Tr Jain & Vk Ohri (2017) for Class 11 Humanities Economics Chapter 10 - Measures Of Central Tendency Median And Mode

Question 1:

Marks of 15 students in their Economics paper are:
6, 9, 10, 12, 18, 19, 23, 23, 24, 28, 37, 48, 49, 53, 60
Find the median marks.

Answer:

S. No.	Marks
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15	6 9 10 12 18 19 23 23 24 28 37 48 49 53 60
N = 15

Median = size of ${(\frac{N + 1}{2})}^{th}$ item
    or, M = size of ${(\frac{15 + 1}{2})}^{th}$ item
    or, M = size of 8^th item
    $\Rightarrow$ M = 23

Therefore, the median marks in the economics paper is 23.

Question 2:

Height of seven students is measured in cm as,
140, 142, 144, 145, 147, 149, 151
Find the median height.

Answer:

S.No	Height (cms)
1 2 3 4 5 6 7	140 142 144 145 147 149 151
N = 7

Median = Size of

{(\frac{N + 1}{2})}^{th}

item
or, M= Size of

{(\frac{7 + 1}{2})}^{th}

item
or, M= size of 4^th item

\Rightarrow

M= 145

Hence, the median height is 145 cm.

Question 3:

Weight of eight students in kg is noted as,
71, 72, 64, 68, 70, 76, 73, 75
Find the median weight.

Answer:

Arranging the series in the ascending order

S. No.	Weight
1 2 3 4 5 6 7 8	64 68 70 71 72 73 75 76
N = 8

Median = \frac{Size of (\frac{N}{2}) th item + Size of (\frac{N}{2} + 1) th item}{2} or, M = \frac{Size of (\frac{8}{2}) th item + Size of (\frac{8}{2} + 1) th item}{2} or, M = \frac{Size of 4 th item + Size of 5 th item}{2} or, M = \frac{71 + 72}{2} = \frac{143}{2} \Rightarrow M = 71.5 k g s

Hence, the median weight is 71.5 kgs.

Question 1:

Find out median of the following series:

Size	20	25	30	35	40	45	50	55
Frequency	14	18	33	30	20	15	13	7

Answer:

Size	Frequency	Cumulative Frequency
20 25 30	14 18 33	14 14 + 18 = 32 32 + 33 = 65
35	30	65 + 30 = 95
40 45 50 55	20 15 13 7	95 + 20 = 115 115 + 15=130 130 + 13=143 143 + 7 = 150
	N = 150

Median or M = Size of

{(\frac{N + 1}{2})}^{th}

item
or, M = Size of

{(\frac{15 + 1}{2})}^{th}

item
or, M = Size of 75.5^th item

It shows that median value corresponds to the 75.5th item in the series. This item appears first of all in 95^th cumulative frequency of the series. Therefore, median shall be the value corresponding to the 95^th cumulative frequency, which is 35.

Therefore, the median value will be 35.

Question 2:

Find out median of the following series:

Selling Price (paise)	45	46	47	48	49	50	51	52
Frequency	23	20	42	50	41	12	8	4

Answer:

Selling Price (paise)	Frequency	Cumulative frequency
45 46 47 48 49 50 51 52	23 20 42 50 41 12 8 4	23 23 + 20 = 43 43 + 42 = 85 85 + 50 = 135 135 + 41= 176 176 + 12= 188 188 + 8 = 196 196 + 4 = 200
	N = 200

Median or M = Size of

{(\frac{N + 1}{2})}^{th}

item
or, M = Size of

{(\frac{200 + 1}{2})}^{th}

item
or, M = Size of 100.5^th item

It shows that median value corresponds to the 100.5^thitem in the series. This item appears first of all in 135^th cumulative frequency of the series. Therefore, median shall be the value corresponding to the 135^thcumulative frequency, which is 48 paise.

Therefore, the median value is 48 paise.

Question 1:

Find out median of the following series:

Items	20−30	30−40	40−50	50−60	60−70	70−80	80−90
Frequency	5	10	16	18	12	10	8

Answer:

Items	Frequency	Cumulative Frequency
20 − 30 30 − 40 40 − 50	5 10 16	5 10 + 5 = 15 15 + 16 = 31 (c.f)
50 − 60	18 (f)	31 + 18 = 49
60 − 70 70 − 80 80 − 90	12 10 8	49 + 12 = 61 61 + 10 = 71 71 + 8 = 79
	N = 79

Median or M = Size of

(\frac{N}{2}) th

item
or, M = Size of

(\frac{79}{2}) th

item
or, M = Size of 39.5th item

Hence, the median lies in the class 50-60

M = l_{1} + \frac{\frac{N}{2} - c . f}{f} \times i or, M = 50 + \frac{39.5 - 31}{18} \times 10 or, M = 50 + \frac{8.5}{18} \times 10 or, M = 50 + \frac{85}{18} or, M = 50 + 4.72 \Rightarrow M = 54.72

Hence, the median is of the above series is 54.7

Question 2:

Find out median marks of the following marks distribution for 100 students.

Marks	0−10	10−20	20−30	30−40	40−50
Number of Students	8	30	40	12	10

Answer:

Marks	No. of students	Cumulative frequency
0 − 10 10 − 20	8 30	8 8 + 30 = 38 (c.f.)
20 − 30	40 (f)	38 + 40 = 78
30 − 40 40 − 50	12 10	78 + 12 = 90 90 + 10 = 100
	N = 100

Median or M = Size of

{(\frac{N}{2})}^{th}

item
or, M = Size of

{(\frac{100}{2})}^{th}

item
or, M = Size of 50^th item

Hence, median lies in the class 20-30

Median = l_{1} + \frac{\frac{N}{2} - c . f}{f} \times i or, M = 20 + \frac{50 - 38}{40} \times 10 or, M = 20 + \frac{120}{40} or, M = 20 + 3 \Rightarrow M = 23 m a r k s

Hence, the median marks is 23.

Question 1:

Calculate median of the following data:

Marks (less than)	15	30	45	60	75	90
Number of Students	18	35	62	81	95	100

Answer:

Converting the Cumulative frequency of 'less than' type into a simple frequency distribution.

Marks	Cumulative Frequency	Frequency
0 − 15 15 − 30	18 35 *(c.f)*	18 35 − 18 = 17
(l₁) 30 − 45	62	62−35= 27(f)
45 − 60 60 − 75 75 − 90	81 95 100	81 − 62 = 19 95 − 81 = 14 100 − 95 = 5
		$Σ f = N =$ 100

Median or M = Size of

{(\frac{N}{2})}^{th}

item
or, M = Size of

{(\frac{100}{2})}^{th}

item
or, M = Size of 50^th item

The 50^th item lies in 62^thcumulative frequency of the series. The corresponding class interval, 30-45 is, therefore, the median class interval.

M = l_{1} + \frac{\frac{N}{2} - C . f}{f} \times i or, M = 30 + \frac{50 - 35}{27} \times 15 or, M = 30 + \frac{15}{27} \times 15 or, M = 30 + \frac{225}{27} or, M = 30 + 8.33 \Rightarrow M = 38.33 marks

Hence, the median marks of the above series is 38.33

Question 2:

Find out median in the following series:

Size (less than)	5	10	15	20	25	30	35
Frequency	1	3	13	17	27	36	38

Answer:

Converting the Cumulative series of 'less than' type into a simple frequency distribution.

Size	Cumulative Frequency	Frequency
0 − 5 5 − 10 10 − 15 15 − 20	1 3 13 17(c.f.)	1 3 − 1 = 2 13 − 3 = 10 17 − 13 = 4
(l₁)20 − 25	27	27 − 17 = 10 (f)
25 − 30 30 − 35	36 38	36 − 27 = 9 38 − 36 = 2
		$Σ f = N = 38$

Median = Size of

{(\frac{N}{2})}^{th}

item
or, M = Size of

{(\frac{38}{2})}^{th}

item
or, M = Size of 19^th item

Hence, median lies in class interval 20 − 25

Median or M = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i or, M = 20 + \frac{19 - 17}{10} \times 5 or, M = 20 + \frac{2}{10} \times 5 or, M = 20 + \frac{10}{10} or, M = 20 + 1 \Rightarrow M = 21

Hence, the median value of the above series is 21.

Question 3:

Find out median of the following data-set:

Class Interval	60−69	50−59	40−49	30−39	20−29	10−19
Frequency	13	15	21	20	19	12

Answer:

This is an inclusive series given in the descending order. It should first be converted into an exclusive series and placed in the ascending order.

Class Interval	Frequency	Cumulative Frequency
9.5 − 19.5 19.5 − 29.5	12 19	12 12 + 19 = *31 (c.f.)*
(l₁)29.5 − 39.5	20 (f)	31 + 20 = 51
39.5 − 49.5 49.5 − 59.5 59.5 − 69.5	21 15 13	51 + 21 = 72 72 + 15 = 87 87 + 13 = 100
	N = 100

Median = Size of

{(\frac{N}{2})}^{th}

item
= Size of

{(\frac{100}{2})}^{th}

item
= Size of 50^th item

Hence, median lies in the class interval 29.5 − 39.5

Median or M = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i or, M = 29.5 + \frac{50 - 31}{20} \times 10 or, M = 29.5 + \frac{19}{20} \times 10 or, M = 29.5 + \frac{190}{20} or, M = 29.5 + 9.5 \Rightarrow M = 39

Hence, the median value of the above series is 39.

Question 4:

Calculate median from the following series:

Class Interval	10−20	20−40	40−70	70−120	120−140
Frequency	4	10	26	8	2

Answer:

Class Interval	Frequency	Cumulative Frequency
10 − 20 20 − 40	4 10	4 *4 + 10 = 14 (c.f.)*
(l₁) 40 − 70	26 (f)	14 + 26 = 40
70 − 120 120 − 140	8 2	40 + 8 = 48 48 + 2 = 50
	$Σ f$ =N = 50

Median = Size of

(\frac{N}{2}) th

item
= Size of

(\frac{50}{2}) th

item
= Size of 25th item

Hence, median lies in class interval 40 − 70

Median or M = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i or, M = 40 + \frac{25 - 14}{2} \times 30 or, M = 40 + \frac{11 \times 30}{26} or, M = 40 + 12.69 \Rightarrow M = 52.69

Hence, the median value of the above series is 52.69

Question 5:

Calculate median from the following data-set:

Marks	Number of Students	Marks	Number of Students
1−5 6−10 11−15 16−20 21−25	7 10 16 32 24	26−30 31−35 36−40 41−45	18 10 5 1

Answer:

Coverting the Inclusive series into exclusive series.

Marks	Frequency	Cumulative Frequency
0.5 − 5.5 5.5 − 10.5 10.5 − 15.5	7 10 16	7 7 + 10 = 17 *17 + 16 = 33 (c.f.)*
(l_₁)15.5 − 20.5	32 (f)	33 + 32 = 65
20.5 − 25.5 25.5 − 30.5 30.5 − 35.5 35.5 − 40.5 40.5 − 45.5	24 18 10 5 1	65 + 24 = 89 89 + 18 = 107 107 + 10 = 117 117 + 5 = 122 122 + 1 = 123
	$Σ f =$ N = 123

Median = Size of

{(\frac{N}{2})}^{th}

item
= Size of

{(\frac{123}{2})}^{th}

item
= Size of 61.5^th item

Hence, the median lies in the class 15.5 − 20.5

Median or M = l_{1} + \frac{\frac{N}{2} - c . f}{f} \times i or, M = 15.5 + \frac{61.5 - 33}{32} \times 5 or, M = 15.5 + \frac{28.5 \times 5}{32} or, M = 15.5 + \frac{142.5}{32} or, M = 15.5 + 4.45 \Rightarrow M = 19.95 marks

Hence, the median marks of the above data-set is 19.95

Question 1:

Find out median of the following series, using graphic technique:

Class Interval	40−45	45−50	50−55	55−60	60−65	65−70	70−75	75−80	80−85
Frequency	4	6	8	10	7	6	5	3	1

Answer:

Marks	Cumulative Frequency
Less than 45 Less than 50 Less than 55 Less than 60 Less than 65 Less than 70 Less than 75 Less than 80 Less than 85	4 4 + 6 = 10 10 + 8 = 18 18 + 10 = 28 28 + 7 = 35 35 + 6 = 41 41 + 5 = 46 46 + 3 = 49 49 + 1 = 50

So, median is 58.5.

