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.

Using the Assumed Mean Method with a = 550, the mean daily wages of the workers is calculated to be ₹ 545.20.

In a retail market, fruit vendors were selling mangoes kept in packing boxes. Find the mean number of mangoes kept in a packing box.

After adjusting the discontinuous class intervals to continuous ones (e.g., 49.5-52.5) and using the Step Deviation Method, the mean number of mangoes is approximately 57.19.

NCERT Solutions Class 10 Maths Chapter 13: Statistics

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

By using the formula for the mean and substituting the given mean of 18, the missing frequency f is calculated to be 20.

Q: 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.

Since the class sizes are unequal, the Direct Method is used. The mean number of days a student was absent is calculated to be 12.48 days.

Detailed solutions for Exercise 13.1 (Mean of Grouped Data).

Ex 13.1 Ex 13.2 Ex 13.3

#### **Q1. 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.** | 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 | **Which method did you use for finding the mean, and why?** #### **A1. Solution:** We use the **Direct Method** because the numerical values of $x_i$ and $f_i$ are small. | Number of plants (Class Interval) | Number of houses ($f_i$) | Class Mark ($x_i$) | $f_i x_i$ | | :--- | :---: | :---: | :---: | | 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** | **$\sum f_i = 20$** | | **$\sum f_i x_i = 162$** | Mean ($\bar{x}$) = $\frac{\sum f_i x_i}{\sum f_i}$ $= \frac{162}{20} = 8.1$ **Mean number of plants per house is 8.1.** #### **Q2. Consider the following distribution of daily wages of 50 workers of a factