactly two patient are cured (c) At least two patients are cured.
[(𝐚) 𝟎. 𝟎𝟎𝟑𝟗, (𝐛)𝟎. 𝟐𝟏𝟎𝟗, (𝐜)𝟎. 𝟗𝟒𝟗𝟐]
C Que 8.
Assume that on the average one telephone number out of fifteen called between 1 p.m. and 2 p.m. on weekdays is busy. What is the probability that if 6 randomly selected telephone numbers were called (i) not more than three, (ii) at least three of them would be busy?
[𝟎. 𝟗𝟗𝟗𝟕] , [𝟎. 𝟎𝟎𝟓𝟏]
Nov-16
T Que 9.
A dice is thrown 6 times getting an odd number of success, Find probability
(a) Five success (b) At least five success (c) At most five success.
[(𝐚)𝟑
𝟑𝟐, (𝐛)
𝟕
𝟔𝟒, (𝐜)
𝟔𝟑
𝟔𝟒]
2. Probability Distribution PAGE | 18
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
C Que 10.
A multiple choice test consist of 8 questions with 3 answer to each question (of which only one is correct). A student answers each question by rolling a balanced dice and checking the first answer if he gets 1 or 2, the second answer if he gets 3 or 4 and the third answer if he gets 5 or 6. To get a distinction, the student must secure at least 75% correct answers. If there is no negative marking, what is the probability that the student secures a distinction? [𝐏(𝐗 ≥ 𝟔) = 𝟎. 𝟎𝟏𝟗𝟕]
May-15
C Que 11. Obtain the binomial distribution for which mean is 10 and variance is 5.
[𝐏(𝐗 = 𝒙) = 𝐶𝑥20 (𝟎. 𝟓)𝐱(𝟎. 𝟓)𝟐𝟎−𝐱] Dec-15
H Que 12.
Determine binomial distribution whose mean is 4 and variance is 3 and hence evaluate P(X ≥ 2).
[𝐏(𝐗 = 𝐱) = { 𝑪𝒙𝟏𝟔 (𝟏
𝟒)
𝐱
(𝟑
𝟒)
𝟏𝟔−𝐱
, 𝐗 = 𝟎, 𝟏, … 𝟏𝟔
𝟎 , 𝐨𝐭𝐡𝐞𝐫𝐰𝐢𝐬𝐞
& 𝐏(𝐗 ≥ 𝟐) = 𝟎. 𝟗𝟑𝟔𝟓]
H Que 13.
For the binomial distribution with n = 20, p = 0.35. Find (a) Mean, (b) Variance and (c) Standard deviation.
[(𝐚)𝟕. 𝟎𝟎𝟎𝟎 (𝐛)𝟒. 𝟓𝟓, (𝐜)𝟐. 𝟏𝟑𝟑𝟏]
C Que 14. If the probability of a defective bolt is 0.1 Find mean and standard deviation of the distribution of defective bolts in a total of 400.
[𝛍 = 𝟒𝟎, 𝛔 = 𝟔]
Poisson Distribution
A discrete random variable 𝑋 is said to follow Poisson distribution if it assume only
nonnegative values and its probability mass function is given by
𝐏(𝐗) =𝒆−𝝀𝝀𝒙
𝒙!; 𝒙 = 𝟎, 𝟏, 𝟐, 𝟑, …
Examples Of Poisson Distribution 1. Number of defective bulbs produced by a reputed company.
2. Number of telephone calls per minute at a switchboard.
3. Number of cars passing a certain point in one minute.
4. Number of printing mistakes per page in a large text.
5. Number of persons born blind per year in a large city.
Properties Of Poisson Distribution The Poisson distribution holds under the following conditions.
1. The random variable 𝑥 should be discrete.
2. The number of trials 𝑛 is very large.
3. The probability of success 𝑝 is very small (very close to zero).
4. 𝜆 = 𝑛𝑝 Is finite.
5. The occurrences are rare.
Note
The mean and variance of the Poisson distribution with parameter 𝜆 are defined as
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Exercise - 3
C Que 1.
In a company, there are 250 workers. The probability of a worker remain absent on any one day is 0.02. Find the probability that on a day seven workers are absent.
[𝐏(𝐗 = 𝟕) = 𝟎. 𝟏𝟎𝟒]
C Que 2.
A book contains 100 misprints distributed randomly throughout its 100 pages. What is the probability that a page observed at random contains at least two misprints. Assume Poisson Distribution.
[𝟎. 𝟐𝟔𝟒𝟐]
Dec-15
H Que 3. For Poisson variant X, if P(X = 3) = P(X = 4) then, Find P(X = 0).
[𝐏(𝐗 = 𝟎) = 𝐞−𝟒]
T Que 4.
In a bolt manufacturing company, it is found that there is a small chance
of 1
500 for any bolt to be defective. The bolts are supplied in a packed of 30
bolts. Use Poisson distribution to find approximate number of packs, (a) Containing no defective bolt. (b) Containing two defective bolt, in the consignment of 10000
packets. [(𝐚)𝟗𝟒𝟐𝟎 , (𝐛) 𝟏𝟕]
H Que 5.
In sampling a large number of parts manufactured by a machine, the mean number and of defectives in a sample of 20 is 2. Out of 1000 such samples, how many would be expected to contain exactly two defective parts?
[𝐏(𝐗 = 𝟐) = 𝟎. 𝟐𝟕𝟎]
May-15
C Que 6.
Potholes on a highway can be serious problems. The past experience suggests that there are, on the average, 2 potholes per mile after a certain amount of usage. It is assumed that the Poisson process applies to the random variable “no. of potholes”. What is the probability that no more than four potholes will occur in a given section of 5 miles?
[𝐏(𝐗 ≤ 𝟒) = 𝟎. 𝟎𝟑𝟏𝟓]
May-16
T Que 7.
A car hire firm has two cars, which it hires out day by day. The number of demands for a car on each day is distributed on a Poisson distribution with mean 1.5. Calculate the proportion of days on which neither car is used and proportion of days on which some demand is refused. (e−1.5 =0.2231).
[𝐏(𝐗 = 𝟎) = 𝟎. 𝟐𝟐𝟑𝟏 ; 𝟏 − 𝐏(𝐗 ≤ 𝟐) = 𝟎. 𝟏𝟗𝟏𝟐]
Nov-16
C Que 8.
100 Electric bulbs are found to be defective in a lot of 5000 bulbs. Find the probability that at the most 3 bulbs are defective in a box of 100 bulbs.
[𝐏(𝐗 ≤ 𝟑) = 𝟎. 𝟖𝟓𝟔𝟕]
H Que 9.
Certain mass produced articles of which 0.5% are defective ,are packed in cartons each containing 100. What proportion of cartons are free from defective articles, and what proportion contain 3 or more defectives?
[𝐏(𝐗 = 𝟎) = 𝟔𝟎. 𝟔𝟓, 𝐏(𝐗 ≥ 𝟑) = 𝟎. 𝟎𝟏𝟒𝟒]
2. Probability Distribution PAGE | 20
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
T Que 10.
If a bank receives an average six back cheques per day what are the probability that bank will receive
(i) 4 back cheques on any given day (ii) 10 back cheques on any consecutive day.
[(𝐢)𝐏(𝐗 = 𝟒) = 𝟎. 𝟏𝟑𝟑𝟖 (𝐢𝐢)𝟐𝐏(𝐗 = 𝟏𝟎) = 𝟎. 𝟎𝟖𝟐𝟔]
H Que 11.
Suppose 1% of the items made by machine are defective. In a sample of 100 items find the probability that the sample contains
(i) All good (ii) 1 defective (iii) At least 3 defective
[𝐏(𝐗 = 𝟎) = 𝟎. 𝟑𝟔𝟕𝟗𝐏[𝐗 = 𝟏] = 𝟎. 𝟑𝟔𝟕𝟗
𝐏(𝐱 ≥ 𝟑) = 𝟎. 𝟎𝟖𝟎𝟑
]
2. Probability Distribution PAGE | 21
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Normal Distribution (or Gauss Distribution)
A continuous random variable 𝑿 is said to follow a normal distribution of its probability
density function is given by
𝒇(𝒙) =𝟏
𝝈√𝟐𝝅 𝒆𝐱𝐩 [−
𝟏
𝟐(
𝐱 − 𝛍
𝛔)
𝟐
] ; −∞ < 𝐱 < ∞, 𝛔 > 𝟎
μ = mean of the distribution
σ = Standard deviation of the distribution
Note: μ and σ2 (variance) are called parameters of the distribution.
If X is a normal random variable with mean 𝛍 and standard deviation 𝛔, and if we find
the random variable 𝐙 =𝐗−𝛍
𝛔 with mean 0 and standard deviation 1, then Z in called the
standard (standardized) normal variable.
The probability destiny function for the normal distribution in standard form is given by
𝒇(𝒛) =𝟏
√𝟐𝝅𝒆−
𝟏𝟐
𝒛𝟐
; −∞ < 𝒛 < ∞
For normal distribution
1. p(−∞ ≤ z ≤ ∞) = 1 (Total area)
2. p(−∞ ≤ z ≤ 0) = p (0 ≤ z ≤ ∞) = 0.5
3. p(−z1 ≤ z ≤ 0) = p(0 ≤ z ≤ z1) (∵ Symmetry)
Exercise - 4
C Que 1.
Compute the value of following: I. P(0 ≤ z ≤ 1.43)
II. p(−0.73 ≤ z ≤ 0) III. p(−1.37 ≤ z ≤ 2.02) IV. p(0.65 ≤ z ≤ 1.26) V. p(z ≥ 1.33)
The compressive strength of the sample of cement can be modelled by a normal distribution With a mean 6000 kg/cm2 and standard deviation of 100 kg/cm2. (i) What is the probability that a sample strength is less than 6250 kg/cm2 ? (ii)What is the probability if sample strength is between 5800 and 5900 kg/cm2 ?(iii)What strength is exceeded by 95% of the samples? [𝑷(𝒛 = 𝟐. 𝟓) = 𝟎. 𝟗𝟗𝟑𝟖, 𝑷(𝒛 = 𝟏) = 𝟎. 𝟖𝟒𝟏𝟑, 𝑷(𝒛 = 𝟐) =𝟎. 𝟗𝟕𝟕𝟐, 𝑷(𝒛 = 𝟏. 𝟔𝟓) = 𝟎. 𝟗𝟓]
[(𝟏)𝟎. 𝟗𝟗𝟑𝟖 , (𝟐)𝟎. 𝟏𝟖𝟏𝟓 , (𝟑) 𝟏. 𝟔𝟓 , 𝒙 = 𝟔𝟏𝟔𝟓]
May-16
2. Probability Distribution PAGE | 22
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
C Que 3.