Question 2:

Calculate median value of the following data-set using graphic technique:

Size	0−10	10−20	20−30	30−40	40−50	50−60
Frequency	3	10	20	7	6	4

Answer:

Size	Cumulative Frequency
Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60	3 3 + 10 = 13 13 + 20 = 33 33 + 7 = 40 40 + 6 = 46 46 + 4 = 50

So, median is 26.

Question 3:

Graph the following data in the form of 'less than' and 'more than' ogives; and calculate the median value through the graph:

Marks	0−5	5−10	10−15	15−20	20−25	25−30	30−35	35−40
Number of Students	7	10	20	13	17	10	14	9

Answer:

Estimation of the median

Less than and More than Ogive

Marks	Cumulative Frequency
Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40	7 7 + 10 = 17 17 + 20 = 37 37 + 13 = 50 50 + 17 = 67 67 + 10 = 77 77 + 14 = 91 91 + 9 = 100

Marks	Cumulative Frequency
More than 0 More than 5 More than 10 More than 15 More than 20 More than 25 More than 30 More than 35 More than 40	100 100 − 7 = 93 93 − 10 = 83 83 − 20 = 63 63 − 13 = 50 50 − 17 = 33 33 − 10 = 23 23 − 14 = 9 9 − 9 = 0

So, median is 20.

Question 1:

Find out mode of the following series:
7, 4, 10, 9, 15, 12, 7, 9, 7

Answer:

In order to find the mode in an individual series we first need to arrange the series in an ascending order i.e.
4, 7, 7, 7, 9, 9, 10, 12, 15

An inspection of the series shows that 7 occurs most frequently in the series i.e. 3 times

Hence, Mode (Z) of the above series is 7.

Question 2:

Weight of 50 students is given below. Calculate mode.

Weight (in kg)	48	49	50	51	52	53
Number of Students	4	10	20	11	3	2

Answer:

Weight	No. of student (frequency)
48 49 50 51 52 53	4 10 20 11 3 2

An inspection of the series reveals that 50 is the value with highest frequency of 20.

Hence, 50 is the Mode (Z) of the series.

Question 1:

Calculate mode of the following series:

Class Interval	0−5	5−10	10−15	15−20	20−25	25−30
Frequency	20	24	32	28	20	26

Answer:

Class Interval	Frequency
0 − 5 5 − 10 (l₁)10 − 15 15 − 20 20 − 25 25 − 30	20 24 f₀ 32 f₁ 28 f₂ 20 26

A glance at the series reveal that 10 − 15 is the modal class because it has the maximum frequency i.e. 32. Therefore,the modal value will be:

Z = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 10 + \frac{32 - 24}{2 (32) - 24 - 28} \times 5 or, Z = 10 + \frac{8}{64 - 52} \times 5 or, Z = 10 + \frac{40}{12} or, Z = 10 + 3.33 \Rightarrow Z = 13.33

Hence, the Mode (Z) of the above series is 13.33

Question 2:

Calculate mode, given the following distribution of data:

Class Interval	4−8	8−12	12−16	16−20	20−24	24−28	28−32	32−36	36−40
Frequency	10	12	16	14	10	8	17	5	4

Answer:

Grouping table and subsequent analysis table are as follows:

Grouping table:

Analysis Table:

Column	Class interval corresponding to highest frequencies in the grouping table
Column	4 − 8	8 − 12	12 − 16	16 − 20	20 − 24	24 − 28	28 − 32	32 − 26	36 − 40
I II III IV V VI	✓	✓ ✓ ✓	✓ ✓ ✓ ✓ ✓	✓ ✓ ✓	✓		✓
Total	1	3	5	3	1	0	1	0	0

Analysis table shows that of all the class intervals, 12 − 16 has the highest frequency of 5. It has got the maximum (✓) marks. Accordingly, 12 − 16 is the model class interval of the series.
lower limit of the modal group (l₁)= 12
f₁= frequency of the modal class=16
f₀= frequency of class preceding the modal class (8-12)=12
f₂= frequency of class following the modal class (16-20)=14
i=class interval=16-12=4

Z can be calculated by applying the following formulae:

Z = l_{1} = \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i

or, Z = 12 + \frac{16 - 12}{2 (16) - 12 - 14} \times 4 or, Z = 12 + \frac{4 \times 4}{32 - 26} or, Z = 12 + \frac{16}{6} or, Z = 12 + 2.67 \Rightarrow Z = 14.67

Hence, the mode (Z) of the above series= 14.67.

Question 1:

Calculate mode of the following series:

Class Interval	30−59	60−89	90−119	120−149	150−179	180−209	210−239
Frequency	4	7	12	15	18	6	5

Answer:

Class Interval	Exclusive Class interval	Frequency
30 − 59 60 − 89 90 − 119 120 − 149	29.5 − 59.5 59.5 − 89.5 89.5 − 119.5 119.5 − 149.5	4 7 12 15 f₀
150 − 179	(l₁) 149.5 − 179.5	18 f₁
180 − 209 210 − 239	179.5 − 209.5 209.5 − 239.5	6 f₂ 5

A glance at the above table reveals that 149.5 − 179.5 is the model class interval as it has the highest frequency i.e.18.

