Category: CBSE

CBSE Class 12 Physical Education Notes Chapter 1 Planning in Sports is part of Physical Education Class 12 Notes for Quick Revision. Here we have given NCERT Physical Education Class 12 Notes Chapter 1 Planning in Sports.

Physical Education Class 12 Notes Chapter 1 Planning in Sports

Meaning and Objectives of Planning
Planning is a process of setting objectives and deciding how to accomplish them. Planning in sports is required because there is a definite time frame for achieving the objectives and the resiilts are also known very clearly and in definite terms.
Objectives of planning are given below.

1. Timely achievements of targets.
2. Improving the percentage of success.
3. Finding out the causes of failure.
4. Deciding the appropriate time duration of a programme.
5. Distribution of work according to the ability of the people involved.
6. Remaining within the budget while achieving the required targets.

Various Committees and their Responsibilities

1. Committee for Publicity Its main duty is to advertise the sports events.
2. Transport Committee Its main responsibility is to make necessary arrangements for transportation.
3. Grounds and Equipment Committee This committee is responsible for making the grounds or laying out the track and field. , .
4. Refreshments and Entertainment Committee This committee takes the charge of supplying refreshments and drinks to the guests, officials, competitors etc.
5. Reception Committee The members of this committee are responsible to welcome the chief guests at the opening and closing ceremonies. .
6. Committee on Entries and Programmes This committee sends entry forms to the various institutions early. It also arranges seats for guests and spectators. It sometimes also prepares fixtures of teams participating in the competition.
7. Committee for Officials This committee selects various officials such as referees, judges, umpires etc.
8. Announcement Committee This committee is responsible for making all the announcements during the period of sports events,
9. First Aid Committee This committee provides first aid to the victim or affected sportsman/athlete.

For successful organisation of the sports meet, committees are formed under three heads These are Pre-meet committees like publicity, ground and equipment, reception committee During meet committees like refreshment, transport committee and Post-meet committees like award committee.

Tournaments
A tournament is a competition held among various teams in a particular activity according to a fixed schedule where a winner is decided.
There are various methods for organising a tournament which depends upon various factors such as

• The number of participating teams Facilities of ground, equipment and number of officials available
• Days in which the tournament is to be completed

Importance of Tournaments

1. Development of Sports Skills Various skills of sports are developed in tournaments.
2. Helpful in Selection of Players Good players can be selected easily by observing their performance in the tournament.
3. Development of National and International Integration It not only develops the integration of national and international level but also gives sense of unity and brotherhood.
4. Development of Social Qualities Social traits such as respect, brotherhood, discipline, sympathy, tolerance etc are developed among participants through tournaments.

Types of Tournament

1. Knock-out or Elimination tournament
2. League or Round Robin tournament
3.  Combination tournament
4. Challenge tournament

1. Knock-out or Elimination Tournaments
A single elimination tournament (also called an Olympic system tournament) or a knock-out tournament is a type of elimination tournament where the loser of each bracket is immediately eliminated from the tournament. In such tournaments, if the number of competitors or competing teams are not a multiple of 2 (i.e. 4, 8, 16,..) then some teams may be given a ‘bye’ and some teams are given ‘seeding’.
Seeding is a procedure by which good teams are placed in fixtures in such a way that stronger teams do not meet each other at the very start of a tournament.

Note We will cover in detail the first three tournaments because these three tournaments are included in syllabus.
Bye is a privilege given to a team which is decided generally by seeding it or by draw of lots.
Advantages of Knock-out Tournaments

1. Minimum number of officials are required in organising tournaments.
2. Owing to less number of matches so it takes less time to complete the tournaments.
3. It helps in enhancing the standard of sports.

Disadvantages of Knock-out Tournaments

1. Good teams can get eliminated because of their poor performance even in first round.
2. There are maximum chances of weak teams to enter into the final round.

2. League or Round Robin Tournaments
A league tournament, also called as round robin tournament, is a type of tournament in which each contestant meets all other contestants in turn. In this type of tournament, there has to be sufficient time to complete the tournament.
There are two types of league tournaments

1. Single league
2. Double league

In a single league round robin schedule, each participant plays with every other participant once. If each participant plays with others twicely, this is frequently called a round robin double league tournament.

1. Single League Number of matches \(=\frac { N(N-1) }{ 2 }\) (N = Number of teams)
2. Double League Number of matches = N (N-1) (N = Number of teams)

Fixture
Any tournament whether it is knock-out or league, is arranged according to a set procedure which is known as fixture. For knock-out tournament, the procedure to draw fixture is through bye and seeding. For league tournaments, the procedure to draw fixture is through staircase and cyclic method.

