NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1#

Exercise 13.1 covers three methods of finding the mean of grouped data: Direct Method, Assumed Mean Method, and Step Deviation Method.

Methods to Find Mean of Grouped Data:

  1. Direct Method: $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$
  2. Assumed Mean Method: $\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$, where $d_i = x_i - a$
  3. Step Deviation Method: $\bar{x} = a + \dfrac{\sum f_i u_i}{\sum f_i} \times h$, where $u_i = \dfrac{x_i - a}{h}$

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. #

Answer
Number of plants Number of houses ($f_i$) $x_i$ $f_i x_i$
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
Total 20 162

Using $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$:

$$\bar{x} = \frac{162}{20} = 8.1$$

∴ Mean number of plants per house = 8.1

2. 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. #

Answer (Step Deviation Method)

Taking the assumed mean $a = 150$ and class size $h = 20$, so $u_i = \dfrac{x_i - 150}{20}$:

Daily wages (₹) $f_i$ $x_i$ $u_i$ $f_i u_i$
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 50 -12

Here $a = 150$ and $h = 20$. Using $\bar{x} = a + \dfrac{\sum f_i u_i}{\sum f_i} \times h$, we find the mean:

$$= 150 + \frac{-12}{50} \times 20$$

$$= 150 - 4.8$$

$$= 145.2$$

∴ Mean daily wages = ₹145.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency f. #

Answer
Daily pocket allowance (₹) $f_i$ $x_i$ $f_i x_i$
11-13 7 12 84
13-15 6 14 84
15-17 9 16 144
17-19 13 18 234
19-21 f 20 20f
21-23 5 22 110
23-25 4 24 96
Total 44+f 752+20f

The mean is ₹18, so using $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$:

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

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

$$\Rightarrow 752 + 20f = 792 + 18f$$

$$\Rightarrow 2f = 40$$

$$\Rightarrow f = 20$$

∴ Missing frequency $f =$ 20

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method. #

Answer (Assumed Mean Method)

Taking the assumed mean $a = 75.5$, so $d_i = x_i - 75.5$:

Heart beats/min $f_i$ $x_i$ $d_i$ $f_i d_i$
65-68 2 66.5 -9 -18
68-71 4 69.5 -6 -24
71-74 3 72.5 -3 -9
74-77 8 75.5 0 0
77-80 7 78.5 3 21
80-83 4 81.5 6 24
83-86 2 84.5 9 18
Total 30 12

Here $a = 75.5$. Using $\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$:

$$\bar{x} = 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9$$

∴ Mean heart beats = 75.9 per minute

5. In a retail market, a fruit vendor was selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box. #

Answer (Step Deviation Method)

Taking the assumed mean $a = 57$ and class size $h = 3$, so $u_i = \dfrac{x_i - 57}{3}$:

Mangoes per box $f_i$ $x_i$ $u_i$ $f_i u_i$
50-52 15 51 -2 -30
53-55 110 54 -1 -110
56-58 135 57 0 0
59-61 115 60 1 115
62-64 25 63 2 50
Total 400 25

Here $a = 57$ and $h = 3$. Using $\bar{x} = a + \dfrac{\sum f_i u_i}{\sum f_i} \times h$, we find the mean:

$$= 57 + \frac{25}{400} \times 3$$

$$= 57 + 0.1875$$

$$= 57.19 \text{ (about)}$$

∴ Mean number of mangoes ≈ 57.19

6. 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. #

Answer (Step Deviation Method)

Taking the assumed mean $a = 225$ and class size $h = 50$, so $u_i = \dfrac{x_i - 225}{50}$:

Daily expenditure (₹) $f_i$ $x_i$ $u_i$ $f_i u_i$
100-150 4 125 -2 -8
150-200 5 175 -1 -5
200-250 12 225 0 0
250-300 2 275 1 2
300-350 2 325 2 4
Total 25 -7

Here $a = 225$ and $h = 50$. Using $\bar{x} = a + \dfrac{\sum f_i u_i}{\sum f_i} \times h$, we find the mean:

$$= 225 + \frac{-7}{25} \times 50$$

$$= 225 - 14$$

$$= 211$$

∴ Mean daily expenditure = ₹211

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. Find the mean concentration of SO₂ in the air. #

Answer (Direct Method)
Concentration of SO₂ (ppm) $f_i$ $x_i$ $f_i x_i$
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 0.22 0.44
Total 30 2.96

Using $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$:

$$\bar{x} = \frac{2.96}{30} = 0.099$$

∴ Mean concentration of SO₂ = 0.099 ppm (about 0.10 ppm)

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. #

Answer (Direct Method)
Number of days $f_i$ $x_i$ $f_i x_i$
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
Total 40 499

Using $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$:

$$\bar{x} = \frac{499}{40} = 12.475$$

∴ Mean number of days absent = 12.48 (about)

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. #

Answer (Assumed Mean Method)

Taking the assumed mean $a = 70$, so $d_i = x_i - 70$:

Literacy rate (%) $f_i$ $x_i$ $d_i$ $f_i d_i$
45-55 3 50 -20 -60
55-65 10 60 -10 -100
65-75 11 70 0 0
75-85 8 80 10 80
85-95 3 90 20 60
Total 35 -20

Here $a = 70$. Using $\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$, we find the mean:

$$= 70 + \frac{-20}{35}$$

$$= 70 - 0.571$$

$$= 69.43 \text{ (about)}$$

∴ Mean literacy rate ≈ 69.43%

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