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#### Question 1:

Find out the median.

 S. No. 1 2 3 4 5 6 7 8 9 Marks Obtained 10 12 14 17 18 20 21 30 32

10, 12, 14, 17, 18, 20, 21 30, 32

N = 9

Thus, Median is given by the size of the 5th  item. Therefore, Median of the data so given is 18.

#### Question 2:

Find the value of the median from the following data: 15, 35, 48, 46, 50, 43, 55, 49.

First of all, we need to arrange the data in an ascending order. Thus, the data is presented below as:

15, 35, 43, 46, 48, 49, 50, 55

N = 8
Since the number of items in the series is even, thus the following formula is used to calculate the median.

Thus, Median is 47.

#### Question 3:

Calculate the value of median: 25, 20, 15, 45, 18, 7, 10, 64, 38, 12.

First of all, we need to arrange the data in an ascending order. Thus, the data is presented below as:

7, 10, 12, 15, 18, 20, 25, 38, 45, 64

N = 10
Since the number of items in the series is even, thus the following formula is used to calculate the median.

Thus, Median is 19.

#### Question 4:

From the following series of marks obtained by 10 candidates in an examination. Find the median: 26, 14, 30, 18, 11, 35, 41, 12, 32.

First of all, we need to arrange the data in an ascending order. Thus, the data is presented below as:

11, 12, 14, 18, 22, 26, 30, 32, 35, 41

N = 10
Since the number of items in the series is even, thus the following formula is used to calculate the median.

Thus, Median is 24.

#### Question 5:

Calculate the value of median from the following data:

 Income (₹) 12,00 1,800 5,000 2,500 3,000 1,600 3,500 No. of Persons 12 16 2 10 3 15 7

 Income Number of Person (f) Cumulative Frequency (c.f.) 1200 1600 12 15 12 27 1800 16 43 2500 3000 3500 5000 10  3  7  2 53 56 63 65 N=65

Median = size of ${\left(\frac{N+1}{2}\right)}^{\mathrm{th}}$ item
or, Median
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 33rd is 43 (in the c.f. column), which is corresponding to 1800. ​ Hence, median is 1800.

#### Question 6:

Find out the nedian size from the following:

 X 160 150 152 161 156 f 5 8 6 3 7

 X f Cumulative Frequency (cf) 150 152 8 6 8 14 156 7 21 160 161 5 3 26 29 N=29

Median = size of ${\left(\frac{N+1}{2}\right)}^{\mathrm{th}}$ item
or, Median
Now, we need to locate this item in the column of Cumulative Frequency. The item exceeding 15th is 21 (in the c.f. column), which is corresponding to 156. ​ Hence, median is 156.

#### Question 7:

Find out the median size from the following:

 Size 10−20 20−30 30−40 40−50 Frequency 42 25 58 40

 Size Frequency (f) Cumulative Frequency (c.f.) 10 − 20 20 − 30 42 25 42 67 30 − 40 58 (f) 125 40 − 50 40 165 $\underset{}{N=\sum }f=165$

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{165}{2}\right)}^{th}$ item, which is 82.5th item.
This corresponds to the class interval of 30 − 40, so this is the median class.

#### Question 8:

Find the median of the following data:

 Age 20−25 25−30 30−35 35−40 40−45 45−50 50−55 55−60 No. Persons 50 70 100 180 150 120 70 60

 Age Frequency (f) Cumulative Frequency (c.f.) 20 − 25 25 − 30 30 − 35 50 70 100 50 120 220 (c.f.) 35 − 40 180 (f) 400 40 − 45 45 − 50 50 − 55 55 − 60 150 120 70 60 550 670 740 800 $\underset{}{N=\sum }f=800$

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{800}{2}\right)}^{th}$ item, which is 400th item.
This corresponds to the class interval of 35 − 40, so this is the median class.

#### Question 9:

Calculate the median from the following data:

 Marks Frequency Less than 10 5 Less than 20 20 Less than 30 45 Less than 40 80 Less than 50 100 Less than 60 115 Less than 70 125 Less than 80 130

 Marks Class Interval Frequency (f) Cumulative Frequency (c.f.) Less than 10 Less than 20 Less than 30 0 − 10 10 − 20 20 − 30 5 15 25 5 20 45 c.f Less than 40 30 − 40 (f) 35 80 Less than 50 Less than 60 Less than 70 Less than 80 40 − 50 50 − 60 60 − 70 70 − 80 20 15 10 5 100 115 125 130 $\underset{}{N=\sum }f=130$

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{130}{2}\right)}^{\mathrm{th}}$ item, which is 65th item.
This corresponds to the class interval of 30 − 40, so this is the median class.