Calculating the mode of the series:

Z = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 149.5 + \frac{18 - 15}{2 (18) - 15 - 6} \times 30 or, Z = 149.5 + \frac{3}{36 - 21} \times 30 or, Z = 149.5 + \frac{3 \times 30}{15} or, Z = 149.5 + 6 \Rightarrow Z = 155.5

Hence, the modal value of the above series is 155.5

Question 2:

The following data-set gives mid-values and frequencies. Calculate its mode.

Mid-value	5	10	15	20	25	30	35	40	45
Frequency	7	13	19	24	32	28	17	8	6

Answer:

A series with 'mid-values' is to be expanded as a series with class intervals, given as below:

Mid value	Class interval	Frequency
5 10 15 20	2.5 − 7.5 7.5 − 12.5 12.5 − 17.5 17.5 − 22.5	7 13 19 24 f₀
25	(l₁) 22.5 − 27.5	32 f₁
30 35 40 45	27.5 − 32.5 32.5 − 37.5 37.5 − 42.5 42.5 − 47.5	28 f₂ 17 8 6

As class interval 22.5 − 27.5 is the one with highest frequency. Therefore, this is the modal class interval.

Calculating the mode of the above series:

Z = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 22.5 + \frac{32 - 24}{2 (32) - 24 - 28} \times 5 or, Z = 22.5 + \frac{8}{64 - 52} \times 5 or, Z = 22.5 + \frac{40}{12} or, Z = 22.5 + 3.33 \Rightarrow Z = 25.83

Hence, the Mode (Z) = 25.83

Question 2:

Median and mean weight of the students of a class are 35.83 and 37.06 respectively. Calculate the mode.

Answer:

Given,

Median (M) = 35.83
Mean $(X)$ = 37.06
Mode = Z

We know,
Z = 3M − 2 $(X)$

or, Z = 3(35.83) − 2(37.06)
or, Z = 107.49 − 74.12
$\Rightarrow$ Z = 33.37

Hence, the mode weight of the students of a class is 33.37 kgs.

Question 3:

If median and mean of a distribution are 18.8 and 20.2 respectively, what would be its mode?

Answer:

Given,

Median (M)=18.8
Mean $(X)$ = 20.2
Mode = Z

We know,
Z = 3 Median − 2 Mean

or, Z = 3 (18.8) − 2 (20.2)
or, Z = 56.4 − 40.4
$\Rightarrow$ Z = 16

Hence, mode (Z) of the above distribution is 16.

Question 1:

Present the following information in the form of a histogram and locate the modal value. Give a cross-check to your answer, calculating mode through its standard formula.

Class Interval	0−10	10−20	20−30	30−40	40−50	50−60	60−70
Frequency	4	8	14	20	30	15	6

Answer:

Class Interval	Frequency
0 − 10 10 − 20 20 − 30 30 − 40	4 8 14 20 f₀
(l₁) 40 − 50	30 f₁
50 − 60 60 − 70	15 f₂ 6

Mode (Z) value is 44.

Cross Check
A glance at the above table reveals that 40 − 50 is the modal class interval as it claims the highest frequency i.e. 30

Z = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 40 + \frac{30 - 20}{2 (30) - 20 - 15} \times 10 or, Z = 40 + \frac{10}{60 - 35} \times 10 or, Z = 40 + \frac{100}{25} or, Z = 40 + 4 \Rightarrow Z = 44

Hence, Mode (Z) is 44.

Question 2:

Find out mode of the following distribution, using histogram.

Age	20−25	25−30	30−35	35−40	40−45	45−50
Frequency	50	70	100	180	150	120

Answer:

Age	Frequency
20 − 25 25 − 30 30 − 35 35 − 40 40 − 45 45 − 50	50 70 100 180 150 120

Hence, Mode (Z) is 38.6

Question 3:

Find out mode of the following data with the help of histogram.

Marks	0−10	10−20	20−30	30−40	40−50	50−60
Frequency	6	8	16	25	12	8

Answer:

Marks	Frequency
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60	6 8 16 25 12 8

Mode (Z) is 34.1.

Question 1:

Given below is the data of the age of 9 children of a street. Find the median.
5, 8, 7, 3, 4, 6, 2, 9, 1

Answer:

First of all, we need to arrange the given data in an ascending order. Thus, the data is presented below as:
1, 2, 3, 4, 5, 6, 7, 8, 9
N = 9
Median = size of ${(\frac{N + 1}{2})}^{t h}$ item
or, Median = size of ${(\frac{9 + 1}{2})}^{t h}$ item
Thus, Median is given by the size of 5^th item. Now, let's locate the fifth item in the data (as arranged in the ascending order)
$1, 2, 3, 4, 5, 6, 7, 8, 9$
Therefore, Median of the data so given is 5.

Question 2:

Find the median of the following values:
30, 20, 15, 10, 25, 35, 18, 21, 28, 40, 36

Answer:

First of all, we need to arrange the given data in an ascending order. Thus, the data is presented below as:
10, 15, 18, 20, 21, 25, 28, 30, 35, 36, 40
N = 11
Median = size of ${(\frac{N + 1}{2})}^{t h}$ item
or, Median = size of ${(\frac{11 + 1}{2})}^{t h}$ item
Thus, Median is given by the size of 6^th item. Now, let's locate the sixth item in the data (as arranged in the ascending order)
$10, 15, 18, 20, 21, 25, 28, 30, 35, 36, 40$
Therefore, Median of the data so given is 25.

Question 3:

Find out median of the series of the following table:\

Items	3	4	5	6	7	8
Frequency	6	9	11	14	23	10

Answer:

Items	Frequency (f)	Cumulative Frequency (c.f.)
3 4 5 6 7 8	6 9 11 14 23 10	6 15 26 40 63 73





	N = 73

Median = size of

{(\frac{N + 1}{2})}^{t h}

item
or, Median = size of

{(\frac{73 + 1}{2})}^{t h}

item
or, Median = size of

{(\frac{74}{2})}^{t h}

item = size of 37^th item.
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 37^th is 40 (in the c.f. column), which is corresponding to 6. Hence, median is 6.

Question 4:

Data relating to wages of some workers are given below. Find out median wage.

