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

For a comprehensive understanding of Class 11 Economics and to achieve excellence in board and competitive examinations, NCERT Solutions are crucial. These resources, developed by experts, focus on important chapter concepts and are tailored to the CBSE curriculum, providing vital assistance in your studies.

In these NCERT Solutions for Class 11 Economics, we have discussed all types of NCERT intext questions and exercise questions.

Concepts covered in Class 11 Economics, Statistics for Economics, Chapter- 3 Organisation of Data, are-

  • Introduction
  • Raw data
  • Classification of data
  • Variables: Continuous and discrete
  • What is a frequency distribution?
  • How to prepare a frequency distribution?
  • Should we have equal or unequal-sized class intervals?
  • How many classes should we have?
  • How should we determine the class limits?
  • What should be the size of each class?
  • Adjustment in class interval
  • How should we get the frequency for each class?
  • Loss of information

Our NCERT Solutions for Class 11 Economics feature in-depth explanations that facilitate homework and assignment completion. By gaining a solid understanding of each chapter’s concepts with these solutions, you’ll be ready to secure top marks in your exams. Embark on your journey to academic excellence today!

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3 Answers

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

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

Solution:

(a) the average of the upper-class limit and the lower class limit

The class mid-point refers to the middle value of a class. It is the middle value between the upper class limit and the lower class limit of a class and is computed as

Class Mid-Point = \(\frac{\text{(Lower Class Limit + Upper Class Limit)}}2\)

(ii) The frequency distribution of two variables is known as:

(a) Univariate Distribution
(b) Bivariate Distribution
(c) Multivariate Distribution
(d) None of the above


Solution:

(b) Bivariate Distribution

Bi refers to two and therefore, the frequency distribution of two variables is called a 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

Solution:

(d) the class mid-points 

The class mid-points of each class is used to present the class and thus, it is used in further estimates after the raw data and facts are grouped into classes.

(iv) 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

Solution:

(a) difference between the largest and the smallest observations

The range is the difference between the largest value and the smallest value of the variable. A large range depicts the values of the variable that are widely spread.

2. Can there be any advantage in classifying things? Explain with an example from your daily life.

Solution:

Classification means organising and arranging the similar data or homogenous data into groups. Classification of objects saves the valuable cost in terms of time, money and effort. Classification is done to group the homogenous things. For example, one classify his/her wardrobe into different types of dresses as per the occasions. They put school uniform, party wears, night wears and casual daily wears separately. This will aids in an orderly and systematic arrangement of clothes and one can easily find out the clothes they want at a specific time without much searching. Hence, it is clear that classification not only saves time but also saves labour and aids to produce the desired outcomes.

3. What is a variable? Distinguish between a discrete and a continuous variable.

Solution:

A characteristic that takes different values at different time intervals and in different situations is known as variable. It always keeps changing. Different variables changes based upon the way they vary. They are further divided into two parts, i.e.,

S. No. Discrete Variable Continuous Variable
(i) A discrete variable deals only with whole numbers. A continuous variable deals with any numerical value.
(ii) Discrete variables only takes the finite value and does not deals with the intermediate value between them. Continuous variables take any conceivable value or intermediate value.
(iii) For examples, number of residents in a colony, number of workers in a factory, number of students in a class etc. For examples, weight, height, distance, percentage scored in a test etc.

4. Explain the ‘exclusive’ and ‘inclusive’ methods used in classification of data.

Solution:

In the exclusive method, the classes are formed such that the lower class limit of one class becomes the upper class limit of the preceding class. Therefore, the continuation of the data is maintained. Also, in this method, the upper class limit of a class is not included, but the lower class limit is included in the interval.

In simple terms, if an observation is exactly the same as of the upper class limit, then it will not be included in that class but will be included in the next class. In contrast, if an observation is exactly the same as of the lower class limit, then it will be included in that class only. For example, if the class intervals are 0-10, 10-20, 20¬30 and so on, then a value of 20 will be included in the interval 20-30 and not in the interval 10-20.

Inclusive Method is just the opposite of Exclusive Method. The inclusive method does include the upper class limit also. Therefore, both class limits are included in the class interval. For example, if the class intervals are 0-10, 11-20, 21-30 and so on, then a value of 20 will be included in the interval 11-20.