#### Question 10:

Find the median form the following:

 Marks (More than) 0 10 20 30 40 50 60 70 No. of Students 100 92 78 44 32 18 15 13

 Class Interval Frequency (f) Cumulative Frequency (c.f.) 0 − 10 10 − 20 8 14 8 22 (c.f.) 20 − 30 34 (f) 56 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 12 14 3 2 13 68 82 85 87 100 $N={\sum }_{}^{}f=100$

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item, which is 50th item.
This corresponds to the class interval of 20 − 30, so this is the median class.

#### Question 11:

Compute median from the following data:

 Mid-values 37.5 42.5 47.5 52.5 57.5 No. of Students 30 20 15 13 22

For the calculation of median, the given mid-values must be converted into class intervals using the following 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. In this manner, we obtain the following distribution.

 Mid Values Class Interval Frequency (f) Cumulative Frequency (c.f.) 37.5 42.5 35 − 40 40 − 45 30 20(f) 30(c.f.) 50 47.5 45 − 50 15 65 52.5 57.5 50 − 55 55 − 60 13 22 78 100 N = ∑f =100

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e.${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item, which is 50th item.
This corresponds to the class interval of (40 45), so this is the median class.

#### Question 12:

Calculate median from the following figures:

 Class-intervals 10−29 30−39 40−49 50−59 59−60 60−69 Frequencies 12 19 20 21 15 13

Note that the given distribution is in the form of inclusive class intervals. For the calculation of mode, first the class intervals must be converted into exclusive form using the following formula.

Value of lower limit of one class  Value of upper limit of the preceeding class2
The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.

 Inclusive Class Interval Exclusive Class Interval Frequency (f) Cumulative Frequency (c.f.) 10 − 19 20 − 29 9.5 − 19.5 19.5 − 29.5 12 19 12 31 (c.f.) 30 − 39 29.5 − 39.5 20 (f) 51 40 − 49 50 − 59 60 − 69 39.5 − 49.5 49.5 − 59.5 59.5 − 69.5 21 15 13 72 87 100 $N={\sum }_{}^{}f=100$

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e.${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item, which is 50th item.
This corresponds to the class interval of (29.5 39.5), so this is the median class.

#### Question 13:

Calculate median from following data:

 X Below 10 10−20 20−30 30−40 40−50 50 and Above Frequency 3 7 15 9 6 4

 X Frequency (f) Cumulative Frequency (c.f.) 0 − 10   10 − 20 3 7 3 10 (c.f.) 20 − 30 15 (f) 25 30 − 40 40 − 50  50 − 60 9 6 4 34 40 44 $N={\sum }_{}^{}f=44$

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e.${\left(\frac{44}{2}\right)}^{\mathrm{th}}$ item, which is 22nd item.
This corresponds to the class interval of (20 30), so this is the median class.

#### Question 14:

From the following data, compute median:

 Marks 0−10 10−20 20−40 40−60 60−80 80−100 No. of Students 8 10 22 25 10 5

 Class Interval Frequency (f) Cumulative Frequency   (c.f.) 0 − 10 10 − 20 8 10 8 18 (c.f.) 20 − 40 22 (f) 40 40 − 60 60 − 80   80 − 100 25 10 5 65 75 80 N = ∑f =80

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e.${\left(\frac{80}{2}\right)}^{\mathrm{th}}$ item, which is 40th item.
This corresponds to the class interval of (20 40), so this is the median class.

#### Question 15:

An Incomplete distribution is given below:

 Variables 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Total Frequency 12 30 ? 65 ? 25 18 229
You are given that the median value is 46. Using the median formula, fill up the missing Frequency.

Given, Median = 46
Let the missing frequencies be f1 & f2.

 Class Interval Frequency (f) Cumulative Frequency   (c.f.) 10 − 20 20 − 30 30 − 40 12 30 f1 12 42 42 + f1 (c.f.) 40 − 50 65(f) 107 + f1 50 − 60 60 − 70 70 − 80 f2 25 18 107 + f1 + f2 132 + f1 + f2 229 N = ∑f =229

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e.${\left(\frac{229}{2}\right)}^{\mathrm{th}}$ item, which is 114.5th item.
This corresponds to the class interval of (40 50) as median is 46.