Fixture is the process of arranging the teams in systematic order in various groups for competition in a physical activity. In other words, it is the set-up of various teams for competitive matches where they play in a systematic order as per the fixture schedule. The success of a tournament depends upon planning of suitable fixture. Tournaments are played in various forms of fixtures.

Fixtures in Knock-out Tournament
In a knock-out tournament, the fixtures are drawn by the following methods
1. Seeding Method
Seeding is a procedure by which good teams are placed in fixtures in such a way that stronger teams do not meet each other at the very start of a tournament. It is done to overcoine the drawbacks of a single knock-out tournament. Seeding can be done only if the standards of the teams are known before the start of the tournament. The draw may result between the strong competitors at early level competition, so they are ‘seeded’ to prefvent this.

Thus, the top competitors will not meet until the quarter final, semi final or final round. Seeding is used to ensure that-players or teams of recognised outstanding abilltytlo not confront each other in the early rounds. It represents the tournament committee’s subjective rating of the various players, and chances of winning the tournament.

Special Seeding
It is a method of seeding in which the players or teams directly participate in the quarter final or semi-final matches thus avoiding their participation in the initial rounds.

2. Bye Method
Bye means the avoiding of playing a match in the first round of the tournament. It is a privilege given to a team which is decided generally by seeding it or by draw of lots. Some teams may get bye in first round, by which they get promoted to higher round competition.

This may be given as a reward for their some previous achievements. Byes can be applied equally to single person competitions and team sports as well as to single game eliminations and best of series elimination. Byes are given in 1st round only. The number of byes that should be given in a tournament is decided by finding the difference between the number of teams and the next power of two.
Example : (N – 1) if 20 teams are participating, then 19 matches (20 – 1) =19 will be played.

Method of Drawing Fixtures in Knock-out Tournament
In order to draw fixtures in a knock-out tournament, following things are essential to calculate

1. Total Number of Matches It means the total number of matches that will be played during the entire tournament. The number of matches to be played is calculated by subtracting 1 from total number of teams (N – 1).
Suppose total teams are 11 so matches to be played during the tournament are 10 (11 – 1).

2. Number of Rounds It means the total number of rounds that should be played during the tournament. Rounds will include the initial rounds, quarter finals, semi finals and finals. If the number of teams is a power of 2, then number of rounds will exactly be multiple of 2 up to that number.
Example :

If number of teams is not a power of 2, then number of rounds will be equal to the multiple of next power of two.
Example :

The next power of 2 after 21 is 32 which is 25.

3. Number of teams in each half All the teams are divided into two halves for the sake of convenience. If the number of teams is even, then the number of teams in upper and lower half will be equal i.e Total teams 20 then apply the formulae \(\frac { N }{ 2 }\)
Number of teams in upper half
\(\frac { N }{ 2 } =\frac { 20 }{ 2 }=10\)
Number of teams in lower half
\(\frac { N }{ 2 } =\frac { 20 }{ 2 } =10\)
If the number of tearms is odd ,then apply\(\frac { N+1 }{ 2 }\) for upper half and \(\frac { N-1 }{ 2 }\)
Example : Total teams = 15
Number of teams in upper half
\(\frac { N+1 }{ 2 } =\frac { 15+1 }{ 2 } =8\)
Number of teams in lower half
\(\frac { N-1 }{ 2 } =\frac { 15-1 }{ 2 } =8\)

4. Number of byes Teams getting the byes do not play in the initial round. They are given by subtracting the number of teams by the next power of 2.
Example 1 : Total number of teams =19
Next power of two after 19 = 32
Power of two – 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32
therefore number of byes will be 32 – 19 =13

Example 2 : Total number of teams =12
Next power of two after 12 = 16
therefore number of byes 16 – 12 = 4
Hence, 13 teams in example 1 and 4 teams in example 2 will not play the match in 1st round.

5. Number of byes in upper and lower half
For calculating how many teams in upper and lower half will get byes, the method is as follows
Number of byes in upper half = \(\frac { Nb-1 }{ 2 }\)
Number of byes in lower half= \(\frac { Nb+1 }{ 2 }\)
Total number of byes = 13
Number of byes in upper half= \(\frac { 13-1 }{ 2 }=6\)
Number of byes in lower half= \(\frac { 13+1 }{ 2 }=7\)

6. Method of fixing byes The number of the teams are placed in upper and lower half according to the draw of lots or using other methods. Then following procedure is adopted for fixing the byes.
The last team of lower half gets first bye.
The first team of upper half gets second bye.
The first team of lower half gets third bye.
The last team of upper half gets fourth bye.
The second last team of lower half gets fifth bye.
Like this the order continues.