Wages (₹)	20−30	30−40	40−50	50−60	60−70
Number of Workers	25	12	15	13	5

Answer:

Wages	No. of Workers (F)	Cumulative Frequency (c.f.)
20 − 30 30 − 40 40 − 50 50 − 60 60 − 70	25 12 (f) 15 13 5	25 37 52 65 70




	N = ∑f = 70

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{70}{2})}^{t h}

item, which is 35^th item.
This corresponds to the class interval of 30 − 40, so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = l_{1} + \frac{\frac{70}{2} - c . f .}{f} \times i or, Median = 30 + \frac{35 - 25}{12} \times 10 or, Median = 30 + \frac{100}{12} Thus, Median = 30 + 8.33 = 38.33

Question 5:

The following table expresses the age of eight students. Find the median age.

S. No.	1	2	3	4	5	6	7	8
Age (Years)	18	16	14	11	13	10	9	2

Answer:

This is an individual series, hence, first of all, we need to arrange the data in the ascending order.

S.No.	Ages
1 2 3 4 5 6 7 8	2 9 10 11 13 14 16 18

Here, the number of observation is 8 (even).

Median = \frac{size of {(\frac{N}{2})}^{t h} item + size of {(\frac{N}{2} + 1)}^{t h} item}{2} = \frac{size of {(\frac{8}{2})}^{t h} item + size of {(\frac{8}{2} + 1)}^{t h} item}{2} = \frac{size of 4^{t h} item + size of 5^{t h} item}{2} = \frac{11 + 13}{2} = 12 Hence, Median = 12 years

Question 6:

Number of persons living in a house is reported to be as under 500 houses in a village. Find the median number of persons in a house in the village.

Number of Persons in a House	1	2	3	4	5	6	7	8	9	10
Number of Houses	26	113	120	95	60	42	21	14	5	4

Answer:

No. of Persons in a House	No. of Houses (f)	Cumulative Frequency *(c.f.)*
1 2 3 4 5 6 7 8 9 10	26 113 120 95 60 42 21 14 5 4	26 139 259 354 414 456 477 491 496 500









	∑f = 500

Median = size of

{(\frac{N + 1}{2})}^{t h}

item
or, Median = size of

{(\frac{500 + 1}{2})}^{t h}

item
or, Median = size of 250.5^th item
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 250.5^th is 259 (in the c.f. column), which is corresponding to 3.
Hence, median is 3.

Question 7:

Find out median of the following series:

Size	15	20	25	30	35	40
Frequency	10	15	25	5	5	20

Answer:

Size	Frequency (f)	Cumulative Frequency (c.f.)
15 20 25 30 35 40	10 15 25 5 5 20	10 25 50 55 60 80





	N = 80

Median = size of

{(\frac{N + 1}{2})}^{t h}

item
or, Median = size of

{(\frac{80 + 1}{2})}^{t h}

item
or, Median = size of 40.5^th item
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 40.5^th is 50 (in the c.f. column), which is corresponding to 25.
Hence, median is 25.

Question 8:

Distribution of marks obtained by 100 students of a class is given below. Find out the median marks.

Marks	0	5	10	15	20	25	30	35	40	45
Number of Students	4	6	15	5	8	12	28	14	3	5

Answer:

Marks	No. of Student (f)	Cumulative Frequency (c.f.)
0	4	4
5	6	10
10	15	25
15	5	30
20	8	38
25	12	50
30	28	78
35	14	92
40	3	95
45	5	100
	N = 100

Median = size of

{(\frac{N + 1}{2})}^{t h}

item
or, Median = size of

{(\frac{100 + 1}{2})}^{t h}

item
or, Median = size of 50.5^th item
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 50.5^th is 78 (in the c.f. column), which is corresponding to 30.
Hence, median is 30.

Question 9:

Find out median wage rate from the following data-set:

Wage Rate (₹)	5−15	15−25	25−35	35−45	45−55	55−65
Number of Workers	4	6	10	5	3	2

Answer:

Wages	No. of Worker (f)	Cumulative Frequency *(c.f.)*
5 − 15 15 − 25 25 − 35 35 − 45 45 − 55 55 − 65	4 6 10 5 3 2	4 10 20 25 28 30
	N = ∑f = 30

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{30}{2})}^{t h}

item, which is 15^th item.
This corresponds to the class interval of (25 − 35), so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = l_{1} + \frac{\frac{30}{2} - c . f .}{f} \times i or, Median = 25 + \frac{15 - 10}{10} \times 10 or, Median = 25 + 5 Thus, Median is 30

Question 10:

Find out median of the following series:

Wage Rate (₹)	25−30	20−25	15−20	10−15	5−10	0−5
Number of Workers	5	10	20	5	8	2

Answer:

Wages	No. of Worker (f)	Cumulative Frequency (c.f.)
0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30	2 8 5 20 (f) 10 5	2 10 15 (c.f.) 35 45 50
	N = ∑f = 50

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{50}{2})}^{t h}

item, which is 25^th item.
This corresponds to the class interval of (15 − 20), so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = l_{1} + \frac{\frac{50}{2} - c . f .}{f} \times i or, Median = 15 + \frac{25 - 15}{20} \times 5 or, Median = 15 + \frac{50}{20} Thus, Median = 15 + 2.5 = 17.5

Question 11:

Calculate the median from the following series:

Age (Years)	55−60	50−55	45−50	40−45	35−40	30−35	25−30	20−25
Number of Students	7	13	10	15	30	33	28	14

Answer:

Age	No. of students (f)	Cumulative Frequency *(c.f.)*
20 − 25 25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 50 − 55 55 − 60	14 28 33 (f) 30 15 10 13 7	14 42 (c.f.) 75 105 120 130 143 150







	N = ∑f = 150

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{150}{2})}^{t h}

item, which is 75^th item.
This corresponds to the class interval of (30 − 35), so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = l_{1} + \frac{\frac{150}{2} - c . f .}{f} \times i or, Median = 30 + \frac{75 - 42}{33} \times 5 or, Median = 30 + \frac{33}{33} \times 5 Thus, Median = 35

Question 12:

50 students of economics, secured the following marks in an examination:

Marks	20−25	25−30	30−35	35−40	40−45	45−50	50−55	55−60	60−65	65−70
Students	6	3	7	4	6	4	2	8	3	7

Calculate median.

