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Page No 9.74:

Question 1:

Find out the median.

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

Answer:

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

N = 9
Median=N+12th itemMedian=9+12th=5th item
Thus, Median is given by the size of the 5th  item. Therefore, Median of the data so given is 18.

Page No 9.74:

Question 2:

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

Answer:

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

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

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

Median=Size of (N2)th item + Size of (N2+1)th item2              =Size of (82)th item + Size of (82+1)th item2              =Size of 4th item + Size of 5th item2              = 46+482=47

Thus, Median is 47.



Page No 9.75:

Question 3:

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

Answer:


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

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

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

Median=Size of (N2)th item + Size of (N2+1)th item2              =Size of (102)th item + Size of (102+1)th item2              =Size of 5th item + Size of 6th item2              = 18+202=19

Thus, Median is 19.

Page No 9.75:

Question 4:

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

Answer:

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

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

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

Median=Size of (N2)th item + Size of (N2+1)th item2              =Size of (102)th item + Size of (102+1)th item2              =Size of 5th item + Size of 6th item2              = 22+262=24

Thus, Median is 24.

Page No 9.75:

Question 5:

Calculate the value of median from the following data:

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

Answer:

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

Median = size of N+12th item
or, Median =65+12=33rd item
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 33rd is 43 (in the c.f. column), which is corresponding to 1800. ​ Hence, median is 1800.

Page No 9.75:

Question 6:

Find out the nedian size from the following:

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

Answer:

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


Median = size of N+12th item
or, Median =29+12=15th item
Now, we need to locate this item in the column of Cumulative Frequency. The item exceeding 15th is 21 (in the c.f. column), which is corresponding to 156. ​ Hence, median is 156.

Page No 9.75:

Question 7:

Find out the median size from the following:

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

Answer:

Size Frequency
(
f)
Cumulative Frequency
(
c.f.)
10 − 20
20 − 30
42
25
42
67
30 − 40 58 (f) 125
40 − 50 40 165
  N=f=165  

Median class is given by the size of N2th item, i.e.1652th item, which is 82.5th item.
This corresponds to the class interval of 30 − 40, so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+1652-c.f.f×ior, Median=30+82.5-6758×10or, Median=30+2.67=32.67Thus, Median=32.67

Page No 9.75:

Question 8:

Find the median of the following data:

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

Answer:

Age Frequency (f) Cumulative Frequency
(
c.f.)
20 − 25
25 − 30
30 − 35
50
70
100
50
120
220 (c.f.)
35 − 40 180 (f) 400
40 − 45
45 − 50
50 − 55
55 − 60
150
120
70
60
550
670
740
800
  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+5=40Thus, Median=40

Page No 9.75:

Question 9:

Calculate the median from the following data:

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

Answer:

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

Median class is given by the size of N2th item, i.e.1302th item, which is 65th 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+1302-c.f.f×ior, Median=30+65-4535×10or, Median=30+5.71=35.71Thus, Median=35.71



Page No 9.76:

Question 10:

Find the median form the following:

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

Answer:

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


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 20 − 30, so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+1002-c.f.f×ior, Median=20+50-2234×10or, Median=20+8.235=28.23Thus, Median=28.23

Page No 9.76:

Question 11:

Compute median from the following data:

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

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



Median class is given by the size of N2th item, i.e.1002th item, which is 50th 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+1002-c.f.f×ior, Median=40+50-3020×5or, Median=40+5=45Thus, Median=45

Page No 9.76:

Question 12:

Calculate median from the following figures:

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

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.
 