7. Method of Seeding It is done to prevent the strong teams from competing with each other in the initial round. Therefore the strong teams are placed in the position where they will get bye i.e. placing the strong teams on the last position of lower half or first position of upper half.
Example : Total no. of teams = 6
Next power of two after 6 = 8
No. of byes 8 – 6 =2
Seeding = Last of lower half and first of upper half.
Number of matches 6 – 1 = 5

Examples
Example 1 A fixture of 19 teams on knock-out basis.

Total matches = N – 1 = 19 – 1 = 18
Number of teams in upper half
\(\frac { N+1 }{ 2 }=\frac { 19+1 }{ 2 }=\frac { 20 }{ 2 }=10\) teams
Number of teams in lower half
\(\frac { N-1 }{ 2 } =\frac { 19-1 }{ 2 } =\frac { 18 }{ 2 }=9\) teams
Total number of byes
= Next power of two – Total number of teams = 32 – 19=13 byes
Number of byes in upper half
\(\frac { Totalnumberofbyes-1 }{ 2 } \)
i.e\(\left[ \frac { Nb-1 }{ 2 } \right] =\frac { 13-1 }{ 2 } =\frac { 12 }{ 2 } =6byes\)
Number of byes in lower half
\(=\frac { Totalnumberofbyes+1 }{ 2 } \)
\(\left[ \frac { Nb+1 }{ 2 } \right] =\frac { 13+1 }{ 2 } =\frac { 14 }{ 2 } =7byes\)

Example 2 A fixture of 12 teams where 4 teams getting special seeding.
Total number of teams =12
Number of teams m upper half = \(\frac { N }{ 2 } =\frac { 12 }{ 2 } =6\)
Number of teams in lower half = \(\frac { N }{ 2 }\)= 6
Teams setting special seeding = 4
Number of matches = 12 – 1 = 11

Fixture

Team number 1,2, 11 and 12 do not play the match initially as they get special seeding.

• Winner of each bracket enters the next round while losers are eliminated. For example, Team 3 and 4 play match number 1. If 3 wins, then 4 is eliminated from the tournament.
• If the number of teams/ players is odd, then number of teams in upper half and lower half each is \(\frac { N+1 }{ 2 }\)
• Teams getting special seeding do not play initial rounds just like byes

Fixtures in League Tournaments
The three methods used for drawing up fixtures are

1. Staircase method
2. Cyclic method
3. Tabular method

Note We will cover in detail the first two methods because these two methods are included in syllabus.

1. Staircase Method
In this method, the fixtures are made like a staircase. It is arranged in sequential form, there is no bye, no problem of odd or even and therefore it is easiest to arrange. An example will illustrate this method.
Example : Draw up a fixture of 9 teams on Round Robin basis using staircase method.
The number of teams = 9
Thus, number of matches
\(\frac { N\left( N-1 \right) }{ 2 } =\frac { 9\left( 9-1 \right) }{ 2 } =\frac { 9\times 8 }{ 2 } =36matches\)
The fixture is given below

2. Cyclic Method
In this method, if the number of teams is even (i.e. 4, 6, 8, ….), team 1 is fixed on the top right side and the other teams move in clockwise direction down one side of a rectangle and up on the other side. In such a case, the number of rounds will be N-1 if there are N teams participating. However, if the number of teams is odd (i.e. 3, 5, 7,….), a ‘bye’ is fixed at the top and all teams follow it in sequence. In such a case, the number of rounds will be N.

Example 1 Fixture of 6 teams on cyclic method.
Total number of teams = 6
Total number of matches = \(\frac { N\left( N-1 \right) }{ 2 } =\frac { 6\left( 6-1 \right) }{ 2 } =15\)
Total number of rounds = N – 1 = 5

Example 2 Fixture of 5 teams on cyclic method.
Total number of teams = 5
Total number of matches = \(\frac { N\left( N-1 \right) }{ 2 } =\frac { 5\left( 5-1 \right) }{ 2 } =10\)
Total number of rounds = 5

Deciding the Winner
The method to decide the winner in the tournaments is by calculating the percentage of matches won.
\(percentage=\frac { MatchesWon }{ MatchesPlayed } \times 100\)

1.5 Intramurals
The word intramural is derived from the Latin words intra and muros, which means ‘within walls’. The teams compete within the walls of the institution e.g. inter-class or inter-house tournament within an institution.
The objectives of intramurals are as follows

• To provide incentive, motivation and opportunity to learn various skills.
• To develop sportsman spirit among students.
• To help in recreation of the students.
• To improve-social interaction among the students.

