Get complete, step-by-step solutions for NCERT Class 10 Maths Chapter 13 Exercise 13.1 on finding the Mean of Grouped Data. Solutions cover all three methods: Direct Method (Q.1), Assumed Mean Method (Q.2, Q.3, Q.8), and Step-Deviation Method (Q.4, Q.5, Q.6, Q.7, Q.9). Learn how to identify the most suitable method based on the magnitude of $x_i$ and $f_i$ values. Includes finding the missing frequency ($f$) when the mean is given (Q.3) and handling distributions with unequal class intervals (Q.8). Master the calculation of mean for real-world data like wages, heartbeats, and literacy rates.
This exercise focuses on calculating the mean ($\bar{x}$) of grouped data distributions using suitable methods like the Direct Method, Assumed Mean Method, or Step-Deviation Method.
1. Mean Number of Plants per House (Direct Method)
1. A survey was conducted by a group of students as a part of their environment awareness program, 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 (Class Interval, C.I.) | Number of houses (Frequency, fi) | Class Mark (xi) | fixi |
| 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 | $\mathbf{\sum f_i = 20}$ | $\mathbf{\sum f_i x_i = 162}$ |
Calculation (Direct Method):
$$\text{Mean} (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1$$
The mean number of plants per house is 8.1.
Method Used and Why:
The Direct Method was used because the numerical values of the class marks ($x_i$) and frequencies ($f_i$) are small. This minimizes the calculation effort without needing simplification steps like deviation or step-deviation.
2. Mean Daily Wages (Assumed Mean Method)
Consider the following distribution of daily wages of 50 workers of a factory.
| Daily wages (in Rs.) | 500-520 | 520-540 | 540-560 | 560-580 | 580-600 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Let’s use the Assumed Mean Method since the class marks are large, but the class size ($h=20$) is constant.
Assume mean, $A = 550$ (Class mark of 540-560).
| Daily wages (in ₹, C.I.) | Number of workers (fi) | Class Mark (xi) | Deviation (di=xi−A) | fidi |
| 500 – 520 | 12 | 510 | -40 | -480 |
| 520 – 540 | 14 | 530 | -20 | -280 |
| 540 – 560 | 8 | 550 = A | 0 | 0 |
| 560 – 580 | 6 | 570 | 20 | 120 |
| 580 – 600 | 10 | 590 | 40 | 400 |
| Total | $\mathbf{\sum f_i = 50}$ | $\mathbf{\sum f_i d_i = -240}$ |
Calculation (Assumed Mean Method):
$$\text{Mean} (\bar{x}) = A + \frac{\sum f_i d_i}{\sum f_i}$$
$$\bar{x} = 550 + \frac{-240}{50} = 550 – 4.8 = 545.20$$
The mean daily wages of the workers is ₹ 545.20.
3.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 | 23-35 |
| Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
The class size ($h=2$) is constant. Let’s use the Assumed Mean Method.
Assume mean, $A = 18$ (Class mark of 17-19, which simplifies calculations).
| Daily pocket allowance (C.I.) | Number of children (fi) | Class Mark (xi) | Deviation (di=xi−A) | fidi |
| 11 – 13 | 7 | 12 | -6 | -42 |
| 13 – 15 | 6 | 14 | -4 | -24 |
| 15 – 17 | 9 | 16 | -2 | -18 |
| 17 – 19 | 13 | 18 = A | 0 | 0 |
| 19 – 21 | $f$ | 20 | 2 | $2f$ |
| 21 – 23 | 5 | 22 | 4 | 20 |
| 23 – 25 | 4 | 24 | 6 | 24 |
| Total | $\mathbf{\sum f_i = 44 + f}$ | $\mathbf{\sum f_i d_i = 2f – 40}$ |
Calculation (Using Mean Formula):
$$\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}$$
$$18 = 18 + \frac{2f – 40}{44 + f}$$
$$0 = \frac{2f – 40}{44 + f}$$
$$2f – 40 = 0$$
$$2f = 40$$
$$f = 20$$
The missing frequency ($f$) is 20.
4. Thirty women were examined in a hospital by a doctor, and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats 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 | 2 |
The class size ($h=3$) is constant. Let’s use the Step-Deviation Method as the frequencies are not extremely small.
