TR Jain & Vk Ohri (2017) Solutions for Class 11 Humanities Economics Chapter 10 Measures Of Central Tendency Median And Mode are provided here with simple step-by-step explanations. These solutions for Measures Of Central Tendency Median And Mode are extremely popular among class 11 Humanities students for Economics Measures Of Central Tendency Median And Mode Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the TR Jain & Vk Ohri (2017) Book of class 11 Humanities Economics Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s TR Jain & Vk Ohri (2017) Solutions. All TR Jain & Vk Ohri (2017) Solutions for class 11 Humanities Economics are prepared by experts and are 100% accurate.

#### 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.

 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 ${\left(\frac{N+1}{2}\right)}^{\mathrm{th}}$ item
or, M = size of ${\left(\frac{15+1}{2}\right)}^{\mathrm{th}}$ item
or, M = size of 8th item
$⇒$  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.

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

Median = Size of ${\left(\frac{N+1}{2}\right)}^{\mathrm{th}}$ item
or, M= Size of ${\left(\frac{7+1}{2}\right)}^{\mathrm{th}}$ item
or, M= size of 4th item
$⇒$ 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.

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

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

 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 ${\left(\frac{N+1}{2}\right)}^{\mathrm{th}}$ item
or,  M = Size of ${\left(\frac{15+1}{2}\right)}^{\mathrm{th}}$ item
or,  M = Size of 75.5th item

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

 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 ${\left(\frac{N+1}{2}\right)}^{\mathrm{th}}$ item
or, M = Size of ${\left(\frac{200+1}{2}\right)}^{\mathrm{th}}$ item
or, M = Size of 100.5th item

It shows that median value corresponds to the 100.5th item in the series. This item appears first of all in 135th cumulative frequency of the series. Therefore, median shall be the value corresponding to the 135th cumulative 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

 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 $\left(\frac{N}{2}\right)\mathrm{th}$ item
or, M = Size of $\left(\frac{79}{2}\right)\mathrm{th}$ item
or, M  = Size of 39.5th item

Hence, the median lies in the class 50-60

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

 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 ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item
or,   M = Size of ${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item
or,   M = Size of 50th item

Hence, median lies in the class 20-30

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

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 $\Sigma f\mathit{=}N\mathit{=}$100

Median or M = Size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item
or,   M = Size of ${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item
or,   M = Size of 50th item

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

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

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 $\mathrm{\Sigma }f=N=38$

Median = Size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item
or, M = Size of ${\left(\frac{38}{2}\right)}^{\mathrm{th}}$ item
or, M = Size of 19th item

Hence, median lies in class interval 20 − 25

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

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 ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item
= Size of ${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item
= Size of 50th item

Hence, median lies in the class interval 29.5 − 39.5

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

 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 $\Sigma f$=N = 50

Median = Size of $\left(\frac{N}{2}\right)\mathrm{th}$ item
= Size of $\left(\frac{50}{2}\right)\mathrm{th}$ item
= Size of 25th item

Hence, median lies in class interval 40 − 70

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

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.) (l₁)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 $\Sigma f=$N = 123

Median = Size of ${\left(\frac{\mathrm{N}}{2}\right)}^{\mathrm{th}}$ item
= Size of ${\left(\frac{123}{2}\right)}^{\mathrm{th}}$ item
= Size of 61.5th item

Hence, the median lies in the class 15.5 − 20.5

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

 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

 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

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

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

 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

 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:

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

Grouping table and subsequent analysis table are as follows:

Grouping table:

Analysis Table:

 Column Class interval corresponding to highest frequencies in the grouping table 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:

$\mathrm{Z}={l}_{1}=\frac{{f}_{1}-{f}_{0}}{2{f}_{1}-{f}_{0}-{f}_{2}}×i\phantom{\rule{0ex}{0ex}}$

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

 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:

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

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:

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.