In a photographic process, the developing time of prints may be looked upon as a random variable having the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 seconds. Find the probability that it will take (i)Anywhere from 16.00 to 16.50 sec to develop one of the prints; (ii)At least 16.20 sec to develop one of the prints; (iii)At most 16.35 sec to develop one of the prints. [𝑷(𝒛 = 𝟏. 𝟖𝟑) = 𝟎. 𝟗𝟔𝟔𝟒, 𝑷(𝒛 = 𝟎. 𝟔𝟔) = 𝟎. 𝟕𝟒𝟓𝟒, 𝑷(𝒛 = 𝟎. 𝟓𝟖) =𝟎. 𝟕𝟏𝟗𝟎]
[𝑷(𝟏𝟔 < 𝑿 < 𝟏𝟔. 𝟓) = 𝟎. 𝟗𝟓𝟔𝟓
𝑷[𝑿 ≥ 𝟏𝟔. 𝟐𝟎] = 𝟎. 𝟕𝟒𝟕𝟓𝑷(𝑿 ≤ 𝟏𝟔. 𝟑𝟓) = 𝟎. 𝟕𝟏𝟗𝟎
]
May-16
C Que 4.
A sample of 100 dry battery cell tested and found that average life of 12 hours and standard deviation 3 hours. Assuming the data to be normally distributed what percentage of battery cells are expected to have life
(i) More than 15 hour (ii) Less than 6 hour (iii) Between 10 & 14 hours
[𝟏𝟓. 𝟖𝟕%, 𝟐. 𝟐𝟖%, 𝟒𝟗. 𝟕𝟐%]
T Que 5.
In AEC company, the amount of light bills follows normal distribution with standard deviation 60. 11.31% of customers pay light-bill less than Rs.260. Find average amount of light bill.
[𝛍 = 𝟑𝟑𝟐. 𝟔𝟎]
C Que 6.
In examination, minimum 40 marks for passing and 75 marks for distribution are required. In this examination 45% student passed and 9% obtained distribution. Find average Marks and standard deviation of this distribution of marks.
[𝛍 = 𝟑𝟔. 𝟒, 𝛔 = 𝟐𝟖. 𝟖]
T Que 7.
Weights of 500 students of college is normally distributed with average weight 95 lbs & σ =7.5 Find how many students will have the weight between 100 and 110.
[𝟏𝟏𝟒]
C Que 8.
Distribution of height of 1000 soldiers is normal with Mean 165cms and standard deviation 15cms how many soldiers are of height
(i) Less than 138cms (ii) More than 198cms (iii) Between 138 and 198cms
[(𝐢) = 𝟑𝟔 , (𝐢𝐢) = 𝟏𝟒 , (𝐢𝐢𝐢) = 𝟗𝟓𝟎 ]
H Que 9. In a normal distribution 31% of the items are below 45 and 8% are above 64. Determine the mean and standard deviation of this distribution
[𝛍 = 𝟒𝟗. 𝟗𝟕𝟒, 𝛔 = 𝟗. 𝟗𝟓]
T Que 10.
The breaking strength of cotton fabric is normally distributed with E(x) =16 and σ(x) = 1. The fabric is said to be good if x ≥ 14 what is the probability that a fabric chosen at random is good ?
[𝟎. 𝟗𝟕𝟕𝟐]
2. Probability Distribution PAGE | 23
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Standard Normal ( Z ) Table , Area between 0 and Z
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Introduction
Descriptive statistics are used to describe the basic features of the data in a study. They provide simple summaries about the sample and the measures. Together with simple graphics analysis, they form the basis of virtually every quantitative analysis of data.
Descriptive Statistics are used to present quantitative descriptions in a manageable form. In a research study we may have lots of measures. Or we may measure a large number of people on any measure. Descriptive statistics help us to simplify large amounts of data in a sensible way. Each descriptive statistic reduces lots of data into a simpler summary. For instance, his single number describes a large number of discrete events. Or, consider the scourge of many students, the Grade Point Average (GPA). This single number describes the general performance of a student across a potentially wide range of course experiences.
Every time you try to describe a large set of observations with a single indicator you run the risk of distorting the original data or losing important detail. The GPA doesn't tell you whether the student was in difficult courses or easy ones, or whether they were courses in their major field or in other disciplines. Even given these limitations, descriptive statistics provide a powerful summary that may enable comparisons across people or other units.
Definition: Variable
A variable is any characteristics, number, or quantity that can be measured or counted. A variable may also be called a data. Age, sex, business income and expenses, country of birth, capital expenditure, class grades, eye color.
Type of Variable
A discrete variable is a variable whose value is obtained by counting.
E.g. Number of students present, number of red marbles in a jar, number of heads when flipping three coins, students’ grade level
A continuous variable is a variable whose value is obtained by measuring.
E.g. Height of students in class, weight of students in class, time it takes to get to school, distance traveled between classes.
Note
A random variable is a variable whose value is a numerical outcome of a random phenomenon. A random variable is denoted with a capital letter, the probability distribution of a random variable 𝑋 tells what the possible values of 𝑋 are and how probabilities are assigned to those values, a random variable can be discrete or continuous.
3. Descriptive Statistics PAGE | 26
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Definition: Data
Data is a set of values of qualitative or quantitative variables.
Types of data
On basis of Property
o Qualitative Data: Data includes property of object. E.g. Cast, residence area, etc.
o Numerical Data: Data includes numbers. E.g. height, weight, age, etc.
On the basis of Numbers
o Univariate Data: If one variable is required for observation, than it is called
Univariate data. e.g. {𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛}. o Bivariate Data: If two variable is required for observation, than it is called
Univariate data. e.g. {(𝑥1, 𝑦1) , (𝑥2, 𝑦2) , (𝑥3, 𝑦3) , … , (𝑥n, 𝑦n)}. o Trivariate Data: If three variable is required for observation, than it is
o Multivariate Data: If more than one variable is required to describe the
data, than it is called Multivariate data.
In this chapter, we discuss about analysis of Univariate Data.
Univariate Analysis
Univariate analysis involves the examination across cases of one variable at a time. There are three major characteristics of a single variable that we tend to look at:
Distribution ( Data ) Central Tendency Dispersion
Distribution
Distribution of a statistical data set (or a population) is a listing or function showing all the possible values (or intervals) of the data and how often they occur.
Type of distribution (Data) :
Ungrouped data
E.g. Marks of AEM of 10 students are 10,25,26,35,03,08,19,29,30,18.
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Grouped data
Discrete Frequency Distribution
E.g. Data of Students using Library during exam time.
No. reading hours(𝑥𝑖)
1 2 3 4
No. of hostel students(𝑓𝑖)
4 7 10 8
Range of Data
Difference of Maximum frequency and Minimum frequency is called range of
data. In above example, Range of data = 10 − 4 = 6 .
Continuous Frequency Distribution
E.g. Data of Students using Library during exam time.
No. reading hours (𝑥𝑖)
0 − 2 3 − 5 6 − 8 9 − 11
No. of hostel students (𝑓𝑖)
11 7 8 0
Class
I. Exclusive Class: If classes of frequency distributions are 0 − 2,2 − 4,4 − 6, … such classes are called Exclusive Classes.
II. Inclusive Class: If classes of frequency distributions are 0 − 2,3 − 5,6 − 8, … such classes are called Inclusive Classes.
Lower Boundary & Upper Boundary
In Class 𝑥𝑖 – 𝑥𝑖+1, Lower Boundary is 𝑥𝑖and Upper boundary is 𝑥𝑖+1 .
Mid-Point
It is defined as Lower Boundary + Upper Boundary
2.
For Continuous frequency distribution, Mid-Point of class is as 𝑥𝑖 of class.
Class Interval/ Class Length
It is defined as |Upper boundary – Lower Boundary|
3. Descriptive Statistics PAGE | 28
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Central Tendency
The central tendency of a distribution is an estimate of the "center" of a distribution of values. There are three major types of estimates of central tendency:
Arithmetic Mean Median Mode
Arithmetic Mean(�̅�)
The Mean or Average is probably the most commonly used method of describing central tendency. To compute the mean all you do is add up all the values and divide by the number of values.
Arithmetic Mean for Ungrouped data
If data is 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 , than Arithmetic Mean �̅� is defined as below.
Direct Method:
�̅� =𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑛
𝑛=
1
𝑛⋅ ∑ 𝑥𝑖
𝑛
𝑖=1
Assumed Mean Method:
Suppose 𝑥𝑖 is the biggest data among data 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛.LetA = 𝑥𝑖 .
Arithmetic Mean is defined as below.
�̅� = 𝐴 +1
𝑛⋅ ∑ 𝑑𝑖
𝑛
𝑖=1
; 𝑊ℎ𝑒𝑟𝑒 𝑑𝑖 = A − 𝑥𝑖
Generally, highest 𝑥𝑖is the choice of A.
Arithmetic Mean for Grouped data
Arithmetic Mean for following grouped data is defined as below.
Direct Method:
�̅� =𝑥1𝑓1 + 𝑥2𝑓2 + 𝑥3𝑓3 + ⋯ + 𝑥𝑛𝑓𝑛
𝑛=
1
𝑛⋅ ∑ 𝑥𝑖𝑓𝑖
𝑛
𝑖=1
Assumed Mean Method:
Suppose 𝑥𝑖 is the biggest data among data 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛. Let A = 𝑥𝑖 . Arithmetic Mean is defined as below.
�̅� = A +1
𝑛⋅ ∑ 𝑑𝑖𝑓𝑖
𝑛
𝑖=1
; 𝑊ℎ𝑒𝑟𝑒 di = A − 𝑥𝑖
Generally, highest 𝑥𝑖is the choice of A.
3. Descriptive Statistics PAGE | 29
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Step Deviation Method:
Suppose 𝑥𝑖 is the biggest data among data 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 with class length 𝑐.
Let 𝐴 = 𝑥𝑖 . Arithmetic Mean is defined as below.
�̅� = 𝐀 +𝒄
𝒏⋅ ∑ 𝒖𝒊𝒇𝒊
𝒏
𝒊=𝟏
; 𝑾𝒉𝒆𝒓𝒆 𝒖𝒊 =𝐀 − 𝒙𝒊
𝒄
Note:
Step deviation method is used for only continuous frequency distribution.
Exercise – 1
C Que.1
Find mean of temperature recorded in degree centigrade during a week in May, 2015, where the temperature recorded are 38.2, 40.9, 39, 44, 39.6, 40.5, 39.5.
[𝟒𝟎. 𝟐𝟒]
Nov-16
H Que.2
Find the arithmetic mean of the following frequency distribution.
𝑋 1 2 3 4
𝐹 4 5 2 1
[𝟐]
Jun-16
T Que.3
Find arithmetic mean from the following table.
𝑥 35 45 55 60 75 80
𝑓 12 18 10 6 3 11
[𝟓𝟒. 𝟎𝟖]
Nov-16
C Que.4
The marks obtained by 100 students of two classes in mathematics paper
Find the mean of the marks obtained by the students.
[𝟓𝟗. 𝟐𝟏]
3. Descriptive Statistics PAGE | 30
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
C Que.5
Find the mean of the data given below by all the three methods :
Class 0
− 10
10
− 20
20
− 30
30
− 40
40
− 50
50
− 60
60
− 70
Frequency 4 8 3 20 3 4 8
[𝟑𝟓. 𝟖]
T Que.6
The mean of the following frequency distribution is 16 , find the missing
frequency. Where, X=class, F=frequency
X 0
− 4
4
− 8
8
− 12
12
− 16
16
− 20
20
− 24
24
− 28
28
− 32
32
− 36
F 6 8 17 23 16 15 ? 4 3
[𝟖]
Mode ( Z ) :
The Mode is the most frequently occurring value in the set. To determine the mode, you might again order the observations in numerical order, and then count each one. The most frequently occurring value is the mode.