Intramural competitions are conducted by the intramural committee of the institution.
Significance of intramurals are as follows

• It improves the sense of citizenship and adult living. ,
• It helps in selection of teams for participation in extramural tournaments.
• It brings out the hidden talent of the students.
• It develops leadership qualities in students.
• It creates recreational atmosphere among the teachers and students.

1.6 Extramurals
This encloses the activities which are performed outside the walls of the institution or school. It means that students of two or more schools participate in such competitions.
The objectives of extramurals are

• To improve the standard of sports.
• To provide experience to students.
• To develop sportsmanship and fraternity.
• To broaden the base of sports.
• To provide the knowledge of new rules and advanced techniques.

Significance of extramurals are as follows

• They provide recreational opportunities for the campus community. ..
• They give positive experiences, regardless of race, nationality, age and gender.
• They provide an opportunity to develop motor skills which will be beneficial throughout life.
• They develop important values and skills such as self-esteem, social skills, sportsmanship and self-motivation.
• They give an opportunityfor relaxation from strenuous school work and the rapid pace of modern living.

1.7 Specific Sports Programmes
Specific sports programmes are those which are not usually related to competitions. The programmes have various objectives like creating awareness, creating a sense of integration among people and raising funds for charitable organisations. Some of these programmes are Sports day. Run for fun, Health run, Run for specific cause. Run for unity and Run for awareness.

We hope the given CBSE Class 12 Physical Education Notes Chapter 1 Planning in Sports will help you. If you have any query regarding NCERT Physical Education Class 12 Notes Chapter 1 Planning in Sports, drop a comment below and we will get back to you at the earliest.

Statistics for Economics Class 11 NCERT Solutions Chapter 5 Measures of Central Tendency

Question 1.
Which average would be suitable in the following cases?
(i) Average size of readymade garments.
(ii) Average intelligence of students in a class.
(iii) Average production in a factory per shift.
(iv) Average wages in an industrial concern.
(v) When the sum of absolute deviations from average is least.
(vi) When quantities of the variable are in ratios.
(vii) In case of open-ended frequency distribution.
Answer:
(i) Mode Average size of any ready made garments should be the size for which demand is the maximum. Hence, the modal value which represents the value with the highest frequency should be taken as the average size to be produced.
(ii) Median It is the value that divides the series into two equal parts. Therefore, Median will be the best measure for calculating the average intelligence of students in a class as it will give the average intelligence such that there are equal number of students above and below this average. It will not be affected by extreme values.
(iii) Arithmetic Mean The average production in a factory per shift is best calculated by Arithmetic Mean as it will capture all types of fluctuations in production during the shifts.
(iv) Arithmetic Mean Arithmetic Mean will be the most suitable measure. It is calculated by dividing the sum of wages of all the workers by the total number of workers in the industrial concern. It gives a fair idea of average wage bill taking into account all the workers.
(v) Arithmetic Mean The algebraic sum of the deviations of values about Arithmetic Mean is zero. Hence, when the sum of absolute deviations from average is the least, then mean could be used to calculate the average.
(vi) Median Median will be the most suitable measure in case the variables are in ratios as it is least affected by the extreme values.
(vii) Median Median is the most suitable measure as it can be easily computed even in case of open ended frequency distribution and will not get affected by extreme values.

Question 2.
Indicate the most appropriate alternative from the multiple choices provided against each question.
(i) The most suitable average for qualitative measurement is
(a) Arithmetic mean
(b) Median
(c) Mode
(d) Geometric mean
(e) None of these
Answer:
(b) Median is the most suitable average for qualitative measurement because Median divides a series in two equal parts thus representing the average qualitative measure without being affected by extreme values.

(ii) Which average is affected most by the presence of extreme items?
(a) Median
(b) Mode
(c) Arithmetic Mean
(d) Geometric Mean
(e) Harmonic Mean
Answer:
(c) It is defined as the sum of the values of all observations divided by the number of observations and therefore it is. affected the most by extreme values.

(iii) The algebraic sum of deviation of a set of n values from AM is
(a) n
(b) 0
(c) 1
(d) None of these
Answer:
(b) This is one of the mathematical properties of arithmetic mean that the algebraic sum of deviation of a set of n values from AM is zero.