132 + f1 + f2 + 18 = 229
or, f2 = 229 − 132 − 18 − 33.5
$⇒$  f2 = 45.5

Therefore, f1 is 33.5 and f2 is 45.5.

#### Question 16:

Find the missing frequency form the following distribution, if median is 35 and N = 170.

 Variables 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Frequency 10 20 ? 40 ? 25 15

Given, Median = 35
N=170
Let the missing frequencies be f1 & f2.

 Class Interval Frequency (f) Cumulative Frequency 0 − 10 10 − 20 20 − 30 10 20 f1 10 30 30 + f1 (c.f.) 30 − 40 40 (f) 70 + f1 40 − 50 50 − 60 60 − 70 f2 25 15 70 + f1 + f2 95 + f1 + f2 110 + f1 + f2 N = ∑f =170

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e.${\left(\frac{170}{2}\right)}^{\mathrm{th}}$ item, which is 85th item.
This corresponds to the class interval of (30 40) as median is 35.

110 + f1 + f2 = 170
or, f2 = 170 − 110 − 35
$⇒$  f2 = 25

Therefore, f1 is 35 and f2 is 25.

#### Question 17:

Determine the value of median from the following data with the help of: (i) 'less than' and 'More than' Ogive Method; (ii) 'Less than' Ogive method; (iii) 'More than' Ogive method.

 Marks 0−10 10−20 20−30 30−40 40−50 50−60 No. of Students 10 15 25 30 10 10

 Marks (Class Interval) Students (f) Less than c.f. More than c.f. 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 10 15 25 30 10 10 10 25 50 80 90 100 100 90 75 50 20 10 From the point of intersection, a perpendicular is drawn on the x-axis. This line cuts the x-axis at 30. Hence, the median is 30 marks

Median = 30 marks

#### Question 18:

From the following, determine lower quartile and upper quartile:

 X 12 16 17 21 28 19 30 32

12, 16, 17, 19, 21, 28, 30, 32

N = 8

Hence, lower and upper quartile are 16.25 and 29.5 respectively.

#### Question 19:

Calculate lower and upper quartiles.

 S. No. 1 2 3 4 5 6 7 8 9 10 Marks 18 20 25 17 9 11 23 37 38 42

9, 11, 17, 18, 20, 23, 25, 37, 38, 42
N = 10

Hence, lower and upper quartile are 15.5 and 37.25 respectively.

#### Question 20:

Calculate the median, lower quartile and upper quartile from the following data:

 Marks 58 59 60 61 62 63 64 65 66 No. of Students 2 3 6 15 10 5 4 3 1

 Marks Number of students (f) Cumulative Frequency (c.f.) 58 59 60 61 62 63 64 65 66 2 3 6 15 10 5 4 3 1 2 5 11 26 36 41 45 48 49 N=49

Median

Median = size of ${\left(\frac{N+1}{2}\right)}^{\mathrm{th}}$ item
or, Median
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 25th is 26 (in the c.f. column), which is corresponding to 61. ​
Hence, median is 61.

Lower Quartile

This corresponds to 26 in the cumulative frequency.
Hence, lower quartile is 61 marks.

Upper Quartile

This corresponds to 41 in the cumulative frequency.
Hence, upper quartile is 63 marks.

#### Question 21:

From the following series, calculate lower quartile and upper quartile.

 Variable 5 10 15 20 25 30 35 40 Frequency 16 18 22 21 24 14 11 9

 Variable Frequency (f) Cumulative Frequency (c.f.) 5 16 16 10 18 34 15 22 56 20 21 77 25 24 101 30 14 115 35 11 126 40 9 135 N=135

Lower Quartile

This corresponds to 34 in the cumulative frequency.
Hence, lower quartile is 10.

Upper Quartile

This corresponds to 115 in the cumulative frequency.
Hence, upper quartile is 30.

#### Question 22:

Calculate upper and lower quartile from the following data:

 Variable 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Frequency 10 20 35 40 25 25 15

 Class Interval Frequency (f) Cumulative Frequency (c.f.) 0 − 10 10 − 20 10 20 10 30 20 − 30 35 65 Q1 30 − 40 40 105 40 − 50 25 130 Q3 50 − 60 60 − 70 25 15 155 170 N=170

This lies in the class interval (20-30).