Given,

Median (M) = 35.83
Mean $\left(\overline{)\mathrm{X}}\right)$ = 37.06
Mode = Z

We know,
Z = 3M − 2$\left(\overline{)\mathrm{X}}\right)$

or, Z = 3(35.83) − 2(37.06)
or, Z = 107.49 − 74.12
$⇒$ 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?

Given,

Median (M)=18.8
Mean $\left(\overline{)\mathrm{X}}\right)$ = 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
$⇒$  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

 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

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

 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

 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

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 ${\left(\frac{N+1}{2}\right)}^{th}$ item
or, Median = size of ${\left(\frac{9+1}{2}\right)}^{th}$ item
Thus, Median is given by the size of 5th item. Now, let's locate the fifth item in the data (as arranged in the ascending order)

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

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 ${\left(\frac{N+1}{2}\right)}^{th}$ item
or, Median = size of ${\left(\frac{11+1}{2}\right)}^{th}$ item
Thus, Median is given by the size of 6th item. Now, let's locate the sixth item in the data (as arranged in the ascending order)

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

 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 ${\left(\frac{N+1}{2}\right)}^{th}$item
or, Median = size of ${\left(\frac{73+1}{2}\right)}^{th}$item
or, Median = size of ${\left(\frac{74}{2}\right)}^{th}$ item = size of 37th item.
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 37th 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

 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 ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{70}{2}\right)}^{th}$ item, which is 35th item.
This corresponds to the class interval of 30 − 40, so this is the median class.

#### 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

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).

#### 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

 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 ${\left(\frac{N+1}{2}\right)}^{th}$item
or, Median = size of item
or, Median = size of 250.5th item
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 250.5th 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

 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 ${\left(\frac{N+1}{2}\right)}^{th}$item
or, Median = size of item
or, Median = size of 40.5th item
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 40.5th 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

 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 ${\left(\frac{N+1}{2}\right)}^{th}$item
or, Median = size of item
or, Median = size of 50.5th item
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 50.5th 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

 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 ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{30}{2}\right)}^{th}$ item, which is 15th item.
This corresponds to the class interval of (25 35), so this is the median class.

#### 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

 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 ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{50}{2}\right)}^{th}$ item, which is 25th item.
This corresponds to the class interval of (15 20), so this is the median class.

#### 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

 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 ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{150}{2}\right)}^{th}$ item, which is 75th item.
This corresponds to the class interval of (30 35), so this is the median class.

#### 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.

 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 ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{50}{2}\right)}^{th}$ item, which is 25th item.
This corresponds to the class interval of (40 45), so this is the median class.

#### 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

 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 ${\left(\frac{N}{2}\right)}^{th}$ 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 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

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 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 ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{50}{2}\right)}^{th}$ item, which is 25th item.
This corresponds to the class interval of (25 35), so this is the median class.

#### 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

 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.

#### Question 17:

Calculate mode from the following data:

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

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

 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.

#### 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

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

 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 ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{28}{2}\right)}^{th}$ item, which is 14th item.
This corresponds to the class interval of (30 40), so this is the median class.

#### 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

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.

 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.

#### 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

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.

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.

#### 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

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 97 − 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.

#### 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

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

 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

Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 250.5th 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

 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

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

$\overline{)\mathbf{Mode}}$
By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 45.

#### 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

 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 ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{800}{2}\right)}^{th}$ item, which is 400th item.
This corresponds to the class interval of (35 45), so this is the median class.

#### 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

 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

$\overline{)\mathbf{Median}}$
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{100}{2}\right)}^{th}$ item, which is 50th 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 25.

#### Question 28:

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

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.

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.

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

 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

$\overline{)\mathbf{Median}}$
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{150}{2}\right)}^{th}$ item, which is 75th item.
This corresponds to the class interval of 40 50, so this is the median class.

Mode

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

#### 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.

Given,

Median (M) = 46.67
Mean $\left(\overline{)\mathrm{X}}\right)$ = 45.5
Mode = Z

We know,
Z = 3M − 2$\left(\overline{)\mathrm{X}}\right)$

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

Hence, Mode (Z) of the marks is 49.01

View NCERT Solutions for all chapters of Class 14