TR Jain Vk Ohri 2018 Solutions for Class 11 Commerce 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 Commerce 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 2018 Book of class 11 Commerce 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 2018 Solutions. All TR Jain Vk Ohri 2018 Solutions for class 11 Commerce Economics are prepared by experts and are 100% accurate.

Page No 202:

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 N+12th item
    or, M = size of 15+12th item
    or, M = size of 8th item
      M = 23

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

Page No 202:

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 N+12th item
       or, M= Size of 7+12th item
       or, M= size of 4th item
        M= 145

Hence, the median height is 145 cm.

Page No 202:

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=Size of N2th item+Size of N2+1th item2   or,  M =Size of 82th item+Size of 82+1th item2   or,  M =Size of 4th item+Size of 5th item2   or,  M=71+722=1432    M=71.5 kgs

Hence, the median weight is 71.5 kgs.



Page No 203:

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 N+12th item
          or,  M = Size of 15+12th 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.

Page No 203:

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 N+12th item
            or, M = Size of 200+12th 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.



Page No 205:

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 N2th item
            or, M = Size of 792th item
            or, M  = Size of 39.5th item

Hence, the median lies in the class 50-60
M=l1+N2-c.ff×ior, M=50+39.5-3118×10or, M =50+8.518×10or, M=50+8518or, M=50+4.72M =54.72

Hence, the median is of the above series is 54.7

Page No 205:

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 N2th item
           or,   M = Size of 1002th item
           or,   M = Size of 50th item

Hence, median lies in the class 20-30
Median=l1+N2-c.ff×i  or, M =20+50-3840×10  or, M =20 + 12040  or, M =20 + 3    M= 23 marks

Hence, the median marks is 23.



Page No 210:

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
     Σf=N=100


Median or M = Size of N2th item
          or,   M = Size of 1002th 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.

M=l1+N2-C.ff×ior, M =30+50-3527×15or, M =30+1527×15or, M =30+22527or, M = 30+8.33 M =38.33 marks

Hence, the median marks of the above series is 38.33

Page No 210:

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
    Σf=N=38

Median = Size of N2th item
   or, M = Size of 382th item
   or, M = Size of 19th item

Hence, median lies in class interval 20 − 25

Median or M=l1+N2-c.f.f×i or, M=20+19-1710×5 or, M=20+210×5 or, M=20+1010 or, M=20+1 M=21

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

Page No 210:

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 N2th item
             = Size of 1002th item
             = Size of 50th item

Hence, median lies in the class interval 29.5 − 39.5
Median or M=l1+N2-c.f.f×ior, M =29.5+50-3120×10or, M =29.5+1920×10or, M =29.5+19020or, M =29.5+9.5 M=39

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

Page No 210:

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
  Σf=N = 50  

Median = Size of N2th item
             = Size of 502th item
             = Size of 25th item

Hence, median lies in class interval 40 − 70

Median or M=l1+N2-c.f.f×ior, M=40+25-142×30or, M=40+11×3026or, M=40+12.69M=52.69

Hence, the median value of the above series is 52.69

Page No 210:

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)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
  Σf=N = 123  

Median = Size of N2th item
             = Size of 1232th item
             = Size of 61.5th item

Hence, the median lies in the class 15.5 − 20.5

Median or M=l1+N2-c.ff×ior, M =15.5+61.5-3332×5or, M =15.5+28.5×532or, M =15.5+142.532or, M =15.5+4.45 M =19.95 marks

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



Page No 213:

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.

Page No 213:

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.

Page No 213:

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.



Page No 224:

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.

Page No 224:

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.



Page No 227:

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=l1+f1-f02f1-f0-f2×ior, Z =10+32-24232-24-28×5or, Z =10+864-52×5or, Z =10+4012or, Z =10+3.33 Z =13.33

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

Page No 227:

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
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=l1=f1-f02f1-f0-f2×i
or, Z =12+16-12216-12-14×4or, Z =12+4×432-26or, Z =12+166or, Z =12+2.67 Z=14.67

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



Page No 231:

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=l1+f1-f02f1-f0-f2×ior, Z =149.5+18-15218-15-6×30or, Z =149.5+336-21×30or, Z =149.5+3×3015or, Z =149.5+6 Z =155.5

Hence, the modal value of the above series is 155.5

Page No 231:

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=l1+f1-f02f1-f0-f2×ior, Z =22.5+32-24232-24-28×5or, Z =22.5+864-52×5or, Z =22.5+4012or, Z =22.5+3.33 Z =25.83

Hence, the Mode (Z) = 25.83



Page No 232:

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 − 2X

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

Page No 232:

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 − 2X
  
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.

Page No 232:

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
  Z  = 16

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



Page No 233:

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=l1+f1-f02f1-f0-f2×ior, Z =40+30-20230-20-15×10or, Z =40+1060-35×10or, Z =40+10025or, Z =40+4 Z =44

Hence, Mode (Z) is 44.

Page No 233:

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

Page No 233:

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.



Page No 265:

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 N+12th item
or, Median = size of 9+12th 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)
1, 2, 3, 4,  5 , 6, 7, 8, 9
Therefore, Median of the data so given is 5.

Page No 265:

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 N+12th item
or, Median = size of 11+12th 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)
10, 15, 18, 20, 21,  25 , 28, 30, 35, 36, 40
Therefore, Median of the data so given is 25.

Page No 265:

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 N+12thitem
or, Median = size of 73+12thitem
or, Median = size of 742th 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.

Page No 265:

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 N2th item, i.e.702th item, which is 35th item.
This corresponds to the class interval of 30 − 40, so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+702-c.f.f×ior, Median=30+35-2512×10or, Median=30+10012Thus, Median=30+8.33=38.33

Page No 265:

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=size of N2th item +size of N2+1th item2             =size of 82th item+ size of 82+1th item2             =size of 4th item+size of 5th item2             =11+132=12Hence, Median = 12 years

Page No 265:

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 N+12thitem
or, Median = size of 500 + 12th 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.

Page No 265:

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 N+12thitem
or, Median = size of 80 + 12th 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.

Page No 265:

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 N+12thitem
or, Median = size of 100 + 12th 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.

Page No 265:

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 N2th item, i.e.302th item, which is 15th item.
This corresponds to the class interval of (25 35), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+302-c.f.f×ior, Median=25+15-1010×10or, Median=25+5Thus, Median is 30

Page No 265:

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 N2th item, i.e.502th item, which is 25th item.
This corresponds to the class interval of (15 20), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+502-c.f.f×ior, Median=15+25-1520×5or, Median=15+5020Thus, Median=15+ 2.5=17.5



Page No 266:

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 N2th item, i.e.1502th item, which is 75th item.
This corresponds to the class interval of (30 35), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+1502-c.f.f×ior, Median=30+75-4233×5or, Median=30+3333×5Thus, Median=35

Page No 266:

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 N2th item, i.e.502th item, which is 25th item.
This corresponds to the class interval of (40 45), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+502-c.f.f×ior, Median=40+25-206×5or, Median=40+256Thus, Median=40+4.16=44.16

Page No 266:

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 N2th item, i.e.8002th item, which is 400th item.
This corresponds to the class interval of (35 40), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+8002-c.f.f×ior, Median=35+400-220180×5or, Median=35+180180×5Thus, Median=35+5=40

Page No 266:

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 N2th item, i.e.282th item, which is 14th item.
This corresponds to the class interval of (30 40), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+282-c.f.f×ior, Median=30+14-78×10or, Median=30+78×10Thus, Median=30+8.75=38.75

Page No 266:

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 = Mid-point of one class - Mid-point of preceeding class2
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 N2th item, i.e.502th item, which is 25th item.
This corresponds to the class interval of (25 35), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+502-c.f.f×ior, Median=25+25-1213×10or, Median=25+13013Thus, Median=25+10=35

Page No 266:

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)=l1+f1-f02f1-f0-f2×ior, Z =20+54-46254-46-42×10or, Z=20+8108-88×10or, Z=20+820×10or, Z=20+820×10or, Z=20+8020=24Hence, Mode is 24

Page No 266:

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

Page No 266:

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)=l1+f1-f02f1-f0-f2×ior, Z =30+7-627-6-3×5or, Z=30+114-9×5or, Z=30+15×5or, Z=30+1 = 31Hence, mode (Z) is 31

Page No 266:

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. 



Page No 267:

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 = Lower limit of one class - Upper limit of the preceeding class2
 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)=l1+f1-f02f1-f0-f2×ior, Z =49.5+263-2472263-247-133×10or, Z=49.5+16526-380×10or, Z=49.5+16146×10or, Z=49.5+1.09=50.59Hence, mode is 50.59

Page No 267:

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 = 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.

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)=l1+f1-f02f1-f0-f2×ior, Z =15.5+50-30250-30-40×5or, Z=15.5+20100-70×5or, Z=15.5+10030or, Z=15.5+3.33 = 18.33Hence, mode is 18.33

Page No 267:

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
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.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =60+71-31271-31-52×10or, Z=60+40142-83×10or, Z=60+40059or, Z=60+6.78 = 66.78Hence, Mode is 66.78

Page No 267:

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.

Page No 267:

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  

MeanMean (X¯)=ΣfxΣf=1888500=3.78

MedianMedian = size of N+12th item              = size of 500+12th item              = 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.

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.

Page No 267:

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 N2th item, i.e.1592th item, which is 79.5th item.
This corresponds to the class interval of (30 40), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=30+1592-5045×ior, Median=30+29.545×10or, Median=30+6.55Thus, 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)=l1+f1-f02f1-f0-f2×ior, Z =30+45-30245-30-35×10or, Z=30+1590-65×10or, Z=30+15025= 30+6=36Hence, mode (Z) is 36

Page No 267:

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 N2th item, i.e.8002th item, which is 400th item.
This corresponds to the class interval of (35 45), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=35+8002-220180×5or, Median=35+180180×5or, Median=35+5Thus, Median is 40.

Page No 267:

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

MeanMean (X¯)=ΣfmΣf=4000100=40

Median
Median class is given by the size of N2th item, i.e.1002th item, which is 50th item.
This corresponds to the class interval of (30 40), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+1002-c.f.f×ior, Median=30+50-2725×10or, Median=30+23025Thus, 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)=l1+f1-f02f1-f0-f2×ior, Z =30+25-15225-15-20×10or, Z=30+1050-35×10or, Z=30+10015or, Z=30+6.6=36.6Hence, mode (Z) is 36.6

Page No 267:

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 =2673=89
Hence, median is equal to 89



Page No 268:

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.

Page No 268:

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 =1682=84
Hence, mean is 84.

Page No 268:

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
MeanMean (X¯)=ΣfmΣf=6610150=44.07

Median
Median class is given by the size of N2th item, i.e.1502th item, which is 75th item.
This corresponds to the class interval of 40 50, so this is the median class.
Median=l1+N2-c.f.f×iso, Median=40+1502-5822×10or, Median=40+17022or, Median=40+7.73Thus, 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)=l1+f1-f02f1-f0-f2×ior, Z =50+30-22230-22-28×10or, Z=50+860-50×10or, Z=50+810×10or, Z=50+8=58Hence, mode (Z) is 58.



View NCERT Solutions for all chapters of Class 13