Mode of Ungrouped data:
Most repeated observation among given data is called Mode of Ungrouped data.
Mode of Grouped data:
Discrete Frequency Distribution
The value of variate (variable) corresponding to maximum frequency.
Continuous Frequency Distribution
𝒁 = 𝒍 + (𝒇𝟏 − 𝒇𝟎
𝟐𝒇𝟏 − 𝒇𝟎 − 𝒇𝟐) × 𝒄
Note that, Modal Class is the class with highest frequency.
Where,
𝒍 = Lower boundary of Modal Class
𝒄 = class interval OR class length
𝒇𝟏 = Frequency of the modal class.
𝒇𝟎 = Frequency of the class preceding the modal class.
𝒇𝟐 = Frequency of the class succeeding the modal class.
3. Descriptive Statistics PAGE | 31
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Exercise-2
C Que 1. Find mode of following observation.
7,3,6,8,9,8,8,8,9,8,7,5,8. [𝟖]
H Que 2.
Find mode of following observation.
10,9,21,16,14,18,20,18,14,18,23,16,18,4. [𝟏𝟖]
H Que 3.
A survey conducted on 20 hostel students for their reading hours per day resulted in the following frequency table
Number of reading hours
1 − 3 3 − 5 5 − 7 7 − 9 9 − 11
Number of hostel students
7 2 8 2 1
[𝟔]
C Que 4.
The mark distribution of 30 students at mathematics examination in a class as below
What is mode of the following frequency distribution?
Number of
reading hours 1 2 3 4
Number of hostel
students 4 7 10 8
[3]
Jun-16
T Que 6.
What is mode of the following frequency distribution?
No. of
children 0 1 2 3 4 5 6 7 8 9 10 11 12
No. of
family 97 110 254 95 67 43 21 13 7 6 5 4 2
[𝟐]
3. Descriptive Statistics PAGE | 32
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Median ( M )
The Median is the value found at the exact middle of the set of values. To compute the median is to list all observations in numerical order, and then locate the value in the center of the sample.
Median of Ungrouped Data
Let the total number of observation be𝑛.
𝐈𝐟 𝐧 𝐢𝐬 𝐨𝐝𝐝 𝐧𝐮𝐦𝐛𝐞𝐫, than M = (n + 1
2)
th
observation.
𝐈𝐟 𝐧 𝐢𝐬 𝐞𝐯𝐞𝐧 𝐧𝐮𝐦𝐛𝐞𝐫, 𝐭𝐡𝐚𝐧 𝐌 =(
𝒏
𝟐)
𝐭𝐡
𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧 + (𝒏
𝟐+ 𝟏)
𝐭𝐡
𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧
𝟐.
Median of Grouped Data:
Median of Discrete Grouped Data
Let the total number of observation be 𝑛.
𝐈𝐟 𝐧 𝐢𝐬 𝐨𝐝𝐝 𝐧𝐮𝐦𝐛𝐞𝐫, 𝐭𝐡𝐚𝐧 𝐌 = (𝐧 + 𝟏
𝟐)
𝐭𝐡
𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧.
𝐈𝐟 𝐧 𝐢𝐬 𝐞𝐯𝐞𝐧 𝐧𝐮𝐦𝐛𝐞𝐫, 𝐭𝐡𝐚𝐧 𝐌 =(
𝒏
𝟐)
𝐭𝐡
𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧 + (𝒏
𝟐+ 𝟏)
𝐭𝐡
𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧
𝟐.
Note: Formulae for Ungrouped data and discrete grouped data are same.
Median Of Continuous Grouped Data
𝐌 = 𝒍 + (
𝒏
𝟐− 𝑭
𝒇 ) × 𝒄
Where,
𝑙 = lower boundary point of the Median class
𝑛 = total number of observation (sum of the frequencies)
𝐹 = cumulative frequency of the class preceding the median class.
𝑓 = the frequency of the median class
𝑐 = class interval OR class length
Note
Median class = Class whose commulative frequncy with property min {𝑐𝑓 | 𝑐𝑓 ≥𝑛
2} .
3. Descriptive Statistics PAGE | 33
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Exercise-3
C Que 1. Define Mode and also give the relationship between Mean, Median and
Mode. May-16
Nov-16
T Que 2. Find the median by the data 2,8,4,6,10,12,4,8,14,16.
[11] Nov-16
C Que 3.
Obtain the median for the following distribution:
X 1 2 3 4 5 6 7 8 9 10
Y 10 34 27 24 12 27 20 18 15 30
[𝟐𝟐]
T Que 4.
Find Median of following data
Marks obtained 18 22 30 35 39 42 45 47
Number of
students 4 5 8 8 16 4 2 3
[𝟑𝟕]
H Que 5.
Find Median of following data
Number of students 6 4 16 7 8 2
Marks obtained 20 9 25 50 40 80
[𝟐𝟓]
T Que 6.
A Survey regarding the weights (in kg) of 45 students of class X of a school
was conducted and the following data was obtained:
Find the median weight.
[𝟑𝟖. 𝟕𝟓]
Weight
(in kg) 20 − 25
25
− 30
30
− 35
35
− 40
40
− 45
45
− 50
50
− 55
No. of
students 2 5 8 10 7 10 3
3. Descriptive Statistics PAGE | 34
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
C Que 7.
The following data represents the no. of foreign visitors in a multinational
company in every 10 days during last 2 months. Use the data to the
median.
X: 0
− 10 10 − 20
20
− 30
30
− 40
40
− 50
50
− 60
No. of
visitors f: 12 18 27 20 17 06
[27.4074]
Jun-16
C Que 8.
Obtain the median for the following distribution:
Mid value 15 20 25 30 35 40 45 50 55
Frequency 2 22 19 14 3 4 6 1 1
Cumulative 2 24 43 57 60 64 70 71 72
[𝟐𝟓. 𝟔𝟓𝟕𝟗]
Dec-13
H Que 9.
Obtain the median for the following information:
Marks < 0 < 10 < 20 < 30
students 50 38 20 5
[17.22]
T Que 10.
Obtain the median for the following information:
Daily
wage
50 < 50
− 100
100
− 150
150
− 200
200
− 250
250
− 300
300
− 350
frequency 2 4 7 21 25 20 21
[232]
C Que 11.
Calculate Mean, Median and Mode for the following data
[𝟔𝟑. 𝟖𝟐, 𝟔𝟑. 𝟔𝟔, 𝟔𝟑. 𝟐𝟖𝟓𝟕]
Class 50
− 53
53
− 56
56
− 59
59
− 62
62
− 65
65
− 68
68
− 71
71
− 74
74
− 77
F 3 8 14 30 36 28 16 10 5
Dec-15
3. Descriptive Statistics PAGE | 35
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Dispersion:
Dispersion refers to the spread of the values around the central tendency. There are two common measures of dispersion, the range and the standard deviation.
Range
It is simply the highest value minus the lowest value. In our example distribution, the high value is 36 and the low is 15, so the range is 36 − 15 = 21.
Standard Deviation(𝝈)
It is a measure that is used to quantify the amount of variation or dispersion of a set of data values.
Formula to find Standard Deviation
Method Ungrouped Data Grouped Data
Direct Method
σ = √∑ (xj − x̅)
2Nj=1
N
σ = √∑ fj(xj − x̅)
2Nj=1
N
Actual Mean Method σ = √∑ fixi
2ni=1
∑ fini=1
− (∑ fixi
ni=1
∑ fini=1
)
2
Assumed Mean Method σ = √∑ xi
2ni=1
N− (
∑ xini=1
N)
2
σ = √∑ fidi
2ni=1
∑ fini=1
− (∑ fidi
ni=1
∑ fini=1
)
2
Step Deviation Method σ = √∑ di
2ni=1
N− (
∑ dini=1
N)
2
σ = √∑ fiui
2ni=1
∑ fini=1
− (∑ fiui
ni=1
∑ fini=1
)
2
× c
Variance ( 𝑽(𝑿) )
Variance is the expectation of the squared deviation of a random variable from its mean, and it informally measures how far a set of (random) numbers are spread out from their mean.
Relation between Variance and Standard Deviation is as follow.
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Skewness
It is a measure of the asymmetry of the probability distribution of a real-valued random variable about its mean. The skewness value can be positive or negative, or even undefined.
Negative skew:
The left tail is longer; the mass of
the distribution is concentrated on
the right of the figure.
Positive skew:
The right tail is longer; the mass of
the distribution is concentrated on the left of the figure.
𝐒𝐤𝐞𝐰𝐧𝐞𝐬𝐬 =𝐌𝐞𝐚𝐧 − 𝐌𝐨𝐝𝐞
𝐒𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐝𝐞𝐯𝐢𝐚𝐭𝐢𝐨𝐧 =
�̅� − 𝐙
𝛔
Exercise – 4
T Que 1.
The pH of a solution is measured 8 times by one operator using the same
instrument. She obtains the following data: 7.15, 7.20, 7.18, 7.19, 7.21, 7.16 and
7.18. Calculate the sample Mean, the sample variance and sample standard
deviation.
[𝟕. 𝟏𝟖𝟏𝟒𝟑, 𝟎. 𝟎𝟎𝟎𝟒, 𝟎. 𝟎𝟐]
May-16
C Que 2.
Find the Standard deviation and Skewness of the data given below by all the
three methods :
Class 0
− 10
10
− 20
20
− 30
30
− 40
40
− 50
50
− 60
60
− 70
Frequency 4 8 3 20 3 4 8
[𝟏𝟕. 𝟗𝟖, 𝟎. 𝟎𝟒𝟒𝟒𝟗𝟑]
H Que 3.
Find the Standard deviation and Skewness of the mark distribution of 30
students at mathematics examination in a class as below :
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
T Que 4.
Find standard deviation from the following data.
Class 9 − 11 12 − 14 15 − 17 18 − 20
Frequency 2 3 4 1
[𝟐. 𝟕𝟒𝟗𝟓𝟓]
Jun-16
4. Correlation And Regression PAGE | 38
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Coefficient Of Correlation
Correlation is the relationship that exists between two or more variables. Two variables are said to be correlated if a change in one variable affects a change in the other variable. Such a data connecting two variables is called bivariate data.
When two variables are correlated with each other, it is important to know the amount or extent of correlation between them. The numerical measure of correlation of degree of relationship existing between two variables is called the coefficient of correlation and is denoted by r and it is always lies between −1 and 1.
When 𝑟 = 1, it represents Perfect Direct or Positive Correlation. When 𝑟 = −1 it represents Perfect Inverse or Negative Correlation. When 𝑟 = 0, there is No Linear Correlation or it shows Absence Of Correlation. When the value of r is ±0.9 or ±0.8 etc. it shows high degree of relationship
between the variables and when r is small say ±0.2 or ±0.1 etc, it shows low degree of correlation.