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



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 (29.5 39.5), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=29.5+1002-c.f.f×ior, Median=29.5+50-3120×10or, Median=29.5+9.5=39Thus, Median=39

Page No 9.76:

Question 13:

Calculate median from following data:

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

Answer:

X Frequency
(f)
Cumulative Frequency
(c.f.)
  0 − 10
  10 − 20
3
7
3
10 (c.f.)
 20 − 30 15 (f) 25
30 − 40
40 − 50
 50 − 60
9
6
4
34
40
44
  N=f=44  

Median class is given by the size of N2th item, i.e.442th item, which is 22nd item.
This corresponds to the class interval of (20 30), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=20+442-c.f.f×ior, Median=20+22-1015×10or, Median=20+8=28Thus, Median=28

Page No 9.76:

Question 14:

From the following data, compute median:

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

Answer:

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


Median class is given by the size of N2th item, i.e.802th item, which is 40th item.
This corresponds to the class interval of (20 40), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=20+802-1822×20or, Median=20+2222×20or, Median=20+20=40Thus, Median=40

Page No 9.76:

Question 15:

An Incomplete distribution is given below:

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

Answer:

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

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


Median class is given by the size of N2th item, i.e.2292th item, which is 114.5th item.
This corresponds to the class interval of (40 50) as median is 46.
Median=l1+N2-c.f.f×iso, 46=40+2292-(42 +f1)65×10or, 46=40+114.5-42-f165×10or, 390 =725-10f1or, 10f1= 335 f1 = 33.5 

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

Therefore, f1 is 33.5 and f2 is 45.5.

Page No 9.76:

Question 16:

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

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

Answer:

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

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


Median class is given by the size of N2th item, i.e.1702th item, which is 85th item.
This corresponds to the class interval of (30 40) as median is 35.
Median=l1+N2-c.f.f×iso, 35=30+1702-(30+f1)40×10or, 35=30+85-30-f140×10or, 20=55-f1or, f1= 35 f1 = 35 


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

Therefore, f1 is 35 and f2 is 25.



Page No 9.77:

Question 17:

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

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

Answer:

Marks
(Class Interval)
Students
(f)
Less than
c.f.
More than
c.f.
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
10
15
25
30
10
10
10
25
50
80
90
100
100
90
75
50
20
10



From the point of intersection, a perpendicular is drawn on the x-axis. This line cuts the x-axis at 30. Hence, the median is 30 marks

Median = 30 marks

Page No 9.77:

Question 18:

From the following, determine lower quartile and upper quartile:

X 12 16 17 21 28 19 30 32

Answer:

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

N = 8

Lower Quartile:Q1=Size of N+14th itemor, Q1=Size of 8+14th=2.25th itemor, Q1= Size of 2nd item+14Size of 3rd term-Size of 2nd itemor, Q1  =16+1417-16 Q1  =16.25Upper Quartile:Q3=Size of 3N+14th itemor, Q3=Size of 38+14th item or, Q3=6.75th itemor, Q3= Size of 6th item+34( Size of 7th term-Size of 6th item)or, Q3=28+3430-28or, Q3=28+1.5 Q3  =29.5

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

Page No 9.77:

Question 19:

Calculate lower and upper quartiles.

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

Answer:

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

Lower Quartile:Q1=Size of N+14th itemor, Q1=Size of 10+14th=2.75th itemor, Q1= Size of 2nd item+34Size of 3rd term-Size of 2nd itemor, Q1  =11+3417-11 Q1  =15.5Upper Quartile:Q3=Size of 3N+14th itemor, Q3=Size of 310+14th item or, Q3=8.25th itemor, Q3= Size of 8th item+14( Size of 9th term-Size of 8th item)or, Q3=37+1438-37or, Q3=37+0.25 Q3  =37.25
Hence, lower and upper quartile are 15.5 and 37.25 respectively.

Page No 9.77:

Question 20:

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

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

Answer:

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

Median

Median = size of N+12th item
or, Median =49+12=25th item
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 25th is 26 (in the c.f. column), which is corresponding to 61. ​ 
Hence, median is 61.

Lower Quartile

Q1=Size of N+14thitemor,  Q1=Size of 49+14thitem  Q1=12.5th item
This corresponds to 26 in the cumulative frequency.
Hence, lower quartile is 61 marks.