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5. Use the data that relate to monthly household expenditure (in `) on food of 50 households and

1904 1559 3473 1735 2760
2041 1612 1753 1855 4439
5090 1085 1823 2346 1523
1211 1360 1110 2152 1183
1218 1315 1105 2628 2712
4248 1812 1264 1183 1171
1007 1180 1953 1137 2048
2025 1583 1324 2621 3676
1397 1832 1962 2177 2575
1293 1365 1146 3222 1396

(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

(a) less than ₹2,000
(b) more than ₹3,000
(c) between ₹1,500 and ₹2,500

Solution:

(i) Range = Largest Value - Smallest Value

Highest Value = 5090

Lowest Value = 1007

So, Range = 5090 - 1007 = 4083

(ii)

Class Intervals Tally Marks Frequency
1000 – 1500 |||| |||| ||||| 20
1500 – 2000 |||| |||| |||| 13
2000 – 2500 |||| | 06
2500 – 3000 |||| 05
3000 – 3500 || 02
3500 – 4000 | 01
4000 – 45000 || 02
4500 – 5000 00
5000 – 5500 | 01
Total 50

(iii)

(a) Number of households whose monthly expenditure on food is less than ₹2000 = 20 + 13 = 33.

(b) Number of households 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.

6. In a city, 45 families were surveyed for the number of cell phones 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 4
2 4 2 7 4 2 4 3 4 2 0 3 1 4 3

Solution:

Number of Domestic Appliances Number of Households
0 1
1 7
2 15
3 12
4 5
5 2
6 2
7 1
Total 45

7. What is loss of information in classified data?

Solution:

Classification of data helps in summarising the raw data into groups and making it comprehensible and concise. Once, the data are categorised into classes, a personal observation has no importance in further statistical computations.

In a class interval, all the values are expected to be equal to the middle value instead of their actual value that affects considerable loss of information. It not only saves the time but also the energy, that would otherwise be consumed in searching the other things.

8. Do you agree that classified data is better than raw data? Why?

Solution:

The raw data is generally large and uneven in nature and it is very hard to know any meaningful results from them. Classification helps in summarising the raw data and facts into groups. When data of same nature are placed together in the same class, it allows one to understand them easily, evaluate them, analyse them, and make comparison and draw conclusions.

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9. Distinguish between univariate and bivariate frequency distribution.

Solution:

The term “uni” refers to one and therefore the frequency distribution of a one or single variable is known as Univariate Distribution. For example, the frequency distribution of height of students in a school is univariate as it offers the distribution of an one variable, i.e., height. Similarly, “bi” refers to two and therefore the frequency distribution of two variables is known as a Bivariate Frequency Distribution. For example, the frequency distribution of price of tea and demand of the tea is a bivariate distribution.

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

28 17 15 22 29 21 23 27 18 12 7 2 9 4
1. 8 3 10 5 20 16 12 8 4 33 27 21 15
3 36 27 18 9 2 4 6 32 31 29 18 14 13
15 11 9 7 1 5 37 32 28 26 24 20 19 25
19 20 6 9

Solution:

Class Intervals Tally Marks Frequency
0 – 7 |||| |||| |||| |||| 15
8 – 15 |||| |||| |||| |||| 15
16 – 23 |||| |||| |||| 14
24 – 31 |||| |||| | 11
32 – 39 |||| 5
Total 60

11. “The quick brown fox jumps over the lazy dog” Examine the above sentence carefully and note the numbers of letters in each word. Treating the number of letters as a variable, prepare a frequency array for this data.

Solution:

Letter Frequency Tally Marks
1 A 1 |
2 B 15 |
3 C 1 |
4 D 1 |
5 E 3 |||
6 F 1 |
7 G 1 |
8 H 2 ||
9 I 1 |
10 J 1 |
11 K 1 |
12 L 1 |
13 M 1 |
14 N 1 |
15 O 4 |||
16 P 1 |
17 Q 1 |
18 R 2 ||
19 S 1 |
20 t 2 ||
21 U 2 ||
22 V 1 |
23 W 1 |
24 X 1 |
25 Y 1 |
26 Z 1 |

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