Tr Jain Vk Ohri 2018 Solutions for Class 11 Humanities Economics Chapter 4 Organisation Of Data are provided here with simple step-by-step explanations. These solutions for Organisation Of Data are extremely popular among Class 11 Humanities students for Economics Organisation Of Data Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Tr Jain Vk Ohri 2018 Book of Class 11 Humanities Economics Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Tr Jain Vk Ohri 2018 Solutions. All Tr Jain Vk Ohri 2018 Solutions for class Class 11 Humanities Economics are prepared by experts and are 100% accurate.

#### Question 1:

Identify 10 important objects that impact your environment. Classify them, as living and non-living. Is it a quantitative classification? If not, give reasons.

#### Answer:

Objects that impact our environment are:

(i) Factories
(ii) Animals
(iii) Human beings
(iv) Automobiles
(v) Fuel
(vi) Coal
(vii) Soil
(viii) Fertilisers
(ix) Trees
(x) Plastic

The above objects can be classified as:
a) Living- Animals, human beings, trees
b) Non-living- Factories, automobiles, soil, fuel, coal, fertilisers, plastic

No, it is not a quantitative classification as we cannot classify the data on the basis of numerical values into different classes or groups. This is because we cannot judge in numbers the impact of objects on environment.

#### Question 2:

Identify the following variables as discrete and continuous: volume, height, weight, temperature, snowfall, population, crop-production, no. of scooters on the ring road of Delhi, no. of houses demolished by the MCD.

#### Answer:

 Discrete Variable Continuous Variable 1. Population 1. Volume 2. Crop-production 2. Height 3. No. of scooters on the ring road of Delhi 3. Weight 4. No. of houses demolished by the M.C.D 4. Temperature 5. Snowfall

#### Question 1:

Following data relate to the pocket expenses (rupees) of 10 students of Class XI. Arrange them in the ascending and descending orders:

 50, 20, 30, 15, 45, 35, 40, 25, 20, 43

#### Answer:

 Data Arranged in Ascending Order Data Arranged in Descending Order 15 20 20 25 30 35 40 43 45 50 50 45 43 40 35 30 25 20 20 15

#### Question 2:

In a sample investigation, 20 persons are found to have the following money in their pockets as office expenses. Arrange the data in the ascending and descending orders:

 114, 108, 100, 98, 101, 109, 117, 119, 121, 136 131, 136, 143, 156, 169, 182, 195, 207, 219, 255

#### Answer:

 Data Arranged in Ascending Order Data Arranged in Descending Order 98 100 101 108 109 114 117 119 121 131 136 136 143 156 169 182 195 207 219 255 255 219 207 195 182 169 156 143 136 136 131 121 119 117 114 109 108 101 100 98

#### Question 3:

Collect data of weekly expenditure on food by your family. Arrange the data in an order. Also, convert the data into monthly expenditure. Find the total number of observations.

#### Answer:

Data of weekly expenditure on food

 Week Expenditure 1st week 1000 2nd week 2000 3rd week 2100 4th week 2200 5th week 2500 6th week 2700 7th week 2800 8th week 3000

Ascending and Descending order of data

 Ascending order of data Descending order of data 1000 3000 2000 2800 2100 2700 2200 2500 2500 2200 2700 2100 2800 2000 3000 1000

Monthly Expenditure

 Month Expenditure 1st Month 1000 + 2000 + 2100 + 2200 = Rs 7300 2nd Month 2500 + 2700 + 2800 + 3000 = Rs 11,000

Total number of observations = 8

#### Question 1:

In an examination, 25 students secured the following marks:

 23 28 30 32 35 35 36 40 41 43 44 45 45 48 49 52 53 54 56 56 58 61 62 65 68
(i) Arrange these data in the form of a frequency distribution using the following class as intervals:
20−29, 30−39, 40−49, 50−59, and 60−69.
(ii) Arrange the data with cumulative frequencies.

#### Answer:

i) In the form of a frequency distribution, the given data can be arranged as follows.

ii) With cumulative frequencies the given data can be arranged as follows.

 METHOD-1 METHOD-2 Marks No. of Students Marks No. of students Less than 29 Less than 39 Less than 49 Less than 59 Less than 69 0 + 2 = 2 2 + 5 = 7 7 + 8 = 15 15 + 6 = 21 21 + 4 = 25 More than 20 More than 30 More than 40 More than 50 More than 60 25 25 − 2 = 23 23 − 5 = 18 18 − 8 = 10 10 − 6 = 4

#### Question 2:

The following data is of the age of 25 students of Class XI.
Arrange these data in the form of a frequency distribution.

 15 16 16 17 18 18 17 15 15 16 16 17 15 16 15 16 16 18 15 17 17 18 10 16 15

#### Question 3:

Students of Class XI obtained following marks in Economics. Classify the data in the form of individual series, discrete series, continuous series and cumulative frequency series.