Assume mean, $A = 74.5$ (Class mark of 74-77).
| Heartbeats per minute (C.I.) | Number of women (fi) | Class Mark (xi) | Deviation (di=xi−A) | ui=di/h | fiui |
| 65 – 68 | 2 | 66.5 | -9 | -3 | -6 |
| 68 – 71 | 4 | 69.5 | -6 | -2 | -8 |
| 71 – 74 | 3 | 72.5 | -3 | -1 | -3 |
| 74 – 77 | 8 | 75.5 | 0 | 0 | 0 |
| 77 – 80 | 7 | 78.5 | 3 | 1 | 7 |
| 80 – 83 | 4 | 81.5 | 6 | 2 | 8 |
| 83 – 86 | 2 | 84.5 | 9 | 3 | 6 |
| Total | $\mathbf{\sum f_i = 30}$ | $\mathbf{\sum f_i u_i = 4}$ |
Calculation (Step-Deviation Method):
$$\text{Mean} (\bar{x}) = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$
$$\bar{x} = 75.5 + \left(\frac{4}{30}\right) \times 3$$
$$\bar{x} = 75.5 + \frac{4}{10} = 75.5 + 0.4 = 75.9$$
The mean heartbeats per minute is 75.9.
5. 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 according to 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 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Note on Class Intervals: The given class intervals are discontinuous (50-52, 53-55, etc.). Since the actual size of mangoes is continuous, we first need to make them continuous by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit (e.g., $50-52$ becomes $49.5-52.5$). However, the class mark ($x_i$) remains the same even after adjustment, so we can proceed directly using the given intervals.
Let’s use the Step-Deviation Method as the class marks ($x_i$) and frequencies ($f_i$) are large, and the class size ($h=3$) is constant.
Assume mean, $A = 57$ (Class mark of 56-58).
| Number of mangoes (C.I.) | Number of boxes (fi) | Class Mark (xi) | Deviation (di=xi−A) | ui=di/h | fiui |
| 50 – 52 | 15 | 51 | -6 | -2 | -30 |
| 53 – 55 | 110 | 54 | -3 | -1 | -110 |
| 56 – 58 | 135 | 57 = A | 0 | 0 | 0 |
| 59 – 61 | 115 | 60 | 3 | 1 | 115 |
| 62 – 64 | 25 | 63 | 6 | 2 | 50 |
| Total | $\mathbf{\sum f_i = 400}$ | $\mathbf{\sum f_i u_i = 25}$ |
Calculation (Step-Deviation Method):
$$\text{Mean} (\bar{x}) = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$
$$\bar{x} = 57 + \left(\frac{25}{400}\right) \times 3$$
$$\bar{x} = 57 + \frac{1}{16} \times 3 = 57 + \frac{3}{16} = 57 + 0.1875 = 57.19$$
The mean number of mangoes is 57.19.
Method Used and Why:
The Step-Deviation Method was chosen because both the class marks and the frequencies were large, making the calculations using $u_i$ (step deviation) much simpler and reducing the chance of calculation errors.
6. Mean Daily Expenditure (Step-Deviation Method)
6. 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 |
Find the mean daily expenditure on food by a suitable method.
The class size ($h=50$) is constant. Let’s use the Step-Deviation Method.
Assume mean, $A = 225$ (Class mark of 200-250).
| Daily expenditure (in ₹, C.I.) | Number of households (fi) | Class Mark (xi) | Deviation (di=xi−A) | ui=di/h | fiui |
| 100 – 150 | 4 | 125 | -100 | -2 | -8 |
| 150 – 200 | 5 | 175 | -50 | -1 | -5 |
| 200 – 250 | 12 | 225 = A | 0 | 0 | 0 |
| 250 – 300 | 2 | 275 | 50 | 1 | 2 |
| 300 – 350 | 2 | 325 | 100 | 2 | 4 |
| Total | $\mathbf{\sum f_i = 25}$ | $\mathbf{\sum f_i u_i = -7}$ |
Calculation (Step-Deviation Method):
$$\text{Mean} (\bar{x}) = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$
$$\bar{x} = 225 + \left(\frac{-7}{25}\right) \times 50$$
$$\bar{x} = 225 + (-7 \times 2) = 225 – 14 = 211$$
The mean daily expenditure on food is ₹ 211.
7. Mean Concentration of SO₂ (Step-Deviation Method)
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:
| Concentration of SO2 ( 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 SO2 in the air.
