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#### Question 1:

Following are the figures of marks obtained by 40 students. You ae required to arrange them in ascending and in descending order.

 15 18 16 14 10 6 5 3 8 7 22 18 14 19 17 8 6 4 10 3 12 16 15 13 11 10 18 22 14 19 11 18 22 14 25 21 17 8 9 10

Marks in Ascending Order:
3, 3, 4, 5, 6, 6, 7, 8, 8, 8, 9, 10, 10, 10, 10, 11, 11, 12, 13, 14, 14, 14, 14, 15, 15, 16, 16, 17, 17, 18, 18, 18, 18 ,19, 19, 21, 22, 22, 22, 25

Marks in Descending Order:
25, 22, 22, 22, 21, 19, 19, 18, 18, 18, 18, 17, 17, 16, 16, 15, 15, 14, 14, 14, 14, 13, 12, 11, 11, 10, 10, 10, 10, 9, 8, 8, 8, 7, 6, 6, 5, 4, 3, 3

#### Question 2:

Heights (in inches) of 35 students of a class in given below. Classify the following data in a discrete frequency series

 58 60 72 68 58 55 58 72 55 68 66 60 60 55 76 66 58 72 68 60 55 68 55 58 62 66 72 58 60 68 55 58 65 72 68

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

 Height (in inches) Tally Marks Frequency 55 58 60 62 65 66 68 72 $\mathrm{Ι}$ $\mathrm{Ι}$ $\mathrm{ΙΙΙ}$ 6 7 5 1 1 3 6 6 $\sum _{}f=35$

#### Question 3:

The following are the marks of the 30 students in statistics. prepare a frequency distribution taking the class−intervals as 10−20, 20−30 and so on.

 12 33 23 25 18 35 37 49 54 51 37 15 33 42 45 47 55 69 65 63 46 29 18 37 46 59 29 35 45 27

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

 Marks Tally Marks Number of students (Frequency) 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 $\mathrm{ΙΙΙΙ}$ $\mathrm{ΙΙΙΙ}$ $\mathrm{ΙΙΙ}$ 4 5 7 7 4 3 $\sum _{}f=30$

#### Question 4:

Following are the marks (out of 100) obtained by 50 students in statistics:

 70 55 51 42 57 45 60 47 63 53 33 65 39 82 55 64 58 61 65 42 50 52 53 45 45 25 36 59 63 39 65 54 49 54 64 75 42 41 52 35 30 35 15 48 26 20 40 55 46 18
make a frequency distribution taking the first class interval as 0-10.

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

 Marks Tally Marks Number of students (Frequency) 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80  - 90 - $\mathrm{ΙΙ}$ $\mathrm{ΙΙΙ}$ $\mathrm{ΙΙ}$ $\mathrm{Ι}$ 0 2 3 7 12 14 9 2 1 $\sum _{}f=50$

#### Question 5:

Prepare a frequency table taking class intervals 20−24, 25−29, 30−34 and so on, from the following data:

 21 20 55 39 48 46 36 54 42 30 29 42 32 40 34 31 35 37 52 44 39 45 37 33 51 53 52 46 43 47 41 26 52 48 25 34 37 33 36 27 54 36 41 33 23 39 28 44 45 38

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

 Class Intervals Tally Marks Frequency 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 - 59 $\mathrm{ΙΙΙ}$ $\mathrm{Ι}$ 3 5 8 11 8 7 7 1 $\sum _{}f=50$

#### Question 6:

From the following data, calculate the lower limit of the first class and upper limit of the last class.

 Daily Wages Less than 120 120−140 140−160 160−180 Above 180 No. of Workers 35 12 10 40 13

In the given question, the size of each class is 20. Thus, maintaining the uniformity, we take the first class as 100-120 and the last class as 180-200. Therefore, the lower limit of the first class-interval is 120 − 20 = 100 and the upper limit of the last class is 180 + 20 = 200.

#### Question 7:

Calculate the missing class-intervals from the following distribution:

 Class-Interval Below 20 20−50 50−90 90−140 More than 140 Frequency 7 12 16 2 4

In the given question, the class-intervals are of different width. Thus, the missing class-intervals cannot be determined.