Types Of Correlations
Correlation is classified into four types:
1. Positive and negative correlations 2. Simple and multiple correlations 3. Partial and total correlations 4. Linear and nonlinear correlations
Positive and Negative Correlations Depending on the variation in the variables, correlation may be positive or negative.
1. Positive Correlation If both the variables vary in the same direction, the correlation is said to be positive. In the other words, if the value of one variable increases, the value of the other variable also increases, or, if the value of one variable decreases, the value of the other variable also decreases, e.g., the correlation between heights and weights of group of persons is a positive correlation.
2. Negative Correlation If both the variables vary in the opposite direction, correlation is said to be negative. In other words, if the value of one variable increases, the value of the other variable also decreases, or, if the value of one variable decreases, the value of the other variable also increases, e.g., the correlation between the price and demand of a commodity is a negative correlation.
Simple and Multiple Correlation Depending upon the study of the number of variables, correlation may be simple or
multiple.
1. Simple Correlation When only two variables are studied, the relationship is described as simple correlation, e.g., the quantity of money and price level, demand and price, etc.
4. Correlation And Regression PAGE | 39
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
2. Multiple correlation When more than two variables are studied, the relationship is described as multiple correlation, e.g., relationship of price, demand, and supply of a commodity.
Partial and Total Correlation Multiple correlation may be either partial or total.
1. Partial Correlation
When more than two variables are studied excluding some other variables, the relationship is termed as partial correlation.
2. Total Correlation
When more than two variables are studied, without excluding any variables ,the relationship is termed as total correlation.
Linear and Nonlinear Correlation Depending upon the ratio of change between two variables, the correlation may be
linear or nonlinear.
1.Linear Correlation
If the ratio of change between two variables is constant, the correlation is said to be linear.
2. Nonlinear Correlation
If the ratio of change between two variables is not constant, the correlation is said to be nonlinear.
Methods of Studying Correlation
There are two different methods of studying correlation,
1. Graphic methods,
2. Mathematical methods.
Graphic methods are
1. scatter diagram
2. Simple graph
Mathematical methods are
1. Karl Pearson’s coefficient of correlation
2. Spearman’s rank coefficient of correlation
Karl Pearson’s Product Moment Method
This is most popular and widely used mathematical method. In this method the degree
of correlation between two variables can be measured by the coefficient of correlation
and it gives not only magnitude (degree) of correlation but also its direction.
4. Correlation And Regression PAGE | 40
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Let (x1,y1), (x2,y2), . … … … (x𝑛,y𝑛)be n pairs of observations of two variables X and Y,
then coefficients of correlation (r) between X and Y is defined by
The amount of chemical compound , which were dissolved in 100 grams of water at various temperatures , 𝑥 were recorded as:
x(℃) 15 15 30 30 45 45 60 60
y(grams) 12 10 25 21 31 33 44 39
Find the line of regression of 𝑦 on 𝑥 & estimate the amount of chemical that will dissolve in 100 grams of water at50℃.
[𝒀 = 𝟎. 𝟔𝟕𝑿 + 𝟏. 𝟕𝟓, 𝒀(𝟓𝟎°) = 𝟑𝟓. 𝟐𝟓]
4. Correlation And Regression PAGE | 49
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
H Que 8.
The following table gives the age of tablet machine of certain make & annual maintenance costs. Obtain the regression equation for cost related to age.
Age of machine (years) 2 4 6 8
Maintenance cost (in thousand Rs.) 10 20 25 30
[𝒀 = 𝟑. 𝟐𝟓𝑿 + 𝟓]
T Que 9.
Obtain the line of regression of monthly sales(𝑌) on advertisement expenditure (𝑋) & estimate the monthly sales when the company will spend Rs. 50000 on advertisement, if the data on 𝑌 & 𝑋 are as follows:
Y (in lacs) 74 76 60 68 79 70 71 94
X (in thousand) 43 44 36 38 47 40 41 54
[𝒀 = 𝟏. 𝟕𝟐𝟓𝟒𝑿 + 𝟎. 𝟎𝟐𝟑𝟑, 𝒀(𝟓𝟎) = 𝟖𝟔. 𝟐𝟕𝟑𝟑]
T Que 10.
(i)The following data give the experience of machine operators and their performances rating as given by the number of good parts turned out per 100 piece
Operator 1 2 3 4 5 6
Performance rating(x)
23 43 53 63 73 83
Experience(Y) 5 6 7 8 9 10
Calculate the regression line of performing rating on experience and also estimate the probable performance if an operator has 11 years’ experience.
(ii)Define regression coefficients and give its properties.
[𝒙 = 𝟏𝟏. 𝟒𝟐𝟖𝒚 − 𝟐𝟗. 𝟑𝟖]
(performance rating on experience 𝒚 = 𝟏𝟏 years and 𝒙 = 𝟗𝟔. 𝟑𝟐 )
May-15
T Que 11.
A standard curve passing through the origin was prepared for colorimetric estimation of Salphadiazine. The concentration & absorbances are given below. Find the equation of line.
Concentrain 𝜇𝑔/𝑚𝑙
5 10 15 20 30 40
Absorbance 0.12 0.231 0.362 0.458 0.698 0.888
[𝒀 = 𝟎. 𝟎𝟐𝟐𝟏𝑿 + 𝟎. 𝟎𝟏𝟔𝟕]
4. Correlation And Regression PAGE | 50
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
H Que 12.
Obtain the two lines of regression for the following data:
H Que. 7 x + 4y − z = −5, x + y − 6z = −12,3x − y − z = 4.
[𝟏. 𝟔𝟒𝟕𝟕,−𝟏. 𝟏𝟒𝟎𝟖, 𝟐. 𝟎𝟖𝟒𝟓] May-15
8. Linear Algebric Equation PAGE | 93
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
H Que. 8 x + y + 2z = 4,3x + y − 3z = −4,2x − 3y − 5z = −5
[𝟏, −𝟏, 𝟐] May-15
C Que. 9
The following system of equations was generated by applying mess current law to the circuit. Use Gauss Elimination method to find the current in the circuit. 10𝑥1 + 𝑥2 + 𝑥3 = 12, 2𝑥1 + 2𝑥2 + 10𝑥3 =14, 2𝑥1 + 10𝑥2 + 𝑥3 = 15
The following system of equations was generated by applying mess current law to the circuit. Use Gauss Elimination method to find the current in the circuit. 2𝐼1 − 𝐼2 + 3𝐼3 = 8, − 𝐼1 + 2𝐼2 + 𝐼3 = 4, 3𝐼1 +𝐼2 − 4𝐼3 = 0
[𝟐, 𝟐, 𝟐]
June-16
H Que. 12 Solve the following system of equations by Gauss elimination method: 3𝑥 + 𝑦 − 𝑧 = 3; 2𝑥 − 8𝑦 + 𝑧 = −5; 𝑥 − 2𝑦 + 9𝑧 = 8.
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Gauss Jacobi Method
This method is applicable to the system of equations in which leading diagonal elements of the coefficient matrix are dominant (large in magnitude) in their respective rows.
Consider the system of equations.
𝐚𝟏𝐱 + 𝐛𝟏𝐲 + 𝐜𝟏𝐳 = 𝐝𝟏
𝐚𝟐𝐱 + 𝐛𝟐𝐲 + 𝐜𝟐𝐳 = 𝐝𝟐
𝐚𝟑𝐱 + 𝐛𝟑𝐲 + 𝐜𝟑𝐳 = 𝐝𝟑
Where co-efficient matrix 𝐀 must be diagonally dominant,
|𝐚𝟏| ≥ |𝐛𝟏| + |𝐜𝟏|
|𝐛𝟐| ≥ |𝐚𝟐| + |𝐜𝟐|
|𝐜𝟑| ≥ |𝐚𝟑| + |𝐛𝟑| …… (𝟏)
And the inequality is strictly greater than for at least one row.
Solving the system (𝟏) for 𝐱, 𝐲, 𝐳 respectively,we obtain
𝐱 =𝟏
𝐚𝟏(𝐝𝟏 − 𝐛𝟏𝐲 − 𝐜𝟏𝐳)
𝐲 =𝟏
𝐛𝟐(𝐝𝟐 − 𝐚𝟐𝐱 − 𝐜𝟐𝐳)
𝐳 =𝟏
𝐜𝟑(𝐝𝟑 − 𝐚𝟑𝐱 − 𝐛𝟑𝐲)…… (𝟐)
We start with 𝐱𝟎 = 𝟎, 𝐲𝟎 = 𝟎 & 𝐳𝟎 = 𝟎 in equ.(𝟐)
∴ 𝐱𝟏 =𝟏
𝐚𝟏(𝐝𝟏 − 𝐛𝟏𝐲𝟎 − 𝐜𝟏𝐳𝟎)
∴ 𝐲𝟏 =𝟏
𝐛𝟐(𝐝𝟐 − 𝐚𝟐𝐱𝟎 − 𝐜𝟐𝐳𝟎)
∴ 𝐳𝟏 =𝟏
𝐜𝟑(𝐝𝟑 − 𝐚𝟑𝐱𝟎 − 𝐛𝟑𝐲𝟎)
Again substituting these value x1, y1, z1 in Eq. (2), the next approximation is obtained.
This process is continued till the values of 𝐱, 𝐲, 𝐳 are obtained to desired degree of accuracy.
8. Linear Algebric Equation PAGE | 96
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Gauss Seidel Method
This is a modification of Gauss-Jacobi method. In this method we replace the approximation by the corresponding new ones as soon as they are calculated.
Consider the system of equations.
𝐚𝟏𝐱 + 𝐛𝟏𝐲 + 𝐜𝟏𝐳 = 𝐝𝟏
𝐚𝟐𝐱 + 𝐛𝟐𝐲 + 𝐜𝟐𝐳 = 𝐝𝟐
𝐚𝟑𝐱 + 𝐛𝟑𝐲 + 𝐜𝟑𝐳 = 𝐝𝟑
Where co-efficient matrix 𝐀 must be diagonally dominant,
|𝐚𝟏| ≥ |𝐛𝟏| + |𝐜𝟏|
|𝐛𝟐| ≥ |𝐚𝟐| + |𝐜2|
|𝐜𝟑| ≥ |𝐚𝟑| + |𝐛𝟑| …… (𝟏)
And the inequality is strictly greater than for at least one row.
Solving the system (𝟏) for 𝐱, 𝐲, 𝐳 respectively,we obtain
𝐱 =𝟏
𝐚𝟏(𝐝𝟏 − 𝐛𝟏𝐲 − 𝐜𝟏𝐳)
𝐲 =𝟏
𝐛𝟐(𝐝𝟐 − 𝐚𝟐𝐱 − 𝐜𝟐𝐳)
𝐳 =𝟏
𝐜𝟑(𝐝𝟑 − 𝐚𝟑𝐱 − 𝐛𝟑𝐲)…… (𝟐)
We start with 𝐱𝟎 = 𝟎, 𝐲𝟎 = 𝟎 & 𝐳𝟎 = 𝟎 in equ.(𝟐)
∴ 𝐱𝟏 =𝟏
𝐚𝟏(𝐝𝟏 − 𝐛𝟏𝐲𝟎 − 𝐜𝟏𝐳𝟎)
Now substituting 𝐱 = 𝐱𝟏 & 𝐳 = 𝐳𝟎 in the second equ. Of (𝟐)
∴ 𝐲𝟏 =𝟏
𝐛𝟐(𝐝𝟐 − 𝐚𝟐𝐱𝟏 − 𝐜𝟐𝐳𝟎)
Now substituting 𝐱 = 𝐱𝟏 & 𝐲 = 𝐲1 in the third equ. Of (𝟐)
∴ 𝐳𝟏 =𝟏
𝐜𝟑(𝐝𝟑 − 𝐚𝟑𝐱𝟏 − 𝐛𝟑𝐲𝟏)
This process is continued till the values of 𝐱, 𝐲, 𝐳 are obtained to desired degree of accuracy.