Question 3.
Comment whether the following statements are true or false.
(i) The sum of deviation of items from median is zero.
(ii) An average alone is not enough to compare series.
(iii) Arithmetic mean is a positional value.
(iv) Upper quartile is the lowest value of top 25% of items.
(v) Median is unduly affected by extreme observations.
Answer:
(i) False
This mathematical property applies to the arithmetic mean and not to median.
(ii) True
Average is not enough to compare the series as it does not explain the extent of deviation of different items from the central tendency and the difference in the frequency of values. These are measured by measures of dispersion and kurtosis.
(iii) False
Median is a positional value.
(iv) True
The upper quartile also called the third quartile, has 75 % of the items below it and 25 % of items above it.
(v) False
Arithmetic mean is unduly affected by extreme observations.

Question 4.
If the arithmetic mean of the data given below is 28, find (a) the missing frequency and (b) the median of the series

Answer:
(a) Let the missing frequency br f₁.
Arithmetic Mean = 28

or 2240 -2100 = 35f₁ = 28f₁
or 140 = 7f₁
f₁ = 20
Hence, the missing frequency is 20.
(b)

So, the Median class = Size of \(\left(\frac{N}{2}\right)\)th item = 50th term.
50th item lies in the 57th cumulative frequency and the corresponding class interval is 20-30.

Question 5.
The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

Answer:

N = 10
\(\overline{X}=\frac{\Sigma X}{N}=\frac{2400}{10}=240\)
Arithmetic Mean = ₹ 240

Question 6.
Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.

Answer:

Question 7.
The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

Answer:

So, the median class = Size of \(\left(\frac{N}{2}\right)\) th item = 190 item
190th lies in the 129 th cumulative frequency and the corresponding class interval is 200-300.

Median size of land holdings = 241.22 acres

Question 8.
The following series relates to the daily income of workers employed in a firm. Compute (a) highest income of lowest 50% workers, (b) minimum income earned by the top 25% workers and (c) maximum income earned by lowest 25% workers.

Answer:

(a) Highest income of lowest 50% workers will be given by the median. Σf = N = 65
Median class = Size of \(\left(\frac{N}{2}\right)\)th item = Size of \(\left(\frac{65}{2}\right)\)th item=325 th item
32.5th item lies in the 50th cumulative frequency and the corresponding class interval is 24.5 – 29.5.

(b) Minimum income earned by top 25% workers will be given by the lower quartile Q₁.
Class interval of Q₁ = \(\left(\frac{N}{4}\right)\)th item
= \(\left(\frac{65}{4}\right)\)th item = 1625th item
16.25th item lies in the 30th cumulative frequency and the corresponding class interval is 19.5 – 24.5

(c) Maximum income earned by lowest 25% workers will be given by the upper quartile Q₃.
Class interval of Q₃ = \(\left(\frac{N}{4}\right)\)th item
= 3\(\left(\frac{65}{4}\right)\)th item
= 3 × 1625th item
= 48.75th item
48.75th item lines in 50th item and the corresponding class interval is 24.5-29.5.

Question 9.
The following table gives production yield in kg per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode production yield.

Answer:
(i) Mean

(ii) Median

(iii) Mode
Grouping Table

Analysis Table

Statistics for Economics Class 11 NCERT Solutions

NCERT Solutions for Class 11 Economics

Statistics for Economics Class 11 NCERT Solutions Chapter 4 Presentation of Data

Answer: the following questions, 1 to 10, choosing the correct answer.
Question 1.
Bar diagram is a
(a) one-dimensional diagram
(b) two-dimensional diagram
(c) diagram with no dimension
(d) None of these
Answer:
(a) Bar diagrams are one-dimensional diagrams. Though these are represented on a plane of two axis in form of rectangular bars, the width is of no consequence and only the length depicts the frequency.

Question 2.
Data represented through a histogram can help in finding graphically the
(a) mean
(b) mode
(c) median
(d) All of these
Answer:
(b) Histogram gives value of mode of the frequency distribution graphically through the highest rectangle.

Question 3.
Ogives can be helpful in locating graphically the
(a) mode
(b) mean
(c) median
(d) None of these
Answer:
(c) Intersection point of the less than and more than ogives gives the median.

Question 4.
Data represented through arithmetic line graph help in understanding
(a) long term trend
(b) cyclicity in data
(c) seasonality in data
(d) All of the above
Answer:
(a) Arithmetic line graph helps in understanding the trend, periodicity, etc in a long term time series data.

Question 5.
Width of bars in a bar diagram need not be equal. (True/False)
Answer:
False
Bar diagram comprises a group of equispaced and equiwidth rectangular bars for each class or category of data.