Upper Quartile

 Q3=Size of 3N+14thitemor,  Q3=Size of 349+14thitem  Q3=37.5th item

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

Page No 9.77:

Question 21:

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

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

Answer:

 

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


Lower Quartile

Q1=Size of N+14thitemor,  Q1=Size of 135+14thitem  Q1=34th item
This corresponds to 34 in the cumulative frequency.
Hence, lower quartile is 10.

Upper Quartile

 Q3=Size of 3N+14thitemor,  Q3=Size of 3135+14thitem  Q3=102nd item

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

Page No 9.77:

Question 22:

Calculate upper and lower quartile from the following data:

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

Answer:

Class Interval Frequency
(
f)
Cumulative Frequency (c.f.)  
  0 − 10
10 − 20
10
20
10
30

 
20 − 30 35 65 Q1
30 − 40 40 105       
40 − 50 25 130 Q3
50 − 60
60 − 70
25
15
155
170
 
  N=170    


Q1=Size of N4th item=1704th=42.5th item
This lies in the class interval (20-30).

Now,
Q1=l1=N4-c.f.f×ior, Q1 =20+42.5-3035×10Q1=23.57Thus, the value of lower quartile is 23.57.Q3=Size of 31704th=127.5th itemThis lies in the class interval (40-50).Q3=l1=3(N4)-c.f.f×ior, Q3=40+127.5-10525×10Q3=49Thus, the value of upper quartile is 49.

Page No 9.77:

Question 23:

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

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

Answer:

Age
(X)
Tally
Marks
Frequency
(f)
 
15
16

3
3
 
17 4 Modal Class
18 1  
    N=11  

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



Page No 9.78:

Question 24:

Find modal item of the followings set of numbers:

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

Answer:

X Tally
Marks

Frequency
(f)
 
1
2
3
4



1
3
2
1
 
5 6 →Modal Class
6
7

1
1
 
    N=15  

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

Page No 9.78:

Question 25:

From the followind data, calculate the value of mode.

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

Answer:

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

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

Page No 9.78:

Question 26:

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

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

Answer:

(i) Observation Method
 

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

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

(ii) Grouping Method

For the given distribution, the grouping table is as follows.





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

Analysis Table
 
Column 60 62 63 64 65 66 67 68 69 70
I
II
III
IV
V
VI
VII
     























   
        1 4 7 4 2    

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

Page No 9.78:

Question 27:

Find out mode from the following data:

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

Answer:

Class Interval Frequency (f)  
0 − 5
5 − 10
10 − 15
5
7
15


f0
15 − 20 18 f1
20 − 25
25 − 30
30 − 35
35 − 40
16
9
6
3
f2


 


By inspection, we can say that the modal class is (15 – 20) as it has the highest frequency of 18.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =15+18-15218-15-16×5or, Z=15+336-31×5or, Z=15+3= 18Hence, mode (Z) is 18.

Page No 9.78:

Question 28:

Calculate the mode from the following data:

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

Answer:

Class Interval Frequency
(f)
 
10 − 20
20 − 30
30 − 40
8
11
9


f0
40 − 50 25 f1
50 − 60
60 − 70
12
16
f2
 

By inspection, we can say that the modal class is (40 – 50) as it has the highest frequency of 25.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =40+25-9225-9-12×10or, Z=40+1650-21×10or, Z=40+5.517= 45.52Hence, mode (Z) is 45.52.

Page No 9.78:

Question 29:

Calculate the mode from the following data:

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

Answer:

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


By inspection, we can say that the modal class is (30 – 45) as it has the highest frequency of 30.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =30+30-20230-20-24×15or, Z=30+1060-44×15or, Z=30+9.375= 39.375Hence, mode (Z) is 39.375.

Page No 9.78:

Question 30:

Find out mode of the following frequency distribution:

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

Answer:

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

By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 10.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =30+10-8210-8-8×10or, Z=30+24×10or, Z=30+5= 35Hence, mode (Z) is 35.