8. Linear Algebric Equation PAGE | 97
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Exercise-4
C Que. 1 Solve the following system of equations by Gauss-Seidel method. 10𝑥1 + 𝑥2 + 𝑥3 = 6 , 𝑥1 + 10𝑥2 + 𝑥3 = 6 , 𝑥1 + 𝑥2 + 10𝑥3 = 6.
[𝟎. 𝟓, 𝟎. 𝟓, 𝟎. 𝟓]
Jun-10 Dec-15 Nov-16
T Que. 2 Use Gauss seidel method to determine roots of the following equations. 2𝑥 − 𝑦 = 3 , 𝑥 + 2𝑦 + 𝑧 = 3 , −𝑥 + 𝑧 = 3.
[𝟏, −𝟏, 𝟒] Jun-13
H Que. 3 Use Gauss seidel method to find roots of the following equations. 8𝑥 + 𝑦 + 𝑧 = 5 , 𝑥 + 8𝑦 + 𝑧 = 5 , 𝑥 + 𝑦 + 8𝑧 = 5.
[𝟎. 𝟓, 𝟎. 𝟓, 𝟎. 𝟓]
Dec-13 May-15
H Que. 4
State the Direct & iterative method to solve system of linear equations. Arrange following system of equations into diagonally dominant form and solve it using Gauss Seidel method. 10𝑥1 + 𝑥2 + 𝑥3 = 12 , 2𝑥1 + 10𝑥2 + 𝑥3 = 13, 2𝑥1 + 2𝑥2 + 10𝑥3 = 14.
[𝟏, 𝟏, 𝟏]
Dec-10 Dec-15 June-16
C Que. 5
Solve by Gauss-Seidel & Gauss-Jacobi method correct up to two decimal places. 20𝑥 + 2𝑦 + 𝑧 = 30, 𝑥 − 40𝑦 + 3𝑧 = −75,2𝑥 − 𝑦 + 10𝑧 = 30.
[𝟏. 𝟏𝟒, 𝟐. 𝟏𝟑, 𝟐. 𝟗𝟗]
Jun-11
H Que. 6
Solve this system of linear equations using Jacobi’s method in three iterations first check the co-efficient matrix of the following systems is diagonally dominant or not? 20𝑥 + 𝑦 − 2𝑧 = 17, 2𝑥 − 3𝑦 + 20𝑧 = 25, 3𝑥 + 20𝑦 − 𝑧 = −18
[𝟏, −𝟏, 𝟏]
Dec-15
H Que. 7 Solve the following system of equations by Gauss-Seidel method. 20𝑥 + 𝑦 − 2𝑧 = 17, 2𝑥 − 3𝑦 + 20𝑧 = 25, 3𝑥 + 20𝑦 − 𝑧 = −18
[𝟏, −𝟏, 𝟏] Jun-14
C Que. 8 Solve by Gauss-Seidel method correct up to three decimal places. 2𝑥 + 𝑦 + 54𝑧 = 110,27 𝑥 + 6𝑦 − 𝑧 = 85,6𝑥 + 15𝑦 + 2𝑧 = 72.
H Que. 10 Solve by Gauss-Seidel method correct up to three decimal places. 10𝑥 − 5𝑦 − 2𝑧 = 3,4𝑥 − 10𝑦 + 3𝑧 = −3, 𝑥 + 6𝑦 + 10𝑧 = −3.
[𝟎. 𝟑𝟒𝟐, 𝟎. 𝟐𝟖𝟓,−𝟎. 𝟓𝟎𝟓]
T Que. 11
Check whether the following system is diagonally dominant or not. If not, re-arrange the equations so that it becomes diagonally dominant and hence solve the system of simultaneous linear equation by Gauss sidle Method. −100𝑦 + 130𝑧 = 230,−40𝑥 + 150𝑦 − 100𝑧 = 0,60𝑥 − 40𝑦 = 200.
[𝟕. 𝟕𝟖, 𝟔. 𝟔𝟕, 𝟔. 𝟗𝟎]
Dec-12
H Que. 12
Solve the following system of equations using Gauss-Seidel method correct up to three decimal places. 60𝑥 − 4𝑦 + 6𝑧 = 150 ; 2𝑥 + 2𝑦 + 18𝑧 = 30 ; 𝑥 + 17𝑦 − 2𝑧 = 48
[𝟐. 𝟓𝟖𝟎, 𝟐. 𝟕𝟗𝟖, 𝟏. 𝟎𝟔𝟗]
Dec-13
8. Linear Algebric Equation PAGE | 98
N.S.M. - 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
By gauss Seidel method solve the following system upto six iteration 12𝑥1 + 3𝑥2 − 5𝑥3 = 1 ; 𝑥1 + 5𝑥2 + 3𝑥3 = 28 ; 3𝑥1 + 7𝑥2 + 13𝑥3 = 76 Use initial condition (𝑥1 𝑥2 x3) = (1 0 1).
[𝟏, 𝟑, 𝟒]
May-15
H Que. 15 By gauss Seidel method solve the following system 2x + y + 6z = 9 ; 8x + 3y + 2z = 13 ; x + 5y + z = 7
[𝟏, 𝟏, 𝟏] May-15
H Que. 16
State the Direct and iterative methods to solve system of linear equations. Using Gauss-Seidel method, Solve up to 3 iteration 2x1 − x2 = 7 ; −x1 + 2x2 − x3 = 1 ; −x2 + 2x3 = 1
[𝟓. 𝟑𝟏𝟐𝟓, 𝟒. 𝟑𝟏𝟐𝟓, 𝟐. 𝟔𝟓𝟔𝟑]
Dec-15
H Que. 17 Solve the following system of equations by Gauss – Seidel method 10𝑥 − 𝑦 − 𝑧 = 13, 𝑥 + 10𝑦 + 𝑧 = 36, 𝑥 + 𝑦 − 10𝑧 = −35
[𝟐, 𝟑, 𝟒] May-16
C Que. 18 Use Gauss Seidel method to solve 83𝑥 + 11𝑦 − 4𝑧 = 95, 7𝑥 + 52𝑦 + 13𝑧 = 104, 3𝑥 + 8𝑦 + 29𝑧 = 71
[𝟏. 𝟎𝟔, 𝟏. 𝟑𝟕, 𝟏. 𝟗𝟔] May-16
T Que. 19
Using Gauss-Seidel iteration method solve the system of equations: 10𝑥 − 2𝑦 − 𝑧 − 𝑤 = 3, −2𝑥 + 10𝑦 − 𝑧 − 𝑤 = 15, −𝑥 − 𝑦 + 10𝑧 − 2𝑤 = 27, −x − y − 2z + 10w = −9.
[𝟏, 𝟐, 𝟑, 𝟎]
Nov-16
9. Roots Of Non-Linear Equation PAGE | 99
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Introduction
An expression of the form 𝒇(𝒙) = 𝒂𝟎𝒙𝒏 + 𝒂𝟏𝒙𝒏−𝟏 + 𝒂𝟐𝒙𝒏−𝟐 + ⋯ + 𝒂𝒏−𝟏𝒙 + 𝒂𝒏,
where 𝒂𝟎, 𝒂𝟏, 𝒂𝟐 … 𝒂𝒏 are constants and 𝒏 is a positive integer , is called an algebraic
polynomial of degree n if 𝒂𝟎 ≠ 𝟎. The equation 𝒇(𝒙) = 𝟎 is called an algebraic
equation if 𝒇(𝒙) is an algebraic polynomial, e.g. 𝒙𝟑 − 𝟒𝒙 − 𝟗 = 𝟎. If 𝒇(𝒙) contains
functions such as trigonometric, logarithmic, exponential, etc. , then 𝒇(𝒙) is called a
In general, an equation is solved by factorization. But in many cases, the method of
factorization fails. In such cases, numerical methods are used. There are some
methods to solve the equation 𝒇(𝒙) = 𝟎, which are given as follows.
SR.NO. TOPIC NAME
1. Bisection Method
2. Secant Method
3. Regula-Falsi Method (False Position Method)
4. Newton-Raphson Method
Methods to find Roots of equations
Bisection MethodRegula-Falsi Method
Secant Method Newton-Raphson Method
9. Roots Of Non-Linear Equation PAGE | 100
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Bisection Method
Let , an equation which is f(x) = 0
If f(a) > 0 and f(b) < 0 ,Where a and b are consecutive integer, then
x1 =a + b
2
Check f(x1) > 0 OR f(x1) < 0.
If f(x1) > 0,then x2 =x1+b
2
OR
If f(x1) < 0, then we find x2 =a+x1
2.
Check f(x2) > 0 OR f(x2) < 0.
If f(x1) > 0, f(x2) > 0 then x3 =x2+b
2 OR f(x1) < 0, f(x2) > 0 then x3 =
x2+x1
2
OR
If f(x1) > 0, f(x2) < 0 then x3 =x1+x2
2 OR f(x1) < 0, f(x2) < 0 then x3 =
a+x2
2
Processing like this when latest two consecutive values of x are not same.
Exercise-1
C Que.1 Discuss the method Bisection to find the root of an equation 𝑓(𝑥) = 0. Nov-16
C Que.2
Find the positive root of x = cos x correct up to three decimal places
by bisection method.
[𝟎. 𝟕𝟑𝟗]
Jun-10
Jun-14
C Que.3
Explain bisection method for solution of equation. Using this method
find the approximate solution x3 + x − 1 = 0 of correct up to three
decimal points.
[𝟎. 𝟔𝟖𝟑]
Dec-13
H Que.4
Perform three iterations of Bisection method to obtain a root of the
equation 𝑓(𝑥) = cos 𝑥 − 𝑥𝑒𝑥 = 0 in the interval (0.5, 1).
[𝟎. 𝟓𝟑𝟏𝟐]
Nov-16
T Que.5
Perform the five iterations of the bisection method to obtain a root of
the equation f(x) = cos x − xex = 0.
[𝟎. 𝟓𝟑𝟏𝟐]
Nov-10
9. Roots Of Non-Linear Equation PAGE | 101
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
T Que.6
Find root of equation x3 − 4x − 9 = 0, using the bisection method in
four stages.
[𝟐. 𝟔𝟖𝟕𝟓]
Jun-11
H Que.7
Perform the five iteration of the bisection method to obtain a root of
the equation x3 − x − 1 = 0.
[𝟏. 𝟑𝟒𝟑𝟕𝟓]
Nov-11
C Que.8
Find the negative root of x3 − 7x + 3 = 0 bisection method up to
three decimal place.
