In this page we have *NCERT Solutions for Class 10 Maths:Chapter 14 Statistics* for
Exercise 14.1 on pages 270,271 and 272. Hope you like them and do not forget to like , social_share
and comment at the end of the page.

The mean of grouped data can be obtained using

i.Direct method

$M= \frac { \sum f_i x_i}{\sum f_i}$

ii. assumed mean method

$M= a + \frac { \sum f_i d_i}{\sum f_i}$

iii. Step deviation method

$M= a + \frac { \sum f_i u_i}{\sum f_i} \times h$

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.

Which method did you use for finding the mean, and why?

We will use direct method because the numerical value of f

No. of plants(Class interval) |
No. of houses (f_{i}) |
Mid-point (x_{i}) |
f_{i} x_{i} |

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_{i }= 20 |
∑f_{i} x_{i} = 162 |

Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Here The value of mid-point (x

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

u

Daily wages(Class interval) |
Number of workersfrequency (f _{i}) |
Mid-point (x_{i}) |
u_{i }= (x_{i} - 150)/20 |
f_{i}u_{i } |

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_{i }= 50 |
∑ f_{i}u_{i} = -12 |

=150 + (20 × -12/50) = 150 - 4.8 = 145.20

Thus, mean daily wage = Rs. 145.20

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.

Here, Mean= 18

Class interval |
Number of children (f_{i}) |
Mid-point (x_{i}) |
f_{i}x_{i } |

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 |
∑f_{i} = 44+f |
∑ f_{i}x_{i} = 752+20f |

$18= \frac {(752+20f)}{(44+f)}$

$18(44+f) = (752+20f)$

40 = 2f

f = 20

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.

x

Here The value of mid-point (x

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

u

Class Interval |
Number of women (f_{i}) |
Mid-point (x_{i}) |
u_{i} = (x_{i} - 75.5)/h |
f_{i}u_{i} |

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 |

∑f_{i}= 30 |
∑ f_{i}u_{i }= 4 |

= 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

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.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

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 (f_{i}) |
Mid-point (x_{i}) |
d_{i} = x_{i} - A |
f_{i}d_{i} |

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 |

∑ f_{i} = 400 |
∑ f_{i}d_{i} = 75 |

= 57 + (75/400) = 57 + 0.1875 = 57.19

The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Here, assumed mean (A) = 225

Class Interval |
Number of households (f_{i}) |
Mid-point (x_{i}) |
d_{i} = x_{i} - A |
f_{i}d_{i} |

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 |

∑ f_{i} = 25 |
∑ f_{i}d_{i} = -350 |

= 225 + (-350/25) = 225 - 14 = 211

The mean daily expenditure on food is 211

To find out the concentration of SO

Find the mean concentration of SO

Concentration of SO_{2 }(in ppm) |
Frequency (f_{i}) |
Mid-point (x_{i}) |
f_{i}x_{i} |

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 |
∑ f_{i} = 30 |
∑ (f_{i}x_{i}) = 2.96 |

= 2.96/30 = 0.099 ppm

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 |

Class interval |
Frequency (f_{i}) |
Mid-point (x_{i}) |
f_{i}x_{i} |

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 |

∑ f_{i} = 40 |
∑ f_{i}x_{i} = 499 |

$M= \frac {499}{40} = 12.48$ days

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 |

We will use here the step deviation method

Assumed mean =70

H=10

Class Interval |
Frequency (f_{i}) |
(x_{i}) |
d_{i} = x_{i} - a |
u_{i} = d_{i}/h |
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 |

∑ f_{i} = 35 |
∑ f_{i}u_{i} = -2 |

= 70 + (-2/35) ? 10 = 69.42

Download Class 10 Statistics Exercise 14.1 as pdf

**Notes**-
**Assignments** -
**NCERT Solutions**

Class 10 Maths Class 10 Science