#### Question 8:

Convert the following 'more than' cumulative frequency distribution into a 'less than' cumulative frequency distribution

 Class-Interval (More than) 10 20 30 40 50 60 70 80 Frequency 124 119 107 84 55 31 12 2

The 'more than' cumulative frequency distribution can be presented in the form of a 'simple frequency distribution' as follows.

 Class-Interval Frequency 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 124 − 119 = 5 119 − 107 = 12 107 − 84 = 23 84 − 55 = 29 55 − 31 = 24 31 − 12 = 19 12 − 2 = 10                 2

From the above distribution, we can make the 'less than' cumulative frequency distribution as follows.

 Class-Interval Frequency Less then 20 Less then 30 Less then 40 Less then 50 Less then 60 Less then 70 Less then 80 Less then 90 5 5 + 12 = 17 17 + 23 = 40 40 +29 = 69 69 + 24 = 93 93 + 19 = 112 112 + 10 = 122 122 + 2 = 124

#### Question 9:

Convert the following cumulative frequency series into simple frequency series

 Marks No. of Students Less than 20 10 Less than 40 18 Less than 60 25 Less than 80 45 Less than 100 55

The 'less than' cumulative frequency distribution can be presented in the form of a 'simple frequency distribution' as follows.

 Class-Interval Frequency 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 10 18 − 10 = 8 25 − 18 = 7 45 − 25 = 20 55 − 45 = 10

#### Question 10:

Prepare a frequency distribution from the following data:

 Mid-Points 25 35 45 55 65 75 Frequency 6 10 9 12 8 5

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 = 25 – 5 = 20

Upper limit of first class = 25 + 5 = 30

Thus, the first class interval is (20 30). Similarly, we can calculate the remaining class intervals.

Frequency Distribution

 Mid Value Class Interval Frequency 25 35 45 55 65 75 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 6 10 9 12 8 5 ${\sum }_{}^{}f=50$

#### Question 11:

The ages of 20 husbands and wives are given below. Form a two-way frequency table showing the relationship between the ages of husbands and wives with the class-intervals 20−25, 25−30, etc.

 S.No Age of Husband Age of Wife S. No. Age of Husband Age of Wife 1 28 23 11 27 24 2 37 30 12 39 34 3 42 40 13 23 20 4 25 26 14 33 31 5 29 25 15 36 29 6 47 31 16 32 35 7 37 35 17 22 23 8 35 25 18 29 27 9 23 21 19 38 34 10 41 38 20 48 47

The given data can be presented in the form of a bivariate frequency distribution as follows.

 Age of wife (Y) Age of husband (X) Total 20 − 25 25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 20 − 25  5 25 − 30  5 30 − 35   5 35 − 40   3 40 − 45 1 45 − 50 1 Total 3 5 2 6 3 1 20

#### Question 12:

The following data give the points scored in a tennis match by two player X and Y at the end of twenty games:

 (10,12) (2,11) (7,9) (15,19) (17,21) (12,8) (16,10) (14,14) (22,18) (16,4) (15,16) (22,20) (19,15) (7,18) (11,11) (12,18) (10,10) (5,13) (11,7) (10,10)
Taking class intervals as: 5−9, 10−14, 15−19..., for both X and Y, construct a Bivariate Frequency Distribution.

The given data can be presented in the form of a bivariate frequency distribution as follows.

 Player Y Player X Total 0 − 4 5 − 9 10 − 14 15 − 19 20 − 24 0 − 4 1 5 − 9  3 10 − 14    8 15 − 19    6 20 − 24  2 Total 1 3 8 6 2 20

#### Question 13:

In a survey, it was found that 64 families bought milk in the following quantities in a particular month. Prepare a frequency distribution with classes as 5−9, 10−14 etc.

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

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

 Quantity Tally Marks Frequency 5 − 9  6 10 − 14   11 15 − 19   13 20 − 24    18 25 − 29  8 30 − 34 5 35 − 39 3 40 − 44 ${\sum }_{}^{}f=64\phantom{\rule{0ex}{0ex}}$

#### Question 14:

You are given below a mid value series, convert it into a continuous series.

 Mid-Value 15 25 35 45 55 Frequency 8 12 15 10 5

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 = 15 – 5 = 10

Upper limit of first class = 15 + 5 = 20

Thus, the first class interval is (10 20). Similarly, we can calculate the remaining class intervals.

 Mid value Class Interval Frequency 15 25 35 45 55 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 8 12 15 10 5 ${\sum }_{}^{}f=50$

#### Question 15:

For the data given below, prepare a frequency distribution table with classes 100−110, 110-120, etc.