Answer:

Marks	Student (f)	Cumulative Frequency (c.f.)
20 − 25 25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 50 − 55 55 − 60 60 − 65 65 − 70	6 3 7 4 6(f) 4 2 8 3 7	6 9 16 20 (c.f.) 26 30 32 40 43 50
	N = ∑f = 50

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{50}{2})}^{t h}

item, which is 25^th item.
This corresponds to the class interval of (40 − 45), so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = l_{1} + \frac{\frac{50}{2} - c . f .}{f} \times i or, Median = 40 + \frac{25 - 20}{6} \times 5 or, Median = 40 + \frac{25}{6} Thus, Median = 40 + 4.16 = 44.16

Question 13:

Given the following data, find out median:

Age	20−25	25−30	30−35	35−40	40−45	45−50	50−55	55−60
Number of Students	50	70	100	180	150	120	70	60

Answer:

Age	No. of Students (f)	Cumulative Frequency (c.f.)
20 − 25 25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 50 − 55 55 − 60	50 70 100 180 (f) 150 120 70 60	50 120 220 (c.f.) 400 550 670 740 800
	N = ∑f = 800

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{800}{2})}^{t h}

item, which is 400^th item.
This corresponds to the class interval of (35 − 40), so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = l_{1} + \frac{\frac{800}{2} - c . f .}{f} \times i or, Median = 35 + \frac{400 - 220}{180} \times 5 or, Median = 35 + \frac{180}{180} \times 5 Thus, Median = 35 + 5 = 40

Question 15:

Calculate median, given the following data:

Mid-value	20	30	40	50	60	70
Male (c.f.)	12	25	42	46	48	50

c.f. = Cumulative Frequency

Answer:

For the calculation of median, the given mid-values must be converted into class intervals using the following formula.
$Value of adjustment = \frac{Mid - point of one class - Mid - point of preceeding class}{2}$
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 Value	Class Interval	Cumulative Frequency *(c.f.)*	Frequency (f)
20 30 40 50 60 70	15 − 25 25 − 35 35 − 45 45 − 55 55 − 65 65 − 75	12 (c.f.) 25 42 46 48 50	12 25 − 12 = 13 (f) 42 − 25 = 17 46 − 42 = 4 48 − 46 = 2 50 − 48 = 2
			N = ∑f = 50

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{50}{2})}^{t h}

item, which is 25^th item.
This corresponds to the class interval of (25 − 35), so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = l_{1} + \frac{\frac{50}{2} - c . f .}{f} \times i or, Median = 25 + \frac{25 - 12}{13} \times 10 or, Median = 25 + \frac{130}{13} Thus, Median = 25 + 10 = 35

Question 16:

Calculate mode of the following series using the graphic technique. Counter check the modal value with the formula.

Wage	0−10	10−20	20−30	30−40	40−50
Number of Workers	28	46	54	42	30

Answer:

Wages (in Rs)	No. of Workers (f)
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50	28 46 = f₀ 54 = f₁ 42 = f₂ 30

By inspection, we can say that the modal class is 20 – 30 as it has the highest frequency of 54.

Mode (Z) = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 20 + \frac{54 - 46}{2 (54) - 46 - 42} \times 10 or, Z = 20 + \frac{8}{108 - 88} \times 10 or, Z = 20 + \frac{8}{20} \times 10 or, Z = 20 + \frac{8}{20} \times 10 or, Z = 20 + \frac{80}{20} = 24 Hence, Mode is 24

Question 17:

Calculate mode from the following data:

Wages	25	50	75	80	85	90
Number of Workers	4	6	9	3	2	1

Answer:

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, 75 is regarded as the modal value, as it has the highest frequency (of 9 times).
Therefore, mode (Z) is 75

Question 18:

Find out mode from the following data:

Class Interval	5−10	10−15	15−20	20−25	25−30	30−35	35−40
Number of Children	4	5	3	2	6	7	3

Answer:

Class Interval	No. of Children (f)
5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 35 − 40	4 5 3 2 6 = f₀ 7 = f₁ 3 = f₂

By inspection, we can say that the modal class is (30 – 35) as it has the highest frequency of 7.

Mode (Z) = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 30 + \frac{7 - 6}{2 (7) - 6 - 3} \times 5 or, Z = 30 + \frac{1}{14 - 9} \times 5 or, Z = 30 + \frac{1}{5} \times 5 or, Z = 30 + 1 = 31 Hence, mode (Z) is 31

Question 19:

Calculate mode of the following series, using grouping method:

Size	40	44	48	52	56	60	64	68	72	76
Frequency	10	12	14	20	15	20	18	10	8	4

Answer:

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

From the analysis table, it is clear that 60 repeats the maximum number of times. Thus, mode is 60.

Question 14:

Find out median, with the help of the following data:

Price Level	10−20	20−30	30−40	40−50	50−60	60−70
Number of Commodity	2	5	8	4	6	3

Answer:

Price Level	No. of Commodity (f)	Cumulative Frequency (c.f.)
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70	2 5 8(f) 4 6 3	2 7(c.f) 15 19 25 28
	N = ∑f = 28

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{28}{2})}^{t h}

item, which is 14^th item.
This corresponds to the class interval of (30 − 40), so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = l_{1} + \frac{\frac{28}{2} - c . f .}{f} \times i or, Median = 30 + \frac{14 - 7}{8} \times 10 or, Median = 30 + \frac{7}{8} \times 10 Thus, Median = 30 + 8.75 = 38.75

Question 20:

Calculate mode of the following distribution:

Marks	10−19	20−29	30−39	40−49	50−59	60−69	70−79	80−89	90−99
Number of Students	29	87	181	247	263	133	40	9	2

Answer:

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.
$Value of Adjustment = \frac{Lower limit of one class - Upper limit of the preceeding class}{2}$ 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.

