Nm Shah 2018 Solutions for Class 11 Commerce Economics Chapter 1 Organisation Of Data are provided here with simple step-by-step explanations. These solutions for Organisation Of Data are extremely popular among Class 11 Commerce students for Economics Organisation Of Data Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Nm Shah 2018 Book of Class 11 Commerce Economics Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Nm Shah 2018 Solutions. All Nm Shah 2018 Solutions for class Class 11 Commerce Economics are prepared by experts and are 100% accurate.

#### Question 1:

Prepare a statistical table from the following data taking the class width as 7 by inclusive method:

 28 17 15 22 29 21 23 27 18 12 7 2 9 4 6 1 8 3 10 5 20 16 12 8 4 33 27 21 15 9 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

 Class Interval Tally Bars Frequency (f) 0 − 7 15 8 − 15 15 16 − 23 10 24 − 31 10 32 − 39 5 Ʃf = 55

#### Question 2:

Prepare a discrete series from the following data:

 62 50 57 58 51 53 62 64 60 61 60 51 64 55 55 52 60 65 58 60 59 52 63 56 56 58 64 63 62 60

 Value Tally Bar Frequency (f) 50 1 51 2 52 2 53 1 55 2 56 2 57 1 58 3 59 1 60 5 61 1 62 3 63 2 64 3 65 1 Ʃf = 30

#### Question 3:

Arrange the following marks in frequency table taking the lowest class interval as (10-20).

 85 70 36 80 78 51 69 96 73 73 92 64 49 78 85 75 80 42 85 29 72 72 77 65 73 95 53 43 91 84 72 75 57 59 61 70 61 75 85 23

 Marks Tally Bar Frequency (f) 10 − 20 20 − 30 2 30 − 40 1 40 − 50 3 50 − 60 4 60 − 70 5 70 − 80 14 80 − 90 7 90 − 100 4 Ʃf = 40

#### Question 4:

Change the following into continuous series and convert the series into 'less than' and 'more than' cumulative series :

 Marks (mid-values) 5 15 25 35 45 55 No.of students 8 12 15 9 4 2

 Mid Value Class Interval Number of Students (f) 5 15 25 35 45 55 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 8 12 15 9 4 2 Ʃf = 50

Cumulative Series

 Less Than More Than Marks Number of Students Marks Number of Students Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 8 8 + 12 = 20 20 + 15 = 35 35 + 9 = 44 44 + 4 = 48 48 + 2 = 50 More than 0 More than 10 More than 20 More than 30 More than 40 More than 50 50 50 − 8 = 42 42 − 12 = 30 30 − 15 = 15 15 − 9 = 6 6 − 4 = 2

#### Question 5:

Marks obtained by 24 students in English and Statistics in a class  are given below . Prepare two-way frequency distribution table.

 S.No 1 2 3 4 5 6 7 8 9 10 11 12 Marks in English 22 23 23 23 23 24 23 25 22 23 24 24 Marks in Statistics 16 16 18 16 16 17 16 19 16 18 18 17

 S.No 13 14 15 16 17 18 19 20 21 22 23 24 Marks in English 23 25 23 22 27 27 26 28 25 24 23 25 Marks in Statistics 16 17 17 17 15 16 18 19 19 16 17 19

 Marks in English Marks in Statistics 22 23 24 25 26 27 28 Total 15 − − − − − − 1 16   − − − 9 17    − − − 6 18 −  − − − 4 19 − − − - − 4 Total 3 9 4 4 1 2 1 24

#### Question 6:

In a survey ,it was found that 64 families bought milk ( in litres) in the following quantities in a particular month.

 19 16 22 9 22 12 39 19 14 23 6 24 16 18 7 17 20 25 28 18 10 24 20 21 10 7 18 28 24 20 14 23 25 34 22 5 33 23 26 29 13 36 11 26 11 37 30 13 8 15 22 21 32 21 31 17 16 23 12 9 15 27 17 21

Convert the above data in a frequency distribution making classes of 5-9, 10-14 and so on.

 Quantity of Milk (kg) Tally Bar Frequency (f) 5 − 9 7 10 − 14 10 15 − 19 13 20 − 24 18 25 − 29 8 30 − 34 5 35 − 39 3 Ʃf = 64

#### Question 7:

The marks obtained by 20 students in Statistics and Economics are given below . Prepare a bivariate frequency distribution.

 Marks in Statistics 10 11 10 11 11 14 12 12 13 10 13 12 11 12 10 14 14 12 13 10 Marks in Economics 20 21 22 21 23 23 22 21 24 25 24 23 22 23 22 22 24 20 24 23

 Marks in Economics Marks in Statistics 20 21 22 23 24 25 Total 10 −  − 5 11 −   − − 4 12    − − 5 13 − − − − − 3 14 − −   3 Total 2 3 5 5 4 1 20

#### Question 8:

Prepare 'less than' and 'more than' cumulative frequency distributions of the following date:

 Wages (₹) : 140-150 150-160 160-170 170-180 180-190 190-200 No. of workers : 5 10 20 9 6 2

Less than Cumulative Frequency Distribution

 Marks Number  of Students Less than 150 Less than 160 Less than 170 Less than 180 Less than 190 Less than 200 5 5 + 10 = 15 15 + 20=35 35 + 9 = 44 44 + 6 = 50 50 + 2 = 52

More than Cumulative Frequency Distribution

 Marks Number of Students More than 140 More than 150 More than 160 More than 170 More than 180 More than 190 52 52 − 5 = 47 47 − 10 = 37 37 − 20 = 17 17 − 90 = 8 8 − 6 = 2

#### Question 9:

Find out the frequency distribution and 'more than' cumulative frequency table:

 Price (₹) below : 10 20 30 40 50 60 Quantity(kg) : 17 22 29 37 50 60

The frequency distribution for the given data is as follows.

 Price Class Interval Quantity (f) Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 17 22 − 17 = 5 29 − 22 = 7 37 − 29 = 8 50 − 37 = 13 60 − 50 = 10

More than Cumulative Frequency Distribution

 More than Quantity (f) More than 0 More than 10 More than 20 More than 30 More than 40 More than 50 60 60 − 17 = 43 43 − 5 = 38 38 − 7 = 31 31 − 8 = 23 23 − 13 = 10

#### Question 10:

If class mid-points in a frequency distribution  of a group of persons are: 125, 132, 139, 146, 153, 160, 167, 174, 181 pounds, find:
(a) size of the class intervals, and
(b) the class boundaries.

From the mid-values the class size is calculated using the following formula.
Class Size = Mid value of one class - Mid value of the preceding class
Class Size = 132 - 125 = 7
Now, the upper and lower limit for each of the class intervals is calculated as follows.

 Mid Value Class Interval 125 132 139 146 153 160 167 174 181 121.5 − 128.5 128.5 − 135.5 135.5 − 142.5 142.5 − 149.5 149.5 − 156.5 156.5 − 163.5 163.5 − 170.5 170.5 − 177.5 177.5 − 184.5

#### Question 11:

Prepare a continuous series from the following data:

 Mid values : 125 135 145 155 165 Frequency : 10 9 12 8 6

From the mid-values the class size is calculated using the following formula.
Class Size = Mid value of one class - Mid value of the preceding class
Class Size = 135 - 125 = 10
Now, the upper and lower limit for each of the class intervals is calculated as follows.

 Mid Value Class Interval f 125 135 145 155 165 120 − 130 130 − 140 140 − 150 150 − 160 160 − 170 10 9 12 8 ​6

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