Question 6.
Width of rectangles in a histogram should essentially be equal. (True/False)
Answer:
False
If the class intervals are of equal width, the area of the rectangles are proportional to their respective frequencies and width of rectangles will be equal. However, sometimes it is convenient or necessary to use varying width of class intervals and hence unequal width of rectangles.

Question 7.
Histogram can only be formed with continuous classification of data. (True/False)
Answer:
True
A histogram is never drawn for a discrete variable/data. If the classes are not continuous they are first converted into continuous classes.

Question 8.
Histogram and column diagram are the same method of presentation of data. (True/False)
Answer:
False
Histogram is a two dimensional diagram drawn for continuous data and the rectangles do not have spaces in between while column diagram is one dimensional with space in between every column (bar).

Question 9.
Mode of a frequency distribution can be known graphically with the help of histogram. (True/False)
Answer:
True
Histogram gives value of mode of the frequency distribution graphically through the highest rectangle.

Question 10.
Median of a frequency distribution cannot be known from the ogives. (True/False)
Answer:
False
Intersection-point of the less than and more than ogives gives the median.

Question 11.
What kind of diagrams are more effective in representing the following?
(a) Monthly rainfall in a year
(b) Composition of the population of Delhi by religion
(c) Components of cost in a factory
Answer:
(a) The monthly rainfall in a year can be best represented by a bar diagram as only one variable i.e., monthly rainfall is to be presented diagrammatically. The rainfall is plotted on Y-axis in the corresponding month that is plotted on the X-axis.
(b) Composition of the population of Delhi by religion can be represented by a component bar diagram. A component bar diagram shows the bar and its sub-divisions into two or more components. Thus, the total population can be sub divided in terms of religion and presented through a component bar diagram.
(c) Different components of cost in a factory can most effectively be depicted through a pie chart. The circle represents the total cost and various components of costs are shown by different portions of the circle drawn according to percentage of total cost each component covers.

Question 12.
Suppose you want to emphasise the increase in the share of urban non-workers and lower level of urbanisation in India as shown in Example 4.2. How would you do it in the tabular form?
Answer:
Share of urban workers and non workers in India

	Location
Sex	Worker in urban (in crore)	Non-worker in urban (in crore)	Total
Male	50	70	120
Femal	25	50	75
Total	75	120	195

Question 13.
How does the procedure of drawing a histogram differ when class intervals are unequal in comparison to equal class intervals in a frequency table?
Answer:
A histogram is a set of rectangles with bases as the intervals between class boundaries (along X-axis) and with areas proportional to the class frequency. If the class intervals are of equal width, the area of the rectangles are proportional to their respective frequencies.

However, sometimes it is convenient or at times necessary, to use varying width of class intervals. For graphical representation of such data, height for area of a rectangle is the quotient of height i.e., frequency and base i.e., width of the class interval. When intervals are equal, all rectangles have the same base and area can conveniently be represented by the frequency of the interval.

But, when bases vary in their width, the heights of rectangles are to be adjusted to yield comparable measurements by dividing class frequency by width of the class interval instead of absolute frequency. This gives us the frequency density for the purpose of comparison.
Thus \(\frac{\text { Frequency density }}{(\text { Height of rectangle })}=\frac{\text { Class Frequency }}{\text { Width of the class interval }}\)

Question 14.
The Indian Sugar Mills Association reported that, ‘sugar production during the first fortnight of December, 2001 was about 3,87,000 tonnes, as against 3,78,000 tonnes during the same fortnight last year (2000). The off-take of sugar from factories during the first fortnight of December, 2001 was 2,83,000 tonnes for internal consumption and 41,000 tonnes for exports as against 1,54,000 tonnes for internal consumption and nil for exports during the same fortnight last season.’
(i) Present the data in tabular form.
(ii) Suppose you were to present these data in diagrammatic form which of the diagrams would you use and why?
(iii) Present these data diagrammatically.
Answer:
(i) Data in tabular form.
Sugar Production in India

	Total Production (tonnes)	Off-take for Internal Consumption (tonnes)	Off-take for Exports (tonnes)
December 2000	378000	154000	—
December 2001	387000	283000	41000

(ii) The data can effectively be presented diagrammatically using the multiple bar diagram. This is because multiple bar diagrams are used for comparing two or more sets of data for different years or classes, etc.

Question 15.
The following table shows the estimated sectoral real growth rates (percentage change over the previous year) in GDP at factor cost.