Class Interval	Exclusive Class Interval	Frequency (f)
10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 70 − 79 80 − 89 90 − 99	9.5 − 19.5 19.5 − 29.5 29.5 − 39.5 39.5 − 49.5 49.5 − 59.5 59.5 − 69.5 69.5 − 79.5 79.5 − 89.5 89.5 − 99.5	29 87 181 247 = f₀ 263 = f₁ 133 = f₂ 40 9 2

By inspection, we can say that the modal class is (49.5 – 59.5) as it has the highest frequency of 263.

Mode (Z) = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 49.5 + \frac{263 - 247}{2 (263) - 247 - 133} \times 10 or, Z = 49.5 + \frac{16}{526 - 380} \times 10 or, Z = 49.5 + \frac{16}{146} \times 10 or, Z = 49.5 + 1.09 = 50.59 Hence, mode is 50.59

Question 21:

Find out mode, given the following information:

Size	6−10	11−15	16−20	21−25	26−30
Frequency	20	30	50	40	10

Answer:

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.
$Value of Adjustment = \frac{Value of lower limit of one class - Value of upper limit of the preceeding class}{2}$
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.

Size	Exclusive Class Interval	Frequency (f)
6 − 10 11 − 15 16 − 20 21 − 25 26 − 30	5.5 − 10.5 10.5 − 15.5 15.5 − 20.5 20.5 − 25.5 25.5 − 30.5	20 30 f₀ 50 f₁ 40 f₂ 10

By inspection, we can say that the modal class is (15.5 – 20.5) as it has the highest frequency of 50.

Mode (Z) = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 15.5 + \frac{50 - 30}{2 (50) - 30 - 40} \times 5 or, Z = 15.5 + \frac{20}{100 - 70} \times 5 or, Z = 15.5 + \frac{100}{30} or, Z = 15.5 + 3.33 = 18.33 Hence, mode is 18.33

Question 22:

Calculate mode from the following data:

Wages (₹)	Number of Workers
Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80	15 35 60 84 96 127 198 250

Answer:

To calculate the value of mode, we first convert the given less than cumulative frequency distribution into a simple frequency distribution as follows.

Wages	No. of Wages (f)
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	15 35 − 15 = 20 60 − 35 = 25 84 − 60 = 24 96 − 84 = 12 127 − 96 = 31(f₀) 198 − 127 = 71 (f₁) 250 − 198 = 52 (f₂)

By inspection, we can say that the modal class is 60 – 70 as it has the highest frequency of 71.

Mode (Z) = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 60 + \frac{71 - 31}{2 (71) - 31 - 52} \times 10 or, Z = 60 + \frac{40}{142 - 83} \times 10 or, Z = 60 + \frac{400}{59} or, Z = 60 + 6.78 = 66.78 Hence, Mode is 66.78

Question 23:

Calculate mode from the following series:

Size	1	2	3	4	5	6	7	8	9	10
Frequency	8	6	10	12	20	12	5	3	2	4

Answer:

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, 5 is regarded as the modal value, as it has the highest frequency (of 20 times).
Therefore, mode (Z) is 5.

Question 24:

Calculate the mean, median and mode of the number of persons per house in a village with the help of the following information:

Number of Persons per House	1	2	3	4	5	6	7	8	9	10
Number of Houses	26	113	120	95	60	42	21	14	5	4

Answer:

No. of Persons per House (X)	No. of Houses (f)	fx	Cumulative Frequency (c.f.)
1 2 3 4 5 6 7 8 9 10	26 113 120 95 60 42 21 14 5 4	26 226 360 380 300 252 147 112 45 40	26 139 259 354 414 456 477 491 496 500
	N = ∑f = 500	∑fx = 1888

Mean Mean (\bar{X}) = \frac{Σ f x}{Σ f} = \frac{1888}{500} = 3.78

Median Median = size of {(\frac{N + 1}{2})}^{t h} item = size of {(\frac{500 + 1}{2})}^{t h} item = size of 250 . 5^{t h} item

Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 250.5^th is 259 (in the c.f. column), which is corresponding to 3. Hence, median is 3.

Mode

The data given in the question is a discrete series. Therefore, using the inspection method, we can say that 3 is mode of the given series. This is becasue 3 has the highest frequency of 120 times.
Therefore, Mode (Z) is 3.

Question 25:

Calculate the median and mode from the following data:

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Students	2	18	30	45	35	20	6	3

Answer:

Marks	No. of Workers (f)	Cumulative Frequency (c.f.)
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	2 18 30 = f₀ 45 = f₁ 35 = f₂ 20 6 3	2 20 50 95 130 150 156 159
	N = ∑f = 159

Median

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{159}{2})}^{t h}

item, which is 79.5^th item.
This corresponds to the class interval of (30 − 40), so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = 30 + \frac{\frac{159}{2} - 50}{45} \times i or, Median = 30 + \frac{29.5}{45} \times 10 or, Median = 30 + 6.55 Thus, Median = 36.55

Mode

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

Mode (Z) = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 30 + \frac{45 - 30}{2 (45) - 30 - 35} \times 10 or, Z = 30 + \frac{15}{90 - 65} \times 10 or, Z = 30 + \frac{150}{25} = 30 + 6 = 36 Hence, mode (Z) is 36

Question 26:

Calculate the median value, given the following statistical information:

Age	20−25	25−30	30−35	35−40	40−45	45−50	50−55	55−60
Number of Students	50	70	100	180	150	120	70	60

Answer:

Age	No. of Students (f)	Cumulative Frequency (c.f.)
20 − 25 25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 50 − 55 55 − 60	50 70 100 180 (f) 150 120 70 60	50 120 220 (c.f.) 400 550 670 740 800
	∑f = 800

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{800}{2})}^{t h}

item, which is 400^th item.
This corresponds to the class interval of (35 − 45), so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = 35 + \frac{\frac{800}{2} - 220}{180} \times 5 or, Median = 35 + \frac{180}{180} \times 5 or, Median = 35 + 5 Thus, Median is 40 .