Now,

#### Question 23:

Age of 11 students of class XI is given below. Find the modal age by: (i) Observation Method; (ii) Frequency distribution Method.

 Age (in years) 15 16 16 17 15 16 18 17 15 17 17

 Age (X) Tally Marks Frequency (f) 15 16  3 3 17 4 Modal Class 18 1 N=11

Since 17 occurs the highest number of times in the series i.e. 4 times.
Hence, Mode = 17

#### Question 24:

Find modal item of the followings set of numbers:

 2 5 2 3 5 5 6 4 5 3 5 2 5 7 1

 X Tally Marks Frequency (f) 1 2 3 4    1 3 2 1 5 6 →Modal Class 6 7  1 1 N=15

Since 5 occurs the highest number of times in the series i.e. 6 times.
Hence, Mode = 5

#### Question 25:

From the followind data, calculate the value of mode.

 Salary (in ₹) 2,000 2,100 2,400 2,900 3,100 3,300 No. of Workers 3 5 10 19 8 4

 Salary (X) Frequency (f) 2000 2100 2400 3 5 10 2900 19 Modal Class 3100 3300 8 4

The data given in the question is a discrete series. Therefore, using the inspection method of ascertaining mode, we know that the item which repeats itself the maximum number of times is regarded as the mode of the given series. In this manner, 2900 is regarded as the modal value, as it has the highest frequency (of 19 times).
Therefore, mode (Z) is 2900.

#### Question 26:

Compute the mode from the following by: (i) Observation Method; (ii) Grouping Method.

 Hight (in inches) 60 62 63 64 65 66 67 68 69 70 No. of Persons 5 13 18 20 21 30 23 12 4 2

(i) Observation Method

 Height (X) No. of persons (f) 60 62 63 64 65 5 13 18 20 21 66 30 Modal Class 67 68 69 70 23 12 4 2

Using the observation method of ascertaining mode, we know that the item which repeats itself the maximum number of times is regarded as the mode of the given series. In this manner,66 is regarded as the modal value, as it has the highest frequency (of 30 times).
Therefore, mode (Z) is  66 inches.

(ii) Grouping Method

For the given distribution, the grouping table is as follows. On the basis of this grouping table, an analysis table is prepared. For each column of the grouping table, we analyse which item/group of items correspond to the highest frequency.

Analysis Table

 Column 60 62 63 64 65 66 67 68 69 70 I II III IV V VI VII ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ 1 4 7 4 2

From the analysis table, it is clear that 66 repeats the maximum number of times. Thus, mode is 66.

#### Question 27:

Find out mode from the following data:

 Class-Interval 0−5 5−10 10−15 15−20 20−25 25−30 30−35 35−40 Frequency 5 7 15 18 16 9 6 3

 Class Interval Frequency (f) 0 − 5 5 − 10 10 − 15 5 7 15 f0 15 − 20 18 f1 20 − 25 25 − 30 30 − 35 35 − 40 16 9 6 3 f2

By inspection, we can say that the modal class is (15 – 20) as it has the highest frequency of 18.

#### Question 28:

Calculate the mode from the following data:

 Marks 10−20 20−30 30−40 40−50 50−60 60−70 No. of Students 8 11 9 25 12 16

 Class Interval Frequency (f) 10 − 20 20 − 30 30 − 40 8 11 9 f0 40 − 50 25 f1 50 − 60 60 − 70 12 16 f2

By inspection, we can say that the modal class is (40 – 50) as it has the highest frequency of 25.

#### Question 29:

Calculate the mode from the following data:

 Marks(below) 15 30 45 60 75 90 No. of Students 10 30 60 84 90 100

 Class Interval Number of students (c.f) Frequency (f) 0 − 15 15 − 30 10 30 10 30 − 10 = 20 (f0) 30 − 45 60 60 − 30 = 30 (f1) 45 − 60 60 − 75 75 − 90 84 90 100 84 − 60 = 24 (f2) 90 − 84 = 6 100− 90 = 10

By inspection, we can say that the modal class is (30 – 45) as it has the highest frequency of 30.

#### Question 30:

Find out mode of the following frequency distribution:

 Marks (More than) 0 10 20 30 40 50 60 70 80 No. of Students 40 38 33 25 15 7 5 2 0

 Class Interval Number of students (c.f.) Frequency (f) 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 40 38 33 25 15 7 5 2 0 2(=40-38) 5(=38-33)        8(=33-25) (f0)             10(=25-15)  (f1)          8(=15-7)   (f2) 2(=7-5) 3(=5-2) 2(=2-0) 0

By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 10.