Represent the data as multiple time-series graphs.
Answer:

Statistics for Economics Class 11 NCERT Solutions

NCERT Solutions for Class 11 Economics

Statistics for Economics Class 11 NCERT Solutions Chapter 3 Organisation of Data

Question 1.
Which of the following alternatives is true?
(i) The class mid-point is equal to
(a) the average of the upper class limit and the lower class limit
(b) the product of upper class limit and the lower class limit
(c) the ratio of the upper class limit and the lower class limit
(d) None of the above
Answer:
(a) The class mid-point is the middle value of a class. It lies halfway between the lower class limit and the upper class limit of a class and is calculated as
Class Mid-Point or Class Mark = \(\frac{\text { (Upper Class Limit + Lower Class Limit) }}{2}\)

(ii) The frequency distribution of two variables is known as
(a) Univariate Distribution
(b) Bivariate Distribution
(c) Multivariate Distribution
(d) None of the above
Answer:
(b) Bi means two and hence the frequency distribution of two variables is known as Bivariate Distribution.

(iii) Statistical calculation in classified data are based on
(a) the actual values of observations
(b) the upper class limits
(c) the lower class limits
(d) the class mid-points
Answer:
(d) The class mid-points of each class is used to represent the class and therefore, it is used in further calculations after the raw data are grouped into classes.

(iv) Under exclusive method,
(a) the upper class limit of a class is excluded in the class interval
(b) the upper class limit of a class is included in the class interval
(c) the lower class limit of a class is excluded in the class interval
(d) the lower class limit of a class is included in the class interval
Answer:
(a) Under the exclusive method we form classes in such a way that the lower limit of a class coincides with the upper class limit of the previous class. Under the method, the upper class limit is excluded but the lower class limit of a class is included in the interval.

(v) Range is the
(a) difference between the largest and the smallest observations
(b) difference between the smallest and the largest observations
(c) average of the largest and the smallest observations
(d) ratio of the largest to the smallest observation
Answer:
(a) The variation in variable’s value are captured by its range. The range is the difference between the largest and the smallest values of the variable. A large range indicates that the values of the variable are widely spead.

Question 2.
Can there be any advantage in classifying things? Explain with an example from your daily life.
Answer:
Classification refers to arranging or organising similar things into groups or classes. Classification of objects or things saves our valuable time and effort. Classification is done to group things in such a way that each group consists of similar items, e.g., we classify our wardrobe into different types of clothes or dresses according to the occasions on which they are to be worn. We put party wears, school uniform, casual daily wears and night wears separately. This helps us in an orderly arrangement of clothes and we can easily fetch the clothes we want at a particular time without searching through the whole wardrobe. Thus, it is evident that classification saves time and labour and helps to produce the desired results.

Question 3.
What is a variable? Distinguish between a discrete and a continuous variable.
Answer:
A measurable characteristic which takes different values at different points of time and in different circumstance is called a variable as it keeps varying. Different varibles vary differently and depending on the way they vary, they are broadly classified into two types

S.N.	Discrete Variable	Continuous Variable
(i)	A discrete variable can take only whole numbers.	A continuous variable can take any numerical value.
(ii)	Discrete varibles increase in finite jumps from one value to another and cannot take any intermediate value between them.	Continuous variables can take any conceivable value and can be broken into infinite gradations.
(iii)	Examples-number of workers in a factory, number of residents in a colony, etc.	Examples-height, weight, distance, etc.

Question 4.
Explain the ‘exclusive’ and ‘inclusive’ methods used in classification of data.
Answer:
Exclusive Method In this method, the classes are formed in such a way that the upper class limit of one class becomes the lower class limit of the next class. Continuity of the data is maintained in this method. Under this method, the upper class limit is excluded but the lower class limit of a class is included in the interval.

According to this method, an observation that is exactly equal to the upper class limit would not be included in that class but would be included in the next class. On the other hand, if it were equal to the lower class limit then it would be included in that class, e.g., if the class intervals are 0-5, 5-10, 15¬20 and so on, a value of 10 would be included in the 10-15 and not in the interval 5-10.

Inclusive Method The inclusive method does not exclude the upper class limit in a class interval. It inlcludes the upper class in a class. Thus, both class limits are parts of the class interval, e.g., the class intervals of 0-5, 6-10, 11-15, and so on are inclusive.

