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NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.1





In this page we have NCERT Solutions for Class 10 Maths:Chapter 14 Statistics for Exercise 14.1 . Hope you like them and do not forget to like , social_share and comment at the end of the page.
Question 1
A survey was conducted by a group of students as a part of their environment awareness programmes, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of Plants
   0-2   
   2-4   
   4-6   
 6-8   
8-10   
  10-12   
  12-14   
Number of Houses    
1
2
1
5
6
2
3
Which method did you use for finding the mean, and why?
Solution
We will use direct method because the numerical value of fi and xi are small
No. of plants
 (Class interval)  

No. of houses (fi)
Mid-point (xi)
    fi xi
0-2
1
1
1
2-4
2
3
6
4-6
1
5
5
6-8
5
7
35
8-10
6
9
54
10-12
2
11
22
12-14
3
13
39
  ∑f= 20
  ∑fi xi = 162    
NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.1
 Mean(M) = 162/20 = 8.1
 
Question 2
Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.)
100-120   
 120-140   
140-160   
 160-180   
    180-200   
Number of workers    
12
14
8
6
10
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution
Here The value of mid-point (xi) is very large. So, assumed mean or step deviation method would be simple to calculate
Here the class interval is h = 20.
We can see that step deviation method would be best here
Now assumed mean A = 150 and class interval is h = 20.
So, u= (xi - A)/h
 u= (xi - 150)/20
Daily wages
   (Class interval)  

Number of workers
frequency (fi)

Mid-point (xi)
u= (xi - 150)/20
    fiui    
100-120
12
110
-2
-24
120-140
14
130
-1
-14
140-160
8
150
0
0
160-180
6
170
1
6
180-200
10
190
2
20
Total
∑ f= 50
    ∑ fiui = -12  
NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.1
=150 + (20 × -12/50) = 150 - 4.8 = 145.20
Thus, mean daily wage = Rs. 145.20
Question 3
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
Daily pocket Allowance (in c)
11-13
13-15
15-17
17-19
19-21
21-23
23-25
Number of Children
7
6
9
13
f
5
4

Solution
Here, Mean= 18
Class interval
Number of children (fi)
Mid-point (xi)
    fixi    
11-13
7
12
84
13-15
6
14
84
15-17
9
16
144
17-19
13
18 = A
234
19-21
f
20
20f
21-23
5
22
110
23-25
4
24
96
Total
∑fi = 44+f
  ∑ fixi = 752+20f 
NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.1
18= (752+20f)/(44+f)
18(44+f) = (752+20f)
40 = 2f
f = 20
 
Question 4
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Number of heart beats per minute
65-68
68-71
71-74
74-77
77-80
80-83
83-86
Number of Women
2
4
3
8
7
4
3

Solution
x= (Upper limit + Lower limit)/2
Here The value of mid-point (xi) is very large. So, assumed mean or step deviation method would be simple to calculate
Now Class size (h) = 3
We can see that step deviation method would be best here
Now assumed mean A = 75.5 and class interval is h = 3
So, u= (xi - A)/h
 u= (xi – 75.5)/3
 
Class Interval
Number of women (fi)
Mid-point (xi)
ui = (xi - 75.5)/h
fiui
65-68
2
66.5
-3
-6
68-71
4
69.5
-2
-8
71-74
3
72.5
-1
-3
74-77
8
75.5
0
0
77-80
7
78.5
1
7
80-83
4
81.5
3
8
83-86
2
84.5
3
6
  ∑fi= 30
    ∑ fiu= 4
NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.1
= 75.5 + 3×(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9
The mean heart beats per minute for these women is 75.9
 
Question 5
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes per the number of boxes.
Number of Mangoes
50-52
53-55
56-58
59-61
62-64
Number of Boxes
15
110
135
115
25
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution
Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.5 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3
Class Interval
Number of boxes (fi)
Mid-point (xi)
di = xi - A
fidi
49.5-52.5
15
51
-6
90
52.5-55.5
110
54
-3
-330
55.5-58.5
135
57 = A
0
0
58.5-61.5
115
60
3
345
61.5-64.5
25
63
6
150
  ∑ fi = 400
    ∑ fidi = 75
NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.1
= 57 + (75/400) = 57 + 0.1875 = 57.19
 
Question 6
The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure (in c)
100-150
150-200
200-250
250-300
300-350
Number of households
4
5
12
2
2
Find the mean daily expenditure on food by a suitable method.
 
Solution
Here, assumed mean (A) = 225
Class Interval
Number of households (fi)
Mid-point (xi)
di = xi - A
fidi
100-150
4
125
-100
-400
150-200
5
175
-50
-250
200-250
12
225
0
0
250-300
2
275
50
100
300-350
2
325
100
200
  ∑ fi = 25
    ∑ fidi = -350
NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.1
= 225 + (-350/25) = 225 - 14 = 211
The mean daily expenditure on food is 211

 
Question 7
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below
 
Concentration of SO2 (in ppm)
Frequency
0.00-0.04
4
0.04-0.08
9
0.08-0.12
9
0.12-0.16
2
0.16-0.20
4
0.20-0.24
2
 
Find the mean concentration of SO2 in the air.
Solution
Concentration of SO(in ppm)
Frequency (fi)
Mid-point (xi)
fixi
0.00-0.04
4
0.02
0.08
0.04-0.08
9
0.06
0.54
0.08-0.12
9
0.10
0.90
0.12-0.16
2
0.14
0.28
0.16-0.20
4
0.18
0.72
0.20-0.24
2
0.20
0.40
Total
∑ fi = 30
  ∑ (fixi) = 2.96
NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.1
= 2.96/30 = 0.099 ppm
 
Question 8
A class teacher has the following absentee record of 40 students of a class for the whole
term. Find the mean number of days a student was absent.
Number of days
0-6
6-10
10-14
14-20
20-28
28-38
38-40
Number of students
11
10
7
4
4
3
1

Solution
Class interval
Frequency (fi)
Mid-point (xi)
fixi
0-6
11
3
33
6-10
10
8
80
10-14
7
12
84
14-20
4
17
68
20-28
4
24
96
28-38
3
33
99
38-40
1
39
39
  ∑ fi = 40
  ∑ fixi = 499
NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.1
 = 499/40 = 12.48 days
 
Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.

 
Literacy rate (in %)
45-55
55-65
65-75
75-85
85-98
Number of cities
3
10
11
8
3

Solution
We will use here the step deviation method
Assumed mean =70
H=10
 
Class Interval
Frequency (fi)
(xi)
di = xi - a
ui = di/h
fiui
45-55
3
50
-20
-2
-6
55-65
10
60
-10
-1
-10
65-75
11
70
0
0
0
75-85
8
80
10
1
8
85-95
3
90
20
2
6
  ∑ fi  = 35
      ∑ fiui  = -2

= 70 + (-2/35) ? 10 = 69.42
 
 

 
 
Download Statistics Exercise 14.1 as pdf

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