[−𝟐. 𝟖𝟑𝟗]
Jun-12
T Que.9
Use bisection method to find a root of equation
x3 + 4x2 − 10 = 0 in the interval [1,2].Use four iteration.
[𝟏. 𝟑𝟏𝟐𝟓]
Dec-12
H Que.10
Perform three iterations of Bisection method to obtain root of the
equation 2 sin x − x = 0.
[𝟏. 𝟖𝟕𝟓]
May-15
T Que.11
Explain bisection method for solving an equation f(x) = 0.Find the real
root of equation x2 − 4x − 10 = 0 by using this method correct to
three decimal places.
[𝟓. 𝟕𝟒𝟐]
Dec-15
May-16
Method of False Position or Regula-Falsi Method
This is the oldest method to finding the real root of an equation
𝑓(𝑥) = 0 and closely resembles the bisection method.
Here we choose two points 𝑥0 and 𝑥1 such that 𝑓(𝑥0) and 𝑓(𝑥1) are of opposite signs i.e.
the graph of 𝑦 = 𝑓(𝑥) crosses the x-axis between these points. This indicates that a root
lies between 𝑥0 and 𝑥1 and consequently 𝑓(𝑥0)𝑓(𝑥1) < 0.
Equation of the chord joining the points 𝐴[𝑥0, 𝑓(𝑥0)] and 𝐵[𝑥1, 𝑓(𝑥1)] is
𝑦 − 𝑓(𝑥0) =𝑓(𝑥1)−𝑓(𝑥0)
𝑥1−𝑥0(𝑥 − 𝑥0)
9. Roots Of Non-Linear Equation PAGE | 102
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Fig. 2.6
The method consists in replacing the curve AB by means of the chord AB and taking the
point of intersection of the chord with the x-axis as an approximation to the root. So the
abscissa of the point where the chord cuts the x-axis (𝑦 = 0) is given by
𝒙𝟐 = 𝒙𝟎 −𝒙𝟏−𝒙𝟎
𝒇(𝒙𝟏)−𝒇(𝒙𝟎)𝒇(𝒙𝟎) … (1)
Which is an approximation to the root.
If now 𝑓(𝑥0) and 𝑓(𝑥2) are of opposite signs, then the root lies between 𝑥0 and 𝑥2. So
replacing 𝑥1 by 𝑥2 in (1) , we obtain the next approximation 𝑥3. (The root could as well
lie between 𝑥1 and 𝑥2 and we would obtain 𝑥3 accordingly ). This procedure is repeated
till the root is found to the desired accuracy. The iteration process based on (1) is known
as the method of false position and its rate of convergence is faster than that of the
bisection method.
Secant Method
This method is an improvement over the method of false position as it does not require
the condition 𝑓(𝑥0)𝑓(𝑥1) < 0 of that method. Here also the graph of the function 𝑦 = 𝑓(𝑥)
is approximated by a secant line but at each iteration, two most recent approximations to
the root are used to find the next approximation. Also it is not necessary that the interval
must contain the root.
Taking 𝑥0 , 𝑥1 as the initial limits of the interval, we write the equation of the chord joining
these as
𝒚 − 𝒇(𝒙𝟏) =𝒇(𝒙𝟏) − 𝒇(𝒙𝟎)
𝒙𝟏 − 𝒙𝟎(𝒙 − 𝒙𝟏)
9. Roots Of Non-Linear Equation PAGE | 103
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Then the abscissa of the point where it crosses the 𝑋 −axis (𝑦 = 0) is given by
𝒙𝟐 = 𝒙𝟏 −𝒙𝟏−𝒙𝟎
𝒇(𝒙𝟏)−𝒇(𝒙𝟎)𝒇(𝒙𝟏)
Which is an approximation to the root. The general formula for successive
approximations is, therefore, given by
𝒙𝒏+𝟏 = 𝒙𝒏 −𝒙𝒏−𝒙𝒏−𝟏
𝒇(𝒙𝒏)−𝒇(𝒙𝒏−𝟏)𝒇(𝒙𝒏) , 𝑛 ≥ 1.
Note :- If at any iteration 𝑓(𝑥𝑛) = 𝑓(𝑥𝑛−1), this method fails and shows that it does not
converge necessarily. This is drawback of secant method over the method of false
position which always converges
Exercise-2
C Que.1
Find a positive root of 𝑥𝑒𝑥 − 2 = 0 by the method of False
position.
[𝟎. 𝟖𝟓𝟐𝟔]
Nov-16
C Que.2
Apply False Position method to find the negative root of the equation x3 − 2x + 5 = 0 correct to four decimal places.
[−𝟐. 𝟎𝟗𝟒𝟓] May-15
H Que.3
Find a root of the equation x3 − 4x − 9 = 0 using False-position method correct up to three decimal.
[𝟐. 𝟕𝟎𝟔𝟓] Dec-15
9. Roots Of Non-Linear Equation PAGE | 104
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
T Que.4
Find an approximate value of the root the equation 𝑥3 + 𝑥 − 1 = 0
using the method of false position.
[𝟎. 𝟔𝟕𝟐]
Nov-16
H Que.5
Explain False position method for finding the root of the equation f(x) = 0.Use this method to find the root of an equation x = e−x correct up to three decimal places.
[𝟎. 𝟓𝟔𝟕]
Dec-15
Nov-16
T Que.6 Using method of False-position, compute the real root of the equation x𝑙𝑜𝑔10x − 1.2 = 0 correct to four decimals. where(𝑥 > 0)
[𝟐. 𝟕𝟒𝟎𝟐𝟏]
Dec-15
C Que.7 Find the positive solution of f(x) = x − 2sinx = 0 by the secant method, starting form x0 = 2, x1 = 1.9.
[𝟏. 𝟖𝟗𝟓𝟓]
Nov-10
Jun-14
T Que.8 Derive Secant method and solve xex − 1 = 0 correct up to three decimal places between 0 and 1.
[𝟎. 𝟓𝟔𝟕]
Jun-12
H Que.9
Find the real root of the following by secant method. a) x2 − 4x − 10 = 0 (using x0 = 4, x1 = 2,upto six iteration) b) x3 − 2x − 5 = 0 (using x0 = 2, x1 = 3, upto four iteration)
[𝟓. 𝟕𝟒𝟏𝟏, 𝟐. 𝟎𝟗𝟐𝟖]
Nov-16
C Que.10 Use Secant method to find the roots of cos x − xex = 0 correct upto 3 decimal places of decimal. [𝟎. 𝟓𝟏𝟖]
May-15
May-16
Nov-16
Newton-Raphson Method (Newton’s Method)
Let , an equation which is f(x) = 0
If f(a) ∙ f(b) < 0
x0 = a when |f(a)| < |f(b)| OR x0 = b when |f(b)| < |f(a)|. then
𝐱𝐧+𝟏 = 𝐱𝐧 −𝐟(𝐱𝐧)
𝐟′(𝐱𝐧); 𝐧 = 𝟎, 𝟏, 𝟐, 𝟑 …
Where f ′(xn) ≠ 0
Processing like this when latest two consecutive values of x are not same.
Explanation:
Le x1be the root of f(x) = 0 and x0 be an approximation to x1. If h = x1 − x0, then by
Taylor’s Series,
𝐟(𝐱𝟎 + 𝐡) = 𝐟(𝐱𝟎) + 𝐡 𝐟′(𝐱𝟎) +𝐡𝟐
𝟐! 𝐟′′(𝐱𝟎) + ⋯
9. Roots Of Non-Linear Equation PAGE | 105
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Since, x1 = x0 + h is root. f(x1) = f(x0 + h) = 0
If h is chosen too small enough, then we can neglect 2nd , 3rd and higher powers of h.
We have,
𝟎 = 𝐟(𝐱𝟎) + 𝐡 𝐟′(𝐱𝟎) ⟹ 𝐡 = −𝐟(𝐱𝟎)
𝐟′(𝐱𝟎) ; 𝐟′(𝐱𝟎) ≠ 𝟎.
Suppose that, x1 = x0 + h be the better approximation.
⟹ 𝐱𝟏 = 𝐱𝟎 −𝐟(𝐱𝟎)
𝐟′(𝐱𝟎)
By repeating the process,
⟹ 𝐱𝟐 = 𝐱𝟏 −𝐟(𝐱𝟏)
𝐟′(𝐱𝟏)
⟹ 𝐱𝟑 = 𝐱𝟐 −𝐟(𝐱𝟐)
𝐟′(𝐱𝟐)
In general,
𝐱𝐧+𝟏 = 𝐱𝐧 −𝐟(𝐱𝐧)
𝐟′(𝐱𝐧); 𝐧 = 𝟎, 𝟏, 𝟐, 𝟑 …
Where f ′(xn) ≠ 0
This is called Newton-Raphson Formula.
Note
If the function is linear ,( f ′(xn) = 0)then N-R method has to be failed.
Prove that NR procedure is second order convergent.
Or
Rate of convergence of the NR methods.
Proof : Let 𝛼 be exact root of 𝑓(𝑥) = 0 and let 𝑥𝑛,𝑥𝑛+1 be two successive approximations
to the actual root. If ∈𝑛 and ∈𝑛+1 are the corresponding errors then
𝒙𝒏= 𝜶 +∈𝒏
𝒙𝒏+𝟏= 𝜶 +∈𝒏+𝟏
Substituting in equation of ‘NR METHOD FORMULA’
𝜶 +∈𝒏+𝟏 = 𝜶 +∈𝒏 −𝒇(𝜶 +∈𝒏 )
𝒇′(𝜶 +∈𝒏 )
∈𝒏+𝟏=∈𝒏 −𝒇(𝜶 +∈𝒏 )
𝒇′(𝜶 +∈𝒏 )
9. Roots Of Non-Linear Equation PAGE | 106
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
=∈𝒏 −𝒇(𝜶) +∈𝒏 𝒇′(𝜶) +
∈𝒏𝟐
𝟐! 𝒇′′(𝜶) + ⋯
𝒇′(𝜶) +∈𝒏 𝒇′′(𝜶) + ⋯
=∈𝒏 −∈𝒏 𝒇′(𝜶) +
∈𝒏𝟐
𝟐! 𝒇′′(𝜶) + ⋯
𝒇′(𝜶) +∈𝒏 𝒇′′(𝜶)
(∵ 𝒇(𝜶) = 𝟎)
Neglecting the derivative of order higher then two
∈𝒏+𝟏=∈𝒏 −∈𝒏 𝒇′(𝜶) +
∈𝒏𝟐
𝟐! 𝒇′′(𝜶) + ⋯
𝒇′(𝜶) +∈𝒏 𝒇′′(𝜶) + ⋯
=𝟏
𝟐[
∈𝒏𝟐 𝒇′′(𝜶)
𝒇′(𝜶) +∈𝒏 𝒇′′(𝜶)]
=∈𝒏
𝟐
𝟐[
𝒇"(𝜶)𝒇′(𝜶)
𝟏 +∈𝒏 𝒇"(𝜶)𝒇′(𝜶)
]
=∈𝒏
𝟐
𝟐
𝒇"(𝜶)
𝒇′(𝜶)
The last equation shows that the error at stage is proportional to the squares of the error
in the previous stage. Hence, the NR method has a quadratic convergence and the
convergence is of the order 2
Formula for 𝒒𝒕𝒉 root
Let, = 𝑵𝟏
𝒒 , where 𝑞 = 1, 2, 3, … … 𝑛 and N is an natural number.