#### Question 31:

Find out the mode value from the following data:

 Mid-value 15 25 35 45 55 65 75 85 Frequency 5 8 12 16 28 15 3 2

 Mid Value Class Interval Frequency (f) 15 25 35 45 55 65 75 85 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 5   8 12    16 f0     28 f1     15 f2   3   2

By inspection, we can say that the modal class is (50 – 60) as it has the highest frequency of 28.

#### Question 32:

Find out the modal value from the following data:

 Marks 0−9 10−19 20−29 30−39 40−49 50−59 No. of Students 3 7 15 25 10 4

Note that the given distribution is in the form of inclusive class intervals. For the calculation of mode, first the class intervals must be converted into exclusive form using the following formula.

Value of lower limit of one class  Value of upper limit of the preceeding class2
The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.

 Inclusive Class Interval Exclusive Class Interval Frequency (f) 0 − 9 10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 -0.5 − 9.5 9.5 − 19.5 19.5 − 29.5 29.5 − 39.5 39.5 − 49.5 49.5 − 59.5 3 7 15 f0 25 f1 10 f2 4

By inspection, we can say that the modal class is (29.5 – 39.5) as it has the highest frequency of 25.

Hence, mode is 33.5.

#### Question 33:

Calculate mode of the following data:

 X Below 5 5−10 10−15 15−20 20−25 25−30 30−35 35−40 Above 40 Y 20 24 32 28 20 16 34 10 8

 X Frequency (f) Below 5 5 − 10 20     24 f0 10 − 15 32 f1 15 − 20 20 − 25 25 − 30 30 − 35 35 − 40 above 40 28 f2 20 16 34 10  8

By inspection, we can say that the modal class is (10 – 15) as it has the highest frequency of 32.

#### Question 34:

 Variable 0−10 10−20 20−40 40−50 50−70 No. of Students 5 12 40 32 28

In the given question, class intervals are not equal. For the calculation of mode, first the class intervals are made equal and frequencies are adjusted assuming that frequencies are equally distributed. In this manner, we get the following distribution.

 Class Interval Frequency (f) 0 − 10 10 − 20 20 − 30 30 − 40 5  12  20     20 f0 40 − 50 32 f1 50 − 60 60 − 70 14 f2 14

By inspection, we can say that the modal class is (40 – 50) as it has the highest frequency of 32.

#### Question 35:

Compute graphically, the modal value of the given data:

 Age 0−10 10−20 20−30 30−40 40−50 No. of Persons 2 5 7 5 2 Hence, the mode of the series is 25 years.

#### Question 36:

The monthly profits in rupees (in thousands) of 100 shops are given below:

 Marks 10−20 20−30 30−40 40−50 50−60 No. of Students 3 5 9 3 2
Calculate the value of mode by graphical method. Hence, the mode of the series is 34 marks.

#### Question 37:

Find lower quartile, median and upper quartile from the data given below

 Class-interval (More than) 10 20 30 40 50 60 70 Frequency 100 99 96 85 64 31 9

 Class Interval Frequency (f) Cumulative Frequency (c.f.) 10 − 20 20 − 30 30 − 40 40 − 50 1 3 11 21 1   4 15 36 50 − 60 33 (f) 69 60 − 70 70 − 80 22 9 91 100 $\underset{}{N=\sum }f=100$

Lower Quartile

This lies in the class interval (40-50).
Now,

Thus, the value of lower quartile is 44.76.

Median
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item, which is 50th item.
This corresponds to the class interval of 50 − 60, so this is the median class.Uppe

Upper Quartile

This lies in the class interval (60-70).

Thus, the value of upper quartile is 62.72.

#### Question 38:

From the following data of the ages of different persons, determine the modal age.

 Age (in years) 10−20 20−30 30−40 40−50 50−60 60−70 70−80 No. of Persons 4 26 32 10 9 6 3

 Age(in years) (X) Number of Persons (f) 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 4 26 f0 32 f1 10 f2 9 6 3

By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 32.