 125 108 112 126 110 132 136 130 140 155 120 130 136 138 125 111 112 125 140 148 147 137 145 150 142 135 137 132 165 154

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

 Class Interval Telly Marks Frequency 100 − 110 1 110 − 120 4 120 − 130 5 130 − 140  10 140 − 150  6 150 − 160 3 160 − 170 1 $\sum _{}^{}f=30$

#### Question 16:

Prepare a bivariate frequency distribution for the following data for 20 students:

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

The data can be presented in the form of a bivariate distribution as follows.

 Marks in Statistics Marks in Maths 10 11 12 13 14 Total 20  2 21  3 22    5 23    6 24  4 Total 5 4 5 3 3 20

#### Question 17:

In a school, no student has scored less than 25 marks or more than 60  marks in an examination. Their cumulative frequencies are as follows:

 Less than 60 55 50 45 40 35 30 Total frequency 120 115 90 75 65 45 32
Find the frequencies in the class intervals 25−30, 30−35,....55−60.

 Marks Cumulative Frequency Less than 60 Less than 55 Less than 50 Less than 45 Less than 40 Less than 35 Less than 30 120 115 90 75 65 45 32

This can be presented in the form of a simple frequency distribution as follows.

 Marks Number of Students 25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 50 − 55 55 − 60 32 45 − 32 = 13 65 − 45 = 20 75 − 65 = 10 90 − 75 = 15 115 − 90 = 25 120 − 115 = 5

#### Question 18:

Marks scored by 50 students are given below:

 40 45 38 24 46 42 45 18 53 64 45 32 52 54 78 65 52 64 66 43 48 55 50 43 48 20 27 65 37 55 51 55 62 66 38 16 60 58 46 35 72 62 54 58 30 36 43 82 46 53
(a) Arrange the marks in ascending order.
(b) Represent the marks in the form of discrete frequency distribution.
(c) Construct an inclusive frequency distribution with first class as 10−19. Also construct class boundaries.
(d) Construct a frequency distribution with exclusive class-intervals, taking the lowest class as 10−20.
(e) Convert the exclusive series constructed in (d) into 'less than' and 'more than' cumulative frequency distribution.

(a)

 Marks arranged in ascending order 16 18 20 24 27 30 32 35 36 37 38 38 40 42 43 43 43 45 45 45 46 46 46 48 48 50 51 52 52 53 53 54 54 55 55 55 58 58 60 62 62 64 64 65 65 66 66 72 78 82

(b)

Discrete Frequency Distribution
 Marks Frequency (f) 16 18 20 24 27 30 32 35 36 37 38 40 42 43 45 46 48 50 51 52 53 54 55 58 60 62 64 65 66 72 78 82 1 1 1 1 1 1 1 1 1 1 2 1 1 3 3 3 2 1 1 2 2 2 3 2 1 2 2 2 2 1 1 1 $\sum _{}^{}f=50$

(c)
Inclusive Frequency Distribution
 Class Interval (Marks) No. of Students 10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 70 − 79 80 − 89 2 3 7 13 13 9 2 1 $\sum _{}^{}f=50$

(d)
Exclusive Frequency Distribution
 Class Interval No. of Students 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 −90 2 3 7 13 13 9 2 1 $\sum _{}^{}f=50$

(e)
Less than frequency distribution
 Marks Cumulative Frequency Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 Less than 90 2 2 + 3 = 5 5 + 7 = 12 12 + 13 = 25 25 + 13 = 38 38 + 9 = 47 47 + 2 = 49 49 + 1 = 50

More than frequency distribution
 Marks Cumulative Frequency More than 10 More than 20 More than 30 More than 40 More than 50 More than 60 More than 70 More than 80 50 50 − 2 = 48 48 − 3 = 45 45 − 7 = 38 38 − 13 = 25 25 − 13 = 12 12 − 9 = 3 3 − 2 =1

#### Question 19:

From the following data of the ages of different persons, prepare less than and more than cumulative frequency distribution.

 Age (in years) 10−20 20−30 30−40 40−50 50−60 60−70 70−80 No. of Persons 5 12 10 6 4 11 2

 Age (in years) No. of Persons 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 5 12 10 6 4 11 2 $\sum _{}f=50$