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:
- Direct Method: $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$
- Assumed Mean Method: $\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$, where $d_i = x_i - a$
- 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 |
∴ 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)
| Daily wages (₹) | $f_i$ | $x_i$ | $u_i = \frac{x_i-150}{20}$ | $f_i u_i$ |
|---|---|---|---|---|
| 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 | 50 | -12 |
$a = 150$, $h = 20$
$$\bar{x} = 150 + \frac{-12}{50} \times 20 = 150 - 4.8 = 145.2$$∴ 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 |
∴ 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)
| Heart beats/min | $f_i$ | $x_i$ | $d_i = x_i - 75.5$ | $f_i d_i$ |
|---|---|---|---|---|
| 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 |
$a = 75.5$
$$\bar{x} = 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9$$∴ 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)
| Mangoes per box | $f_i$ | $x_i$ | $u_i = \frac{x_i-57}{3}$ | $f_i u_i$ |
|---|---|---|---|---|
| 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 |
$a = 57$, $h = 3$
$$\bar{x} = 57 + \frac{25}{400} \times 3 = 57 + 0.1875 = 57.19$$∴ 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)
| Daily expenditure (₹) | $f_i$ | $x_i$ | $u_i = \frac{x_i-225}{50}$ | $f_i u_i$ |
|---|---|---|---|---|
| 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 |
$a = 225$, $h = 50$
$$\bar{x} = 225 + \frac{-7}{25} \times 50 = 225 - 14 = 211$$∴ 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 |
∴ Mean concentration of SO₂ = 0.099 ppm ≈ 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 |
∴ Mean number of days absent = 12.48
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. #
Answer (Assumed Mean Method)
| Literacy rate (%) | $f_i$ | $x_i$ | $d_i = x_i - 70$ | $f_i d_i$ |
|---|---|---|---|---|
| 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 |
$a = 70$
$$\bar{x} = 70 + \frac{-20}{35} = 70 - \frac{4}{7} = 70 - 0.571 \approx 69.43\%$$∴ Mean literacy rate ≈ 69.43%