The class size ($h=0.04$) is constant. The class marks are small but involve decimals, making the Step-Deviation Method ideal.
Assume mean, $A = 0.10$ (Class mark of 0.08-0.12).
| Concentration of SO2 (in ppm, C.I.) | Frequency (fi) | Class Mark (xi) | Deviation (di=xi−A) | ui=di/h | fiui |
| 0.00 – 0.04 | 4 | 0.02 | -0.08 | -2 | -8 |
| 0.04 – 0.08 | 9 | 0.06 | -0.04 | -1 | -9 |
| 0.08 – 0.12 | 9 | 0.10 = A | 0 | 0 | 0 |
| 0.12 – 0.16 | 2 | 0.14 | 0.04 | 1 | 2 |
| 0.16 – 0.20 | 4 | 0.18 | 0.08 | 2 | 8 |
| 0.20 – 0.24 | 2 | 0.22 | 0.12 | 3 | 6 |
| Total | $\mathbf{\sum f_i = 30}$ | $\mathbf{\sum f_i u_i = -1}$ |
Calculation (Step-Deviation Method):
$$\text{Mean} (\bar{x}) = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$
$$\bar{x} = 0.10 + \left(\frac{-1}{30}\right) \times 0.04$$
$$\bar{x} = 0.10 – \frac{0.04}{30} \approx 0.10 – 0.00133$$
$$\bar{x} = 0.09867$$
The mean concentration of $\text{SO}_2$ in the air is $\mathbf{0.099 \text{ ppm}}$ (rounded to three decimal places).
8. Mean Number of Days Absent (Assumed Mean Method)
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.
| 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 |
Note on Class Intervals: The class intervals are unequal (6, 4, 4, 6, 8, 10, 2). This means we cannot use the Step-Deviation Method, as it requires a constant class size $h$. We must use the Direct or Assumed Mean Method.
Let’s use the Assumed Mean Method.
Assume mean, $A = 17$ (Class mark of 14-20).
| Number of days (C.I.) | Number of students (fi) | Class Mark (xi) | Deviation (di=xi−A) | fidi |
| 0 – 6 | 11 | 3 | -14 | -154 |
| 6 – 10 | 10 | 8 | -9 | -90 |
| 10 – 14 | 7 | 12 | -5 | -35 |
| 14 – 20 | 4 | 17 = A | 0 | 0 |
| 20 – 28 | 4 | 24 | 7 | 28 |
| 28 – 38 | 3 | 33 | 16 | 48 |
| 38 – 40 | 1 | 39 | 22 | 22 |
| Total | $\mathbf{\sum f_i = 40}$ | $\mathbf{\sum f_i d_i = -181}$ |
Calculation (Assumed Mean Method):
$$\text{Mean} (\bar{x}) = A + \frac{\sum f_i d_i}{\sum f_i}$$
$$\bar{x} = 17 + \frac{-181}{40} = 17 – 4.525 = 12.475$$
The mean number of days a student was absent is 12.48 days (rounded to two decimal places).
9. Mean Literacy Rate (Step-Deviation Method)
9. 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 |
The class size ($h=10$) is constant. Let’s use the Step-Deviation Method.
Assume mean, $A = 70$ (Class mark of 65-75).
| Literacy rate (in %, C.I.) | Number of cities (fi) | Class Mark (xi) | Deviation (di=xi−A) | ui=di/h | fiui |
| 45 – 55 | 3 | 50 | -20 | -2 | -6 |
| 55 – 65 | 10 | 60 | -10 | -1 | -10 |
| 65 – 75 | 11 | 70 = A | 0 | 0 | 0 |
| 75 – 85 | 8 | 80 | 10 | 1 | 8 |
| 85 – 95 | 3 | 90 | 20 | 2 | 6 |
| Total | $\mathbf{\sum f_i = 35}$ | $\mathbf{\sum f_i u_i = -2}$ |
Calculation (Step-Deviation Method):
$$\text{Mean} (\bar{x}) = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$
$$\bar{x} = 70 + \left(\frac{-2}{35}\right) \times 10$$
$$\bar{x} = 70 – \frac{20}{35} = 70 – 0.5714… \approx 69.4286$$
The mean literacy rate is 69.43% (rounded to two decimal places).