#### Question 39:

Find mean, median, and mode for the following data

 Classes 0−10 10−20 20−30 30−40 40−50 50−60 Frequencies 4 15 10 7 3 1

 Class Interval Mid Values (m) Frequency (f) Cumulative Frequency (c.f.) fm 0 − 10 5 4(f0) 4 20 10 − 20 15 15(f1) 19 225 → Modal Class 20 − 30 25 10(f2) 29 250 30 − 40 40 − 50 50 − 60 35 45 55 7 3 1 36 39 40 245 135 55 N = ∑f=40 ∑fm =930

$\overline{)\mathbf{Median}}$
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{40}{2}\right)}^{\mathrm{th}}$ item, which is 20th item.
This corresponds to the class interval of (20 30), so this is the median class.

Mode

By inspection, we can say that the modal class is (10 – 20) as it has the highest frequency of 15.

#### Question 40:

Calculate median from the following:

 Marks (less than) 5 10 15 20 25 30 35 40 45 No. of Students 3 11 16 32 58 70 76 79 80

 Class Interval Cumulative Frequency (c.f.) Frequency (f) 0 − 5 5 − 10 10 − 15 15 − 20 3 11 16         32 (c.f.) 3 8 5 16 20 − 25 58 26 (f) 25 − 30 30 − 35 35 − 40 40 − 45 70 76 79 80 12 6 3 1 N=80

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{80}{2}\right)}^{\mathrm{th}}$ item, which is 40th item.
This corresponds to the class interval of 20 − 25, so this is the median class.

#### Question 41:

Calculate the mean, the mode and the median from the following frequency distribution.

 Height (in inches) 60−61 61−62 62−63 63−64 64−65 65−66 66−67 67−68 68−69 Frequency 4 16 8 24 35 18 19 16 10

 Height Mid Values (m) Frequency (f) Cumulative Frequency (c.f.) fm 60 − 61 61 − 62 62 − 63 63 − 64 60.5 61.5 62.5 63.5 4 16   8       24 (f0) 4 20 28        52 (c.f.) 242 984 500 1524 64 − 65 64.5 35 (f1) 87 2257.5 Modal Class 65 − 66 66 − 67 67 − 68 68 − 69 65.5 66.5 67.5 68.5 18 (f2) 19 16 10 105 124 140 150 1179 1263.5 1080 685 N = ∑f=150 ∑fm =9715

$\overline{)\mathbf{Median}}$
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{150}{2}\right)}^{\mathrm{th}}$ item, which is 75th item.
This corresponds to the class interval of (64 65), so this is the median class.

Mode

By inspection, we can say that the modal class is (64 – 65) as it has the highest frequency of 35.

#### Question 42:

In the frequency distribution of 100 students gives below. The number of students corresponding to marks groups 10-20 and 30-40 are missing from the table. However, the median is known to be 23. Find the missing frequencies

 Marks 0−10 10−20 20−30 30−40 40−50 No. of Students 8 ? 40 ? 10

Given, Median = 23
N=100
Let the missing frequencies be f1 and f2.

 Marks Frequency (f) Cumulative Frequency (cf) 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 8 f1 40 f2 10 8 8+f1 48+f1 48+f1+f2 58+f1+f2 N = ∑f =100

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e.${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item, which is 50th item.
This corresponds to the class interval of (20 30) as median is 23.

58 + f1 + f2 = 100
or, f2 = 100 − 58 − 30
$⇒$  f2 = 12

Therefore, f1 is 30 and f2 is 12.

#### Question 43:

Calculate the value of median and mode from the following distribution.

 Mid-point 20 30 40 50 60 70 80 90 Frequency 8 10 15 25 40 20 15 7

For the calculation of median, the given mid-values must be converted into class intervals using the following 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. In this manner, we obtain the following distribution.

 Mid Points Class Interval Frequency (f) Cumulative Frequency (c.f.) 20 30 40 50 15 − 25 25 − 35 35 − 45 45 − 55 8  10  15        25 (f0) 8 18 33         58 (c.f.) 60 55 − 65 40 (f1) 98 70 80 90 65 − 75 75 − 85 85 − 95 20 (f2) 15 7 118 133 140 N = ∑f =140

Median
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e.${\left(\frac{140}{2}\right)}^{\mathrm{th}}$ item, which is 70th item.
This corresponds to the class interval of (55 65), so this is the median class.

Mode

By inspection, we can say that the modal class is (55 65) as it has the highest frequency of 40.

#### Question 44:

From the following data, find the value of mode.