 15 15 18 16 20 21 25 25 15 16 18 22 24 25 20 18 22 24 24 25 25 23 20 15 16 17 19 18 22 22

#### Answer:

Individual series is simply the arrangement of the given data in ascending (or, descending order)

 15 15 15 15 16 16 16 17 18 18 18 18 19 20 20 20 21 22 22 22 22 23 24 24 24 25 25 25 25 25

Discrete series:

Continuous series:

Cumulative frequency series

 Method1 Method2 Marks No. of Students Marks No. of students Less than 15 Less than 19 Less than 23 Less than 27 0 9 + 4 = 13 13 + 9 = 22 22 + 8 = 30 More than 12 More than 16 More than 20 More than 24 30 30 − 4 = 26 26 − 9 = 17 17 − 9 = 8

#### Question 4:

Arrange the following data in the form of an exclusive frequency distribution, using 5−10 as the initial class interval:

 12 36 40 30 28 20 19 10 10 19 27 15 26 10 19 7 45 33 26 37 5 20 11 17 37 30 20

#### Question 5:

Weight of 20 students is given in kilograms. Using class interval of 5, make a frequency distribution.

 30 45 26 25 42 33 15 35 45 45 45 39 42 40 18 35 41 20 36 48

#### Answer:

In the given data, the lowest value is 15, while the highest value is 48. Accordingly, we take (15 - 20) as the initial class interval.

 Weight (kg) Tally Bars No. of Students 15−20 2 20−25 1 25−30 2 30−35 2 35−40 4 40−45 4 45−50 5 Total $\sum f=20$

#### Question 6:

Convert the following data in a simple frequency distribution:

 5 students obtained less than 3 marks 12 students obtained less than 6 marks 25 students obtained less than 9 marks 33 students obtained less than 12 marks

#### Answer:

The given information can be summarised as follows.

 Marks Cumulative Frequency (c.f.) Less than 3 Less than 6 Less than 9 Less than 12 5 12 25 33
This can be presented in the form of a simple frequency distribution as follows.

 Marks Frequency (f) 0 − 3 3 − 6 6 − 9  9 − 12 5 7 (= 12 − 5) 13 (= 25 − 12) 8 (= 33 − 25)

#### Question 7:

In the following statement, take the number of letters in a word as items and number of items a word (of the same size) repeats itself as frequencies. Prepare a discrete series.
"Success in the examination confers no absolute right to appointment unless government is satisfied after such an enquiry as may be considered necessary that the candidate is suitable in all respect for appointment."

#### Answer:

In the given question, the size of the items refers to the number of alphabets in the word. Accordingly, in the given statement there are eight items with size equal to 2, namely, in, no, to, is, an, as, be, is, in. Similarly, there are 5 items with size equal to three, namely, the, may, the, all, for.  In the similar manner, we can find the frequency corresponding to the various items.
The frequency distribution table is prepared in the following manner.

 Size of Item Tally Bars Frequency 2 9 3 5 4 2 5 2 6 1 7 3 8 3 9 3 10 2 11 3 Total $\sum f=33$

#### Question 8:

An economic survey revealed that 30 families in a town incur following expenditure in a day (rupees).

 11 12 14 16 16 17 18 18 20 20 20 21 21 22 22 23 23 24 25 25 26 27 28 28 31 32 32 33 36 38
(i) Convert these data in the form of a frequency distribution, using the following class intervals.
10−14, 15−19, 20−24, 25−29, 30−34 and 35−39.
(ii) How many families spend more than 29 rupees a day?

#### Answer:

i)

ii) Number of families spending more than Rs 29 per day = 4+2 = 6
Thus, there are 6 families that spend more than Rs 29 per day.

Percentage of families spending more than Rs 29

#### Question 9:

From the following data related to the weight of college students in kg, prepare a frequency distribution with a class interval of 10 exclusive and inclusive basis:

 40 92 49 52 69 70 72 42 50 60 63 65 43 48 54 53 53 47 65 82 85 79 50 42 55

#### Answer:

 Exclusive Method Inclusive Method Weight (in kg) No. of Students (f) Weight (in kg) No. of Students (f) 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 90 − 100 7 7 5 3 2 1 40 − 50 51 − 61 62 − 72 73 − 83 84 − 94 95 − 105 9 6 6 2 2 0 ∑f = 25 ∑f = 25

#### Question 10:

Construct the simple frequency distribution from the following data:

 Mid-value 5 15 25 35 45 55 Frequency 2 8 15 12 7 6

#### Answer:

The class interval can be calculated from the mid-points using the following adjustment formula.

The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.

Thus, we add and subtract 5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 5 – 5 = 0

Upper limit of first class = 5 + 5 = 10

Thus, the first class interval is (0 − 10). Similarly, we can calculate the remaining class intervals.

 Mid-value Class-interval Frequency (f) 5 15 25 35 45 55 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 2 8 15 12 7 6 ∑f = 50

#### Question 11:

Classify the following data by taking class interval such that their mid-values are 17, 22, 27, 32 and so on:

 30 30 36 33 42 27 22 41 30 42 30 21 54 36 31 16 40 28 19 17 48 28 48 36 14 37 16 37 17 54 42 41 51 44 32 46 42 31 21 47 25 36 22 52 41 40 40 53

#### Answer:

The class interval can be calculated from the mid-points using the following adjustment formula.

The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.

Thus, we add and subtract 2.5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 12 – 2.5 = 9.5

Upper limit of first class = 12 + 2.5 = 14.5

Thus, the first class interval is (9.5  14.5). Similarly, we can calculate the remaining class intervals.

 Mid-value Class-interval Frequency (f) 12 17 22 27 32 37 42 47 52 9.5 − 14.5 14.5 − 19.5 19.5 − 24.5 24.5 − 29.5 29.5 − 34.5 34.5 − 39.5 39.5 − 44.5 44.5 − 49.5 49.5 − 54.5 1 5 4 4 8 6 11 4 5 ∑f = 48

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