Question 5.
Use the data in Table 3.2 that relate to monthly household expenditure (in ₹) on food of 50 households and
(i) Obtain the range of monthly household expenditure on food.
(ii) Divide the range into appropriate number of class intervals and obtain the frequency distribution of expenditure.
(iii) Find the number of households whose monthly expenditure on food is

• less than ₹ 2,000
• more than ₹ 3,000
• between ₹ 1,500 and ₹ 2,500

Answer:
(i) Range = Largest Value – Smallest Value
Highest Value = 5090
Lowest Value = 1007
So, Range = 5090 – 1007 = 4083

(iii) (a) Number of households whose monthly expenditure on food is less than ₹ 2000
= 20 + 13 = 33
(b) Number of hoseholds whose monthly expenditure on food is more than ₹ 3000
= 2 + 1 + 2 + 0 + 1 = 6
(c) Number of households whose expenditure on food is between ₹ 1500 and ₹ 2500
= 13 + 6= 19

Question 6.
In a city, 45 families were surveyed for the number of domestic appliances they used. Prepare a frequency array based on their replies as recorded below.
1 3 2 2 2 2 1 2 1 2 2 3 3 3 3
3 3 2 3 2 2 6 1 6 2 1 5 1 5 3
2 4 2 7 4 2 4 3 4 2 0 3 1 4 3
Answer:

No. of Domestic Appliances	No. of Households
0	1
1	7
2	15
3	12
4	5
5	2
6	2
7	1
Total	45

Question 7.
What is loss of information’ in classified data?
Answer:
Classification of data as a frequency distribution summarises the raw data making it concise and comprehensible but it does not show the details that are found in raw data. Once, the data are grouped into classes, an individual observation has no significance in further statistical calculations.

All values in a class interval are assumed to be equal to the middle value of the class interval instead of their actual value which causes considerable loss of information. It not only save our time but also our energy, which would otherwise be utilised in searching from entire things.

Question 8.
Do you agree that classified data is better than raw data?
Answer:
The raw data is usually large and fragmented and it is very difficult to draw any meaningful conclusion from them. Classification makes the raw data comprehensible by summarising them into groups. When facts of similar characteristics are placed in the same class, it enables one to locate them easily, analyse them, make comparison and draw inferences.

Question 9.
Distinguish between univariate and bivariate frequency distribution.
Answer:
The term “uni” stands for one and thus the frequency distribution of a single variable is called a Univariate Distribution, e.g., the fequency distribution of age of students in a class is univariate as its gives the distribution of a single variable i.e., age. On the other hand “bi” means two and a Bivariate Frequency Distribution is the frequency distribution of two variables, e.g., the frequency distribution of two varibles, e.g., like price of good and sales of the good is a bivariate distribution.

Question 10.
Prepare a frequency distribution by inclusive method taking class interval of 7 from the following data

Answer:

Statistics for Economics Class 11 NCERT Solutions

NCERT Solutions for Class 11 Economics

Chapter Wise CBSE Class 11 Economics Quick Revision Notes and Key Points in English an Hindi Pdf free download of Indian Economic Development and Statistics for Economics was designed by expert teachers from latest edition of NCERT books to get good marks in board exams. Here we have given NCERT Class 11 Economics Notes.

Indian Economic Development Class 11 Notes

Indian Economic Development Class 11 Notes CBSE Chapter Wise

• Chapter 1 Indian Economy on the Eve of Independence Class 11 Notes
• Chapter 2 Indian Economy 1950 – 1990 Class 11 Notes
• Chapter 3 Liberalisation, Privatisation and Globalisation : An Appraisal Class 11 Notes
• Chapter 4 Poverty Class 11 Notes
• Chapter 5 Rural Development Class 11 Notes
• Chapter 6 Human Capital Formation in India Class 11 Notes
• Chapter 7 Employment: Growth, Informalisation and Related Issues Class 11 Notes
• Chapter 8 Infrastructure Class 11 Notes
• Chapter 9 Environment Sustainable Development Class 11 Notes
• Chapter 10 Comparative Development Experience of India with its Neighbours Class 11 Notes

Statistics for Economics Class 11 Notes

Statistics for Economics Class 11 Notes CBSE Chapter Wise

• Chapter 1 Introduction
• Chapter 2 Collection of Data
• Chapter 3 Organisation of Data
• Chapter 4 Presentation of Data
• Chapter 5 Measures of Central Tendency
• Chapter 6 Measures of Dispersion
• Chapter 7 Correlation
• Chapter 8 Index Numbers

We hope the given Chapter Wise CBSE Class 11 Economics Quick Revision Notes and Key Points in English an Hindi Pdf free download of Indian Economic Development and Statistics for Economics will help you. If you have any query regarding NCERT Class 11 Economics Notes, drop a comment below and we will get back to you at the earliest.

Class 11 Economics NCERT Solutions