 Size 5 10 15 20 25 30 35 40 45 Frequency 2 3 5 7 14 12 8 7 3

 Size Frequency 5 10 15 20 2 3 5 7 25 14 Modal Class 30 35 40 45 12 8 7 3

The data given in the question is a discrete series. Therefore, using the inspection method of ascertaining mode, we know that the item which repeats itself the maximum number of times is regarded as the mode of the given series. In this manner, 25 is regarded as the modal value, as it has the highest frequency (of 14 times).
Therefore, mode (Z) is 25.

#### Question 45:

Calculate the mean, median and mode of the following data:

 Monthly Profit Frequency Less than 10 4 Less than 20 20 Less than 30 35 Less than 40 55 Less than 50 62 Less than 60 67

 Class Interval Cumulative Frequency (c.f.) Frequency (f) Mid Values (m) fm 0 − 10 10 − 20 4      20 (c.f) 4  16 5 15 20 240 20 − 30 35 15 (f0) 25 375 30 − 40 55 20 (f1) 35 700 Modal class 40 − 50 50 − 60 62 67 7 (f2)    5 45 55 315 275 N = ∑f=67 ∑fm =1925

$\overline{)\mathbf{Median}}$
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{67}{2}\right)}^{\mathrm{th}}$ item, which is 33.5th item.
This corresponds to the class interval of (20 30), so this is the median class.

Mode

By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 20.

#### Question 46:

Find the median of the following data:

 Valuable 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Frequency 7 12 18 25 16 14 8

 Variable Frequency (f) Cumulative Frequency (c.f.) 0 − 10 10 − 20 20 − 30 7 12 18 7 19          37 (c.f.) 30 − 40 25 (f) 62 40 − 50 50 − 60 60 − 70 16 14 8 78 92 100 N=100

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item, which is 50th item.
This corresponds to the class interval of (30 40), so this is the median class.

#### Question 47:

Find out mode from the following data:

 Size 0−4 4−6 6−10 10−15 15−20 20−30 30−35 35−40 Frequency 2 4 3 5 2 20 6 8

In the given question, class intervals are not equal. For the calculation of mode, first the class intervals are made equal and frequencies are adjusted assuming that frequencies are equally distributed. In this manner, we get the following distribution.

 Class Interval Frequency (f) 0 − 10 10 − 20 20 − 30 30 − 40 9          7 (f0)        20 (f1)        14 (f2)

By inspection, we can say that the modal class is (20 – 30) as it has the highest frequency of 20.

#### Question 48:

Calculatate the Median and Mode form the following data:

 Marks 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 No. of Students 2 18 30 45 35 20 6 3

 Marks Number of Students (f) Cumulative Frequency (c.f.) 0 − 10 10 − 20 20 − 30 2   18         30 (f0) 2 20 50 (c.f) 30 − 40 45 (f1) 95 40 − 50 50 − 60 60 − 70 70 − 80 35 (f2)   20     6     3 130 150 156 159 N=159

Median

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{159}{2}\right)}^{\mathrm{th}}$ item, which is 79.5th item.
This corresponds to the class interval of (30 40), so this is the median class.

Mode

By inspection, we can say that the modal class is (30 40) as it has the highest frequency of 45.

#### Question 49:

Find out the median of the following frequency distribution:

 Marks 50 55 60 65 70 75 No. of Students 5 7 6 10 5 8

 Marks Number of Students (f) Cumulative Frequency (c.f.) 50 55 60 65 70 75 5 7 6 10 5 8 5 12 18 28 33 41 N= 41

Median = size of ${\left(\frac{N+1}{2}\right)}^{\mathrm{th}}$ item
or, Median
Now, we need to locate this item in the column of Cumulative Frequency. The item exceeding 21st is 28 (in the c.f. column), which is corresponding to 65. ​Hence, median is 65.

#### Question 50:

Calculate mode from the following data:

 Income 15−24 25−34 35−44 45−54 55−64 65−74 No. of Workers 8 10 15 25 40 20

Note that the given distribution is in the form of inclusive class intervals. For the calculation of mode, first the class intervals must be converted into exclusive form using the following formula.

Value of lower limit of one class  Value of upper limit of the preceeding class2
The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.

 Inclusive Class Interval Exclusive Class Interval Frequency (f) 0 − 9 10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 -0.5 − 9.5 9.5 − 19.5 19.5 − 29.5 29.5 − 39.5 39.5 − 49.5 49.5 − 59.5 3 7 15 f0 25 f1 10 f2 4

By inspection, we can say that the modal class is (29.5 – 39.5) as it has the highest frequency of 25.

Hence, mode is 33.5.

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