- Flash back from IX Class
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- Mean for Ungroup Frequency table
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- Mean for group Frequency table
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- Various ways to calculate mean
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- Mode for grouped frequency table
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- Cumulative Frequency chart
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- Median of a grouped data frequency table
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- Empirical Formula between Mode, Mean and Median

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.

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 |

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 |

Mean(M) = 162/20 = 8.1

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 |

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.

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 |

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= (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.

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 |

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.

Number of Mangoes |
50-52 |
53-55 |
56-58 |
59-61 |
62-64 |

Number of Boxes |
15 |
110 |
135 |
115 |
25 |

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.

Daily expenditure (in c) |
100-150 |
150-200 |
200-250 |
250-300 |
300-350 |

Number of households |
4 |
5 |
12 |
2 |
2 |

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

Concentration of SO_{2} (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 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 |

= 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 Statistics Exercise 14.1 as pdf

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