⟹ xq − N = 0 ⟹ f(x) = xq − N
⟹ f ′(x) = qxq−1
By the general formula we get,
9. Roots Of Non-Linear Equation PAGE | 107
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
𝐱𝐧+𝟏 =𝟏
𝒒[(𝒒 − 𝟏)𝒙𝒏 +
𝑵
𝒙𝒏𝒒−𝟏
]
Where 𝑛 = 0, 1, 2 … …
Find the iterative formula for √𝐍 and 𝟏
𝐍 by N-R method.
Formula for √𝐍
Let, 𝐱 = √𝐍 ⟹ 𝐱𝟐 − 𝐍 = 𝟎 ⟹ 𝐟(𝐱) = 𝐱𝟐 − 𝐍
⟹ 𝐟′(𝐱) = 𝟐𝐱
By N-R formula,
𝐱𝐧+𝟏 = 𝐱𝐧 −𝐟(𝐱𝐧)
𝐟′(𝐱𝐧)-
= 𝐱𝐧 −𝐱𝐧
𝟐 − 𝐍
𝟐𝐱𝐧=
𝟐𝐱𝐧𝟐 − 𝐱𝐧
𝟐 + 𝐍
𝟐𝐱𝐧
=𝐱𝐧
𝟐 + 𝐍
𝟐𝐱𝐧=
𝟏
𝟐(𝐱𝐧 +
𝐍
𝐱𝐧)
Formula for 𝟏
𝐍
Let, 𝐱 =𝟏
𝐍⟹
𝟏
𝐱− 𝐍 = −𝟎 ⟹ 𝐟(𝐱) =
𝟏
𝐱− 𝐍
⟹ 𝐟′(𝐱) = −𝟏
𝐱𝟐
By N-R formula,
𝐱𝐧+𝟏 = 𝐱𝐧 −𝐟(𝐱𝐧)
𝐟′(𝐱𝐧)= 𝐱𝐧 −
𝟏
𝐱𝐧−𝐍
−𝟏
𝐱𝐧𝟐
= 𝐱𝐧 + (𝟏
𝐱𝐧− 𝐍) 𝐱𝐧
𝟐 = 𝟐𝐱𝐧 − 𝐍𝐱𝐧𝟐
𝐱𝐧+𝟏 = 𝐱𝐧(𝟐 − 𝐍𝐱𝐧) , where 𝑛 = 0, 1, 2, … …
9. Roots Of Non-Linear Equation PAGE | 108
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Exercise-3
C Que.1 Derive the Newton Raphson iterative scheme by drawing appropriate figure.
Nov-10 May-15
C Que.2 Obtain Newton-Raphson formula from Taylor’s theorem. Jun-12
C Que.3
Explain Newton’s method for solving equation f(x) = 0. Apply this method to find the approximate solution of x3 + x − 1 = 0 correct up to three decimal.
[𝟎. 𝟔𝟖𝟐]
Jun-13
T Que.4
Find the real root of the equation 𝑥4 − 𝑥 − 9 = 0 by Newton-Raphson
method, correct to three decimal places.
[𝟏. 𝟖𝟏𝟑]
Nov-16
H Que.5 Use Newton-Raphson method to find smallest positive root of f(x) = x3 − 5x + 1 = 0 correct to four decimals.
[𝟎. 𝟐𝟎𝟏𝟔𝟒]
Dec-15
C Que.6 Find the positive root of x = cos x correct up to three decimal places by N-R method.
[𝟎. 𝟕𝟑𝟗]
Jun-11
T Que.7 Find to four decimal places, the smallest root of the equation sin x = e−x Using the N-R starting with x0 = 0.6.
[𝟎. 𝟓𝟖𝟖𝟓]
Dec-11
H Que.8 Using Newton-Raphson method find a root of the equation xex = 2 Correct to three decimal places.
[𝟎. 𝟓𝟏𝟖]
Dec-15
C Que.9 Find a root of x4 − x3 + 10x + 7 = 0 correct up to three decimal places between a = −2 & b = −1 by N-R method.
[−𝟏. 𝟒𝟓𝟒]
Jun-12
T Que.10 Find a zero of function f(x) = x3 − cos x with starting point x0 = 1 by NR Method could x0 = 0 be used for this problem ?
[𝟎. 𝟖𝟔𝟓𝟓]
Dec-12
T Que.11 Discuss the rate of convergence of NR Method. Jun-12
H Que.12
Explain Newton’s method for solving equation f(x) = 0.
Apply this method to Find an iterative formula to find √N and hence
find √7 Correct up to three decimal points. [𝟐. 𝟔𝟒𝟔]
Dec-13
T Que.13 Set up a Newton iteration for computing the square root x of a given positive number c and apply it to c = 2.
[𝟏. 𝟒𝟏𝟒𝟐]
Nov -10
C Que.14 Derive iterative formula to find √𝑁. Use this formula to find √28.
[𝟐. 𝟐𝟖𝟗𝟒] Nov-16
H Que.15 Derive an iterative formula to find √N hence find approximate value
of √65 and √3, correct up to three decimal places. [𝟖. 𝟎𝟔𝟐, 𝟏. 𝟕𝟑𝟐]
Dec-14
9. Roots Of Non-Linear Equation PAGE | 109
N.S.M. -2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
T Que.16
Derive Newton Raphson’s formula for finding the cube root of
positive no. N, hence find √123
.
[𝟐. 𝟐𝟖𝟗𝟒
May-16
C Que.17
Derive an iterative formula for finding cube root of any positive number using Newton Raphson method and hence find approximate
value of √583
. [𝟑. 𝟖𝟕𝟎𝟖]
May-15
H Que.18 Find the √10 correct to three decimal places by using Newton-Raphson iterative method.
[𝟑. 𝟏𝟔𝟐𝟑]
Dec-15
C Que.19
Find an iterative formula to find 1
N(Where N is positive number) and
hence evaluate 1
3 ,
1
19 ,
1
23.
[𝟎. 𝟑𝟑𝟑, 𝟎. 𝟎𝟓𝟐𝟔, 𝟎. 𝟎𝟒𝟑𝟓]
10. Numerical methods for O.D.E. PAGE | 110
N.S.M. 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Methods for O.D.E.
Picard's Method
Taylor's Method
Euler's Method(R-K 1st Order Method)
Improved Euler's Method
(R-K 2nd order method)(Heun's method)
Runge-Kutta 4th Order Method
Predictor -Corrector Method
SR. NO. TOPIC NAME
1 Picard's Method
2 Taylor 's Method
3 Euler's Method (R-K 1st Order Method)
4 Improved Euler's Method/Modified Euler's Method
5 Runge-Kutta 4th Order Method
6 Predictor -Corrector Method
10. Numerical methods for O.D.E. PAGE | 111
N.S.M. 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Picard’s Method
If 𝐝𝐲
𝐝𝐱= 𝐟(𝐱, 𝐲) 𝐚𝐧𝐝 𝐲(𝐱𝟎) = 𝐲𝟎 ,then
Picard’s formula
𝐲𝐧 = 𝐲𝟎 + ∫ 𝐟(𝐱,𝐱
𝐱𝟎
𝐲𝐧−𝟏)𝐝𝐱 ; 𝐧 = 𝟏, 𝟐, 𝟑, …
Note
We stop the process, When yn = yn−1, up to the desired decimal places. This method is applicable only to a limited class of equations in which successive
integrations can be performed easily.
Exercise-1
C Que 1. Using Picard’s method solve
dy
dx− 1 = xy with initial condition
y(0) = 1, compute y(0.1) correct to three decimal places. [𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟎𝟓]
T Que 2. Solve
dy
dx= 3 + 2xy , Where y(0) = 1, for x = 0.1 by Picard’s
method. [𝐲(𝟎. 𝟏) = 𝟏. 𝟑𝟏𝟐𝟏]
Jun-12
T Que 3.
Obtain Picard’s second approximation solution of the initial
value problem dy
dx= x2 + y2 for x = 0.4 correct places, given
that y(0) = 0. [𝐲(𝟎. 𝟒) = 𝟎. 𝟎𝟐𝟏𝟒]
C Que 4.
Obtain Picard’s second approximate solution of the initial
value problem 𝑑𝑦
𝑑𝑥=
𝑥2
𝑦2+1, 𝑦(0) = 0
[𝐲𝟐 =𝟏
𝟑𝐱𝟑 −
𝐱𝟗
𝟖𝟏+
𝐱𝟏𝟓
𝟏𝟐𝟏𝟓−]
Nov-16
H Que 5.
Using Picard’s method solve dy
dx= x + y2, y(0) = 1.
[𝐲𝟐 = 𝟏 + 𝐱 +𝟑
𝟐𝐱𝟐 +
𝟐
𝟑𝐱𝟑 +
𝐱𝟒
𝟒+
𝐱𝟓
𝟐𝟎]
10. Numerical methods for O.D.E. PAGE | 112
N.S.M. 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Taylor Series
Taylor’s series expansion
If 𝐝𝐲
𝐝𝐱= 𝐟(𝐱, 𝐲) 𝐚𝐧𝐝 𝐲(𝐱𝟎) = 𝐲𝟎 ,then
𝐲(𝐱) = 𝐲(𝐱𝟎) +(𝐱 − 𝐱𝟎)
𝟏! 𝐲′(𝐱𝟎) +
(𝐱 − 𝐱𝟎)𝟐
𝟐! 𝐲′′(𝐱𝟎) + ⋯
Putting x = x1 = x0 + h ⇒ x − x0 = h
∴ 𝐲(𝐱𝟏) = 𝐲(𝐱𝟎 + 𝐡) = 𝐲(𝐱𝟎) +𝐡
𝟏! 𝐲′(𝐱𝟎) +
𝐡𝟐
𝟐! 𝐲′′(𝐱𝟎) + ⋯
So, 𝐲(𝐱𝟏) = 𝐲𝟏 = 𝐲𝟎 +𝐡
𝟏! 𝐲𝟎
′ +𝐡𝟐
𝟐!𝐲𝟎
′′ + ⋯
In general,
𝐲(𝐱𝐧) = 𝐲𝐧 = 𝐲𝐧−𝟏 +𝐡
𝟏! 𝐲𝐧−𝟏
′ +𝐡𝟐
𝟐! 𝐲𝐧−𝟏
′′ +𝐡𝟑
𝟑! 𝐲𝐧−𝟏
′′′ + ⋯ ; 𝐧 = 𝟏, 𝟐, 𝟑 …
𝐖𝐡𝐞𝐫𝐞, 𝐱𝐧 = 𝐱𝟎 + 𝐧𝐡 ; 𝐧 = 𝟏, 𝟐, 𝟑, …
Here h is step size and n is the number of steps.
Exercise-2
C Que 1.