Page No 9.78:

Question 31:

 Find out the mode value from the following data:

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

Answer:

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


By inspection, we can say that the modal class is (50 – 60) as it has the highest frequency of 28.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =50+28-16228-16-15×10or, Z=50+1225×10or, Z=50+4.8= 54.8Hence, mode (Z) is 54.8.



Page No 9.79:

Question 32:

Find out the modal value from the following data:

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

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.
 

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

 
By inspection, we can say that the modal class is (29.5 – 39.5) as it has the highest frequency of 25.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =29.5+25-15225-15-10×10or, Z=29.5+1050-25×10or, Z=29.5+10025or, Z=29.5+4 = 33.5

Hence, mode is 33.5.

Page No 9.79:

Question 33:

Calculate mode of the following data:

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

Answer:

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

By inspection, we can say that the modal class is (10 – 15) as it has the highest frequency of 32.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =10+32-24232-24-28×5or, Z=10+4012or, Z=10+3.33 = 13.33Hence, mode is 13.33.

Page No 9.79:

Question 34:

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

Answer:

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

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

 
By inspection, we can say that the modal class is (40 – 50) as it has the highest frequency of 32.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =40+32-20232-20-14×10or, Z=40+1264-34×10or, Z=40+12030or, Z=40+4 = 44Hence, mode is 44.

Page No 9.79:

Question 35:

Compute graphically, the modal value of the given data:

Age 0−10 10−20 20−30 30−40 40−50
No. of Persons 2 5 7 5 2

Answer:





Hence, the mode of the series is 25 years.

Page No 9.79:

Question 36:

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

Marks 10−20 20−30 30−40 40−50 50−60
No. of Students 3 5 9 3 2
Calculate the value of mode by graphical method.

Answer:





Hence, the mode of the series is 34 marks.

Page No 9.79:

Question 37:

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

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

Answer:

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

Lower Quartile

Q1=Size of N4th item=1004th=25th item
This lies in the class interval (40-50).
Now,
Q1=l1=N4-c.f.f×ior, Q1 =40+25-1521×10Q1=44.76

Thus, the value of lower quartile is 44.76.

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 50 − 60, so this is the median class.Uppe
Median=l1+N2-c.f.f×iso, Median=l1+1002-c.f.f×ior, Median=50+50-3633×10or, Median=50+4.24=54.24Thus, Median=54.24

Upper Quartile

Q3=Size of 31004th=75th item
This lies in the class interval (60-70).

Q3=l1=3(N4)-c.f.f×ior, Q3=60+75-6922×10Q3=62.72

Thus, the value of upper quartile is 62.72.

Page No 9.79:

Question 38:

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

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

Answer:

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


By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 32.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =30+32-26232-26-10×10or, Z=30+664-36×10or, Z=30+2.14= 32.14Hence, mode (Z) is 32.14 years. 



Page No 9.80:

Question 39:

 Find mean, median, and mode for the following data

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

Answer:

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

MeanMean (X¯)=ΣfmΣf=93040=23.25

Median
Median class is given by the size of N2th item, i.e.402th item, which is 20th item.
This corresponds to the class interval of (20 30), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=20+402-c.f.f×ior, Median=20+20-1910×10or, Median=20+1=21Thus, median=21

Mode

By inspection, we can say that the modal class is (10 – 20) as it has the highest frequency of 15.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =10+15-4215-4-10×10or, Z=10+1130-14×10or, Z=10+11016or, Z=10+6.875=16.875Hence, mode (Z) is 16.875.