Question 27:

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

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Students	5	7	15	25	20	15	8	5

Answer:

Marks	Mid Values (m)	No. of Workers (f)	Cumulative Frequency *(c.f.)*	fm
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	5 15 25 35 45 55 65 75	5 7 15 25 20 15 8 5	5 12 27 52 72 87 95 100	25 105 375 875 900 825 520 375
	15	N = ∑f = 100		∑fm = 4000

Mean Mean (\bar{X}) = \frac{Σ f m}{Σ f} = \frac{4000}{100} = 40

Median

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{100}{2})}^{t h}

item, which is 50^th item.
This corresponds to the class interval of (30 − 40), so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = l_{1} + \frac{\frac{100}{2} - c . f .}{f} \times i or, Median = 30 + \frac{50 - 27}{25} \times 10 or, Median = 30 + \frac{230}{25} Thus, median = 30 + 9.2 = 39.2

Mode

By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 25.

Mode (Z) = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 30 + \frac{25 - 15}{2 (25) - 15 - 20} \times 10 or, Z = 30 + \frac{10}{50 - 35} \times 10 or, Z = 30 + \frac{100}{15} or, Z = 30 + 6.6 = 36.6 Hence, mode (Z) is 36.6

Question 28:

Calculate median in an asymmetrical distribution if mode is 83 and arithmetic mean is 92.

Answer:

Given:
Mode = 83
Mean = 92
Median = ?

Recalling the empirical relationship between mean, median and mode, we can express it algebraically as below.
Mode = 3(Median) – 2(Mean)
Substituting the given values in the formula.

or, 83 = 3 (Median) – 2(92)
or, 3 (Median) = 83 + 184
or, Median $= \frac{267}{3} = 89$
Hence, median is equal to 89

Question 29:

Calculate mode when arithmetic mean is 146 and median is 130.

Answer:

Given:
Mean = 146
Median = 130
Mode = ?

As per the empirical relationship between mean, median and mode:
Mode = 3(Median) – 2(Mean)
Substituting the given values in the formula.

Mode = 3(130) – 2(146)
or, Mode = 390 – 292
Hence, Mode is 98.

Question 30:

If mode is 63 and median is 77, calculate arithmetic mean.

Answer:

Given:
Mode = 63
Median = 77
Mean = ?

As per the empirical relationship between mean, median and mode:
Mode = 3(Median) – 2(Mean)
Substituting the given values in the formula.
63 = 3 (77) – 2(Mean)
or, 2 (Mean) = 231 – 63
Mean $= \frac{168}{2} = 84$
Hence, mean is 84.

Question 31:

Calculate arithmetic mean, median and mode of the following series:

Marks	Number of Students
Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80	12 26 40 58 80 110 138 150

Answer:

Marks	Mid Point (m)	Cumulative Frequency	Frequency (f)	fm
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	5 15 25 35 45 55 65 75	12 26 40 58 80 110 138 150	12 14 (=26 − 12) 14 (=40 − 26) 18 (=58 − 40) 22 (=80 − 58) 30 (=110 − 80) 28 (=138 − 110) 12 (=150 − 138)	60 210 350 630 990 1650 1820 900
			∑f = 150	∑fm = 6610

Mean Mean (\bar{X}) = \frac{Σ f m}{Σ f} = \frac{6610}{150} = 44.07

Median

Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{150}{2})}^{t h}

item, which is 75^th item.
This corresponds to the class interval of 40 − 50, so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = 40 + \frac{\frac{150}{2} - 58}{22} \times 10 or, Median = 40 + \frac{170}{22} or, Median = 40 + 7.73 Thus, median is 47.73

Mode

By inspection, we can say that the modal class is (50 – 60) as it has the highest frequency of 30.

Mode (Z) = l_{1} + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times i or, Z = 50 + \frac{30 - 22}{2 (30) - 22 - 28} \times 10 or, Z = 50 + \frac{8}{60 - 50} \times 10 or, Z = 50 + \frac{8}{10} \times 10 or, Z = 50 + 8 = 58 Hence, mode (Z) is 58 .

Question 1:

Median and mean values of the marks obtained by the students of a class are 46.67 and 45.5 respectively. Find out mode of the marks.

Answer:

Given,

Median (M) = 46.67
Mean $(X)$ = 45.5
Mode = Z

We know,
Z = 3M − 2 $(X)$

or, Z = 3(46.67) − 2(45.5)
or, Z = 140.01 − 91
$\Rightarrow$ Z = 49.01

Hence, Mode (Z) of the marks is 49.01

Tr Jain & Vk Ohri (2017) for Class 11 Humanities Economics Chapter 10 - Measures Of Central Tendency Median And Mode

Page No 184:

Question 1:

Answer:

Page No 184:

Question 2:

Answer:

Page No 184:

Question 3:

Answer:

Page No 185:

Question 1:

Answer:

Page No 185:

Question 2:

Answer:

Page No 187:

Question 1:

Answer:

Page No 187:

Question 2:

Answer:

Page No 192:

Question 1:

Answer:

Page No 192:

Question 2:

Answer:

Page No 192:

Question 3:

Answer:

Page No 192:

Question 4:

Answer:

Page No 192:

Question 5:

Answer:

Page No 195:

Question 1:

Answer:

Page No 195:

Question 2:

Answer:

Page No 195:

Question 3:

Answer:

Page No 206:

Question 1:

Answer:

Page No 206:

Question 2:

Answer:

Page No 209:

Question 1:

Answer:

Page No 209:

Question 2:

Answer:

Page No 213:

Question 1:

Answer:

Page No 213:

Question 2:

Answer:

Page No 214:

Question 2:

Answer:

Page No 214:

Question 3:

Answer:

Page No 215:

Question 1:

Answer:

Page No 215:

Question 2:

Answer:

Page No 215:

Question 3:

Answer:

Page No 237: