03 - Question bank-Statistics - Class 10 - Maths Class 10 Notes | EduRev

Crash Course for Class 10 Maths by Let's tute

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Class 10 : 03 - Question bank-Statistics - Class 10 - Maths Class 10 Notes | EduRev

 Page 1


Direct method 
 
Q.1. A survey was conducted by a group of students as a part of their 
environment awareness programme, 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? 
 
Answer : 
To find the class mark (x i) for each interval, the following relation is used. 
Class mark (x i) =  
x i and f ix i can be calculated as follows. 
Number of plants Number of houses 
(f
i
) 
x
i
 f
i
x
i
 
0 - 2 1 1 1 × 1 = 1 
2 - 4 2 3 2 × 3 = 6 
4 - 6 1 5 1 × 5 = 5 
6 - 8 5 7 5 × 7 = 35 
8 - 10 6 9 6 × 9 = 54 
10 - 12 2 11 2 ×11 = 22 
12 - 14 3 13 3 × 13 = 39 
Total 20  162 
From the table, it can be observed that 
 
Page 2


Direct method 
 
Q.1. A survey was conducted by a group of students as a part of their 
environment awareness programme, 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? 
 
Answer : 
To find the class mark (x i) for each interval, the following relation is used. 
Class mark (x i) =  
x i and f ix i can be calculated as follows. 
Number of plants Number of houses 
(f
i
) 
x
i
 f
i
x
i
 
0 - 2 1 1 1 × 1 = 1 
2 - 4 2 3 2 × 3 = 6 
4 - 6 1 5 1 × 5 = 5 
6 - 8 5 7 5 × 7 = 35 
8 - 10 6 9 6 × 9 = 54 
10 - 12 2 11 2 ×11 = 22 
12 - 14 3 13 3 × 13 = 39 
Total 20  162 
From the table, it can be observed that 
 
Mean, 
 
Therefore, mean number of plants per house is 8.1. 
Here, direct method has been used as the values of class marks (x i) and f i are small. 
 
Assumed Mean Method 
 
Q.1. 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 Rs) 
11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25 
Number of workers 7 6 9 13 f 5 4 
 
Answer : 
To find the class mark (x i) for each interval, the following relation is used. 
 
Given that, mean pocket allowance,  
Taking 18 as assured mean (a), d i and f id i are calculated as follows. 
Daily pocket allowance 
(in Rs) 
Number of children 
f
i
 
Class mark x
i
 d
i
 = x
i
 - 18 f
i
d
i
 
11 - 13 7 12 - 6 - 42 
13 - 15 6 14 - 4 - 24 
15 - 17 9 16 - 2 - 18 
17 - 19 13 18 0 0 
19 - 21 f 20 2 2 f 
Page 3


Direct method 
 
Q.1. A survey was conducted by a group of students as a part of their 
environment awareness programme, 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? 
 
Answer : 
To find the class mark (x i) for each interval, the following relation is used. 
Class mark (x i) =  
x i and f ix i can be calculated as follows. 
Number of plants Number of houses 
(f
i
) 
x
i
 f
i
x
i
 
0 - 2 1 1 1 × 1 = 1 
2 - 4 2 3 2 × 3 = 6 
4 - 6 1 5 1 × 5 = 5 
6 - 8 5 7 5 × 7 = 35 
8 - 10 6 9 6 × 9 = 54 
10 - 12 2 11 2 ×11 = 22 
12 - 14 3 13 3 × 13 = 39 
Total 20  162 
From the table, it can be observed that 
 
Mean, 
 
Therefore, mean number of plants per house is 8.1. 
Here, direct method has been used as the values of class marks (x i) and f i are small. 
 
Assumed Mean Method 
 
Q.1. 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 Rs) 
11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25 
Number of workers 7 6 9 13 f 5 4 
 
Answer : 
To find the class mark (x i) for each interval, the following relation is used. 
 
Given that, mean pocket allowance,  
Taking 18 as assured mean (a), d i and f id i are calculated as follows. 
Daily pocket allowance 
(in Rs) 
Number of children 
f
i
 
Class mark x
i
 d
i
 = x
i
 - 18 f
i
d
i
 
11 - 13 7 12 - 6 - 42 
13 - 15 6 14 - 4 - 24 
15 - 17 9 16 - 2 - 18 
17 - 19 13 18 0 0 
19 - 21 f 20 2 2 f 
21 - 23 5 22 4 20 
23 - 25 4 24 6 24 
Total 
 
  2f – 40 
From the table, we obtain 
 
 
 
Hence, the missing frequency, f, is 20. 
 
Step Deviation Method 
 
Q.1. Consider the following distribution of daily wages of 50 worker of a factory. 
Daily wages (in Rs) 100 - 120 120 - 140 140 - 1 60 160 - 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. 
 
Answer : 
To find the class mark for each interval, the following relation is used. 
 
Page 4


Direct method 
 
Q.1. A survey was conducted by a group of students as a part of their 
environment awareness programme, 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? 
 
Answer : 
To find the class mark (x i) for each interval, the following relation is used. 
Class mark (x i) =  
x i and f ix i can be calculated as follows. 
Number of plants Number of houses 
(f
i
) 
x
i
 f
i
x
i
 
0 - 2 1 1 1 × 1 = 1 
2 - 4 2 3 2 × 3 = 6 
4 - 6 1 5 1 × 5 = 5 
6 - 8 5 7 5 × 7 = 35 
8 - 10 6 9 6 × 9 = 54 
10 - 12 2 11 2 ×11 = 22 
12 - 14 3 13 3 × 13 = 39 
Total 20  162 
From the table, it can be observed that 
 
Mean, 
 
Therefore, mean number of plants per house is 8.1. 
Here, direct method has been used as the values of class marks (x i) and f i are small. 
 
Assumed Mean Method 
 
Q.1. 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 Rs) 
11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25 
Number of workers 7 6 9 13 f 5 4 
 
Answer : 
To find the class mark (x i) for each interval, the following relation is used. 
 
Given that, mean pocket allowance,  
Taking 18 as assured mean (a), d i and f id i are calculated as follows. 
Daily pocket allowance 
(in Rs) 
Number of children 
f
i
 
Class mark x
i
 d
i
 = x
i
 - 18 f
i
d
i
 
11 - 13 7 12 - 6 - 42 
13 - 15 6 14 - 4 - 24 
15 - 17 9 16 - 2 - 18 
17 - 19 13 18 0 0 
19 - 21 f 20 2 2 f 
21 - 23 5 22 4 20 
23 - 25 4 24 6 24 
Total 
 
  2f – 40 
From the table, we obtain 
 
 
 
Hence, the missing frequency, f, is 20. 
 
Step Deviation Method 
 
Q.1. Consider the following distribution of daily wages of 50 worker of a factory. 
Daily wages (in Rs) 100 - 120 120 - 140 140 - 1 60 160 - 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. 
 
Answer : 
To find the class mark for each interval, the following relation is used. 
 
Class size (h) of this data = 20 
Taking 150 as assured mean (a), d i, u i, and f iu i can be calculated as follows. 
Daily wages 
(in Rs) 
Number of workers (f
i
) x
i
 d
i
 = x
i
 – 150 
 
f
i
u
i
 
100 - 120 12 110 - 40 - 2 - 24 
120 - 140 14 130 - 20 - 1 - 14 
140 - 160 8 150 0 0 0 
160 - 180 6 170 20 1 6 
180 - 200 10 190 40 2 20 
Total 50    - 12 
From the table, it can be observed that 
 
 
Therefore, the mean daily wage of the workers of the factory is Rs 145.20. 
 
 
 
 
 
Q.2 :  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 – 95 
Page 5


Direct method 
 
Q.1. A survey was conducted by a group of students as a part of their 
environment awareness programme, 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? 
 
Answer : 
To find the class mark (x i) for each interval, the following relation is used. 
Class mark (x i) =  
x i and f ix i can be calculated as follows. 
Number of plants Number of houses 
(f
i
) 
x
i
 f
i
x
i
 
0 - 2 1 1 1 × 1 = 1 
2 - 4 2 3 2 × 3 = 6 
4 - 6 1 5 1 × 5 = 5 
6 - 8 5 7 5 × 7 = 35 
8 - 10 6 9 6 × 9 = 54 
10 - 12 2 11 2 ×11 = 22 
12 - 14 3 13 3 × 13 = 39 
Total 20  162 
From the table, it can be observed that 
 
Mean, 
 
Therefore, mean number of plants per house is 8.1. 
Here, direct method has been used as the values of class marks (x i) and f i are small. 
 
Assumed Mean Method 
 
Q.1. 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 Rs) 
11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25 
Number of workers 7 6 9 13 f 5 4 
 
Answer : 
To find the class mark (x i) for each interval, the following relation is used. 
 
Given that, mean pocket allowance,  
Taking 18 as assured mean (a), d i and f id i are calculated as follows. 
Daily pocket allowance 
(in Rs) 
Number of children 
f
i
 
Class mark x
i
 d
i
 = x
i
 - 18 f
i
d
i
 
11 - 13 7 12 - 6 - 42 
13 - 15 6 14 - 4 - 24 
15 - 17 9 16 - 2 - 18 
17 - 19 13 18 0 0 
19 - 21 f 20 2 2 f 
21 - 23 5 22 4 20 
23 - 25 4 24 6 24 
Total 
 
  2f – 40 
From the table, we obtain 
 
 
 
Hence, the missing frequency, f, is 20. 
 
Step Deviation Method 
 
Q.1. Consider the following distribution of daily wages of 50 worker of a factory. 
Daily wages (in Rs) 100 - 120 120 - 140 140 - 1 60 160 - 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. 
 
Answer : 
To find the class mark for each interval, the following relation is used. 
 
Class size (h) of this data = 20 
Taking 150 as assured mean (a), d i, u i, and f iu i can be calculated as follows. 
Daily wages 
(in Rs) 
Number of workers (f
i
) x
i
 d
i
 = x
i
 – 150 
 
f
i
u
i
 
100 - 120 12 110 - 40 - 2 - 24 
120 - 140 14 130 - 20 - 1 - 14 
140 - 160 8 150 0 0 0 
160 - 180 6 170 20 1 6 
180 - 200 10 190 40 2 20 
Total 50    - 12 
From the table, it can be observed that 
 
 
Therefore, the mean daily wage of the workers of the factory is Rs 145.20. 
 
 
 
 
 
Q.2 :  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 – 95 
Number of cities 3 10 11 8 3 
 
Answer : 
To find the class marks, the following relation is used. 
 
Class size (h) for this data = 10 
Taking 70 as assumed mean (a), d i, u i, and f iu i are calculated as follows. 
Literacy rate (in %) Number of cities 
f
i
 
x
i
 
d
i
 = x
i
 - 70 
 
f
i
u
i
 
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 
Total 35    - 2 
From the table, we obtain 
 
Therefore, mean literacy rate is 69.43%. 
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