Page No 9.80:

Question 40:

Calculate median from the following:

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

Answer:

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


Median class is given by the size of N2th item, i.e.802th item, which is 40th item.
This corresponds to the class interval of 20 − 25, so this is the median class.
Median=l1+N2-c.f.f×iso, Median=l1+802-c.f.f×ior, Median=20+40-3226×5or, Median=20+1.538=21.538Thus, Median=21.54

Page No 9.80:

Question 41:

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

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

Answer:

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


MeanMean (X¯)=ΣfmΣf=9715150=64.76

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 (64 65), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=64+1502-c.f.f×ior, Median=64+75-5235×1or, Median=64+0.65=64.65Thus, median=64.65

Mode

By inspection, we can say that the modal class is (64 – 65) as it has the highest frequency of 35.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =64+35-24235-24-18×1or, Z=64+1170-42×1or, Z=64+1128or, Z=64+0.39=64.39Hence, mode (Z) is 64.39.

Page No 9.80:

Question 42:

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

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

Answer:


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

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

Median class is given by the size of N2th item, i.e.1002th item, which is 50th item.
This corresponds to the class interval of (20 30) as median is 23.
Median=l1+N2-c.f.f×iso, 23=20+1002-(8+f1)40×10or, 23=20+50-8-f140×10or, 12=42-f1or, f1= 42-12 f1 = 30 


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

Therefore, f1 is 30 and f2 is 12.

Page No 9.80:

Question 43:

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

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

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

Median
Median class is given by the size of N2th item, i.e.1402th item, which is 70th item.
This corresponds to the class interval of (55 65), so this is the median class.
Median=l1+N2-c.f.f×ior, Median=55+70-5840×10or, Median=55+3=58Thus, Median=58

Mode

By inspection, we can say that the modal class is (55 65) as it has the highest frequency of 40.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =55+40-25240-25-20×10or, Z=55+1535×10or, Z=55+4.28= 59.28Hence, mode (Z) is 59.28.

Page No 9.80:

Question 44:

From the following data, find the value of mode.

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

Answer:

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

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

Page No 9.80:

Question 45:

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

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

Answer:

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



MeanMean (X¯)=ΣfmΣf=192567=28.73

Median
Median class is given by the size of N2th item, i.e.672th item, which is 33.5th item.
This corresponds to the class interval of (20 30), so this is the median class.
Median=l1+N2-c.f.f×iso, Median=20+672-c.f.f×ior, Median=20+33.5-2015×10or, Median=20+9=29Thus, median=29

Mode

By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 20.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =30+20-15220-15-7×10or, Z=30+518×10or, Z=30+5018or, Z=30+2.77=32.77Hence, mode (Z) is 32.77.



Page No 9.81:

Question 46:

Find the median of the following data:

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

Answer:

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


Median class is given by the size of 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=30+1002-c.f.f×ior, Median=30+50-3725×10or, Median=30+5.2=35.2Thus, median=35.2

Page No 9.81:

Question 47:

Find out mode from the following data:

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

Answer:

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

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

 
By inspection, we can say that the modal class is (20 – 30) as it has the highest frequency of 20.
Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =20+20-7220-7-14×10or, Z=20+1340-21×10or, Z=20+13019or, Z=20+6.84 = 26.84Hence, mode is 26.84.

Page No 9.81:

Question 48:

Calculatate the Median and Mode form the following data:

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

Answer:

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

Median

Median class is given by the size of 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-c.f.f×ior, Median=30+79.5-5045×10or, Median=30+6.55=36.55Thus, 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+15025or, Z=30+6=36Hence, mode (Z) is 36.

Page No 9.81:

Question 49:

Find out the median of the following frequency distribution:

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

Answer:

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


Median = size of N+12th item
or, Median =41+12=21st item
Now, we need to locate this item in the column of Cumulative Frequency. The item exceeding 21st is 28 (in the c.f. column), which is corresponding to 65. ​Hence, median is 65.

Page No 9.81:

Question 50:

Calculate mode from the following data:

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

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.
 

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

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

Mode (Z)=l1+f1-f02f1-f0-f2×ior, Z =29.5+25-15225-15-10×10or, Z=29.5+1050-25×10or, Z=29.5+10025or, Z=29.5+4 = 33.5

Hence, mode is 33.5.



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