Find by Taylor’s series method the value of y at 𝑥 = 0.1 and 𝑥 =
0.2 to four places of decimal, for 𝑑𝑦
𝑑𝑥= 𝑥2 − 1; 𝑦(0) = 1.
[𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟎𝟎𝟑, 𝐲(𝟎. 𝟐) = 𝟏. 𝟐𝟎𝟐𝟕]
May-16
T Que 2.
Using Taylor series method, find correct four decimal place, the
value of y(0.1), given dy
dx= x2 + y2 and y(0) = 1.
[𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟏𝟏]
Jun-11
C Que 3. Solve y′ = x2 + y2using the Taylor’s series method for the initial condition y(0) = 0.Where 0 ≤ x ≤ 0.4 and h = 0.2.
[𝐲(𝟎. 𝟐) = 𝟎. 𝟎𝟎𝟐𝟕, 𝐲(𝟎. 𝟒) = 𝟎. 𝟎𝟐𝟏𝟒]
H Que 4.
Using Taylor series method, find y(0.1) and y(0.2) correct to four
decimal places, if y(x) satisfies dy
dx= x − y2 , y(0) = 1.
[𝐲(𝟎. 𝟏) = 𝟎. 𝟗𝟏𝟑𝟖, 𝐲(𝟎. 𝟐) = 𝟎. 𝟖𝟓𝟏𝟓]
May-16
H Que 5.
Evaluate y(0.1) correct to four decimal places using Taylor’s
series method if dy
dx= y2 + x , y(0) = 1.
[𝐲(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟔]
May-15
10. Numerical methods for O.D.E. PAGE | 113
N.S.M. 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
T Que 6.
Using Taylor’s series method ,find y(1.1) correct to four decimal
places, given that dy
dx= xy
1
3, y(1) = 1.
[𝐲(𝟏. 𝟏) = 𝟏. 𝟏𝟎𝟔𝟖]
Dec-15
C Que 7.
Using Taylor’s series method, solve 𝑑𝑦
𝑑𝑥= 𝑥2 − 𝑦, 𝑦(0) = 1 at 𝑥 =
0.1, 0.2 𝑎𝑛𝑑 0.3. Also compare the values with exact solution. 𝐲(𝟎. 𝟏) 𝐲(𝟎. 𝟐) 𝐲(𝟎. 𝟑) Taylor’s series
1.1055 0.8242 0.7595
Exact solution
0.9052 0.8213 0.7492
Nov-16
Euler’s Method (RK 1st order method)
If 𝐝𝐲
𝐝𝐱= 𝐟(𝐱, 𝐲) 𝐚𝐧𝐝 𝐲(𝐱𝟎) = 𝐲𝟎 ,then
𝐲𝐧+𝟏 = 𝐲𝐧 + 𝐡 [𝐟(𝐱𝐧, 𝐲𝐧)] ; 𝐧 = 𝟎, 𝟏, 𝟐, …
𝐖𝐡𝐞𝐫𝐞, 𝐱𝐧 = 𝐱𝟎 + 𝐧𝐡 ; 𝐧 = 𝟏, 𝟐, 𝟑, …
here h is step size and n is the number of steps.
Explanation:
Let [a, b] be the interval over which we want to find the solution of
𝐝𝐲
𝐝𝐱= 𝐟(𝐱, 𝐲) ; 𝐚 < 𝐱 < 𝐛 ; 𝐲(𝐱𝟎) = 𝐲𝟎 … (𝟏)
A set of points {(xn, yn)} is generated which are used for an approximation [i. e. y(xn) = yn]
For convenience, we divide [a, b] into n equal subintervals.
⟹ 𝐱𝐧 = 𝐱𝟎 + 𝐧𝐡 ; 𝐧 = 𝟏, 𝟐, … ; 𝐰𝐡𝐞𝐫𝐞, 𝐡 =𝐛−𝐚
𝐧 𝒐𝒓 𝐡 =
𝐱𝐧−𝐱𝟎
𝐧 is called the step size.
Now, y(x) is expand by using Taylor’s series about x = x0 as following
𝐲(𝐱) = 𝐲(𝐱𝟎) +(𝐱 − 𝐱𝟎)
𝟏 !𝐲′(𝐱𝟎) +
(𝐱 − 𝐱𝟎)𝟐
𝟐 !𝐲′′(𝐱𝟎) +
(𝐱 − 𝐱𝟎)𝟑
𝟑 !𝐲′′′(𝐱𝟎) + ⋯ … (𝟐)
𝐖𝐞 𝐡𝐚𝐯𝐞, [𝐝𝐲
𝐝𝐱]
𝐱=𝐱𝟎
= 𝐲′(𝐱𝟎) = 𝐟(𝐱𝟎, 𝐲𝟎) = 𝐟(𝐱𝟎, 𝐲(𝐱𝟎))
By Eq. (2),
𝐲(𝐱𝟏) = 𝐲(𝐱𝟎) +(𝐱𝟏 − 𝐱𝟎)
𝟏 !𝐲′(𝐱𝟎) +
(𝐱𝟏 − 𝐱𝟎)𝟐
𝟐 !𝐲′′(𝐱𝟎) +
(𝐱𝟏 − 𝐱𝟎)𝟑
𝟑 !𝐲′′′(𝐱𝟎) + ⋯
10. Numerical methods for O.D.E. PAGE | 114
N.S.M. 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
Taking, h = x1 − x0
𝐲(𝐱𝟏) = 𝐲(𝐱𝟎) +𝐡
𝟏 !𝐟(𝐱𝟎, 𝐲(𝐱𝟎)) +
𝒉𝟐
𝟐 !𝐲′′(𝐱𝟎) +
𝒉𝟑
𝟑 !𝐲′′′(𝐱𝟎) + ⋯
If the step size is chosen too small enough, then we may neglect the second order term
involving h2 and get
𝐲𝟏 = 𝐲(𝐱𝟏) = 𝐲(𝐱𝟎) +𝐡
𝟏 !𝐟(𝐱𝟎, 𝐲(𝐱𝟎)) = 𝐲𝟎 + 𝐡𝐟(𝐱𝟎, 𝐲𝟎)
Which is called Euler’s Approximation.
The process is repeated and generates a sequence of points that approximate the solution of
curve y = y(x).
The general step for Euler’s method is 𝐲𝐧+𝟏 = 𝐲𝐧 + 𝐡 [𝐟(𝐱𝐧, 𝐲𝐧)] ; 𝐧 = 𝟎, 𝟏, 𝟐, …
𝐖𝐡𝐞𝐫𝐞, 𝐱𝐧 = 𝐱𝟎 + 𝐧𝐡 ; 𝐧 = 𝟏, 𝟐, 𝟑, …
Here h is step size and n is the number of steps.
Exercise-3
C Que 1. Describe Euler’s Method for first order ordinary differential equation.
Dec-10 Jun-12
C Que 2.
Apply Euler’s method to find the approximate solution of dy
dx= x + y with y(0) = 0 and h = 2. Show your calculation up
to five iteration. [𝐲𝟓 = 𝟐𝟑𝟐]
Jun-13
H Que 3. Using Euler’s method solve for y at 𝑥 = 0.1 from
𝑑𝑦
𝑑𝑥= 𝑥 + 𝑦 +
𝑥𝑦. 𝑦(0) = 1, In five steps. 𝒚(𝟎. 𝟏) = 𝟏. 𝟏𝟏𝟐𝟒
May-16
C Que 4.
Derive Euler’s formula for initial value problem dy
dx=
f(x, y); y(x0) = y0. Hence , use it find the value of y for
dy
dx= x + y; y(0) = 1 when x = 0.1, 0.2 with step size h = 0.05.
Also Compare with analytic solution. 𝐲(𝟎. 𝟏) 𝐲(𝟎. 𝟐) Euler’s 𝟏. 𝟏𝟎𝟓𝟎 𝟏. 𝟐𝟑𝟏𝟏 Exact solution 𝟏. 𝟏𝟏𝟎𝟑 𝟏. 𝟐𝟒𝟐𝟖
May-15
10. Numerical methods for O.D.E. PAGE | 115
N.S.M. 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY
T Que 5.
Apply Euler’s method to solve the initial value problem dy
dx= x + y , with y(0) = 0 with choosing h = 0.2 and compute
y1, y2, y3 … … y5 Compare your result with the exact solution. 𝐲(𝟎. 𝟐) 𝐲(𝟎. 𝟒) 𝐲(𝟎. 𝟔) 𝐲(𝟎. 𝟖) 𝐲(𝟏) Euler’s 𝟎 𝟎. 𝟎𝟒 𝟎. 𝟏𝟐𝟖 𝟎. 𝟐𝟕𝟑𝟔 𝟎. 𝟒𝟖𝟖𝟑 Exact solution
𝟎. 𝟎𝟐𝟏𝟒 𝟎. 𝟎𝟗𝟏𝟖 𝟎. 𝟐𝟐𝟐𝟏 𝟎. 𝟒𝟐𝟓𝟓 𝟎. 𝟕𝟏𝟖𝟑
Nov-16
H Que 6.
Using Euler’s method ,find an approximate value of y
corresponding to x = 1 given that dy
dx= x + y and y = 1 when
x = 0. [𝐅𝐨𝐫 𝐡 = 𝟎. 𝟐𝟓, 𝐲𝟒 = 𝟐. 𝟖𝟖𝟐𝟗]
Jun -13
T Que 7.
Use Euler method to find y(0.2) given that dy
dx= y −
2x
y ,
y(0) = 1. (Take h = 0.1) [𝐲(𝟎. 𝟐) = 𝟏. 𝟏𝟗𝟏𝟖]
Jun-11
H Que 8.
Explain Euler’s method for solving first order ordinary differential equation. Hence use this method, find y(2) for dy
dx= x + 2y with y(1) = 1.
[𝐲(𝟐) = 𝟓. 𝟕𝟓]
Dec-15
T Que 9. Solve initial value problem
dy
dx= x√y , y(1) = 1 and hence find
y(1.5) by taking h = 0.1 using Euler’s method. [𝐲(𝟏. 𝟓) = 𝟏. 𝟔𝟕𝟎𝟎]
May-15
T Que 10.
Using Euler’s method, find an approximate value of
corresponding to 𝑥 = 0.3, given that 𝑑𝑦
𝑑𝑥=
𝑦−𝑥
𝑦+𝑥; 𝑦(0) = 1. take
ℎ = 0.1. [𝒚(𝟏) = 𝟏. 𝟐𝟓𝟒𝟒]
Nov-16
H Que 11.
Use Euler’s method to find an approximation value of y at
x = 0.1 for the initial value problem dy
dx= x − y2; y(0) = 1.
[𝐲(𝟎. 𝟏) = 𝟎. 𝟗𝟏𝟑𝟑]
Dec-15
T Que 12.
Using Euler’s method compute 𝑦(0.3) for the initial value problem 𝑦′ = 𝑦2 − 𝑥2, 𝑦(0) = 1 taking the step size ℎ = 0.1.
[𝐲(𝟎. 𝟑) = 𝟏. 𝟑𝟔𝟒𝟖]
June-16
10. Numerical methods for O.D.E. PAGE | 116
N.S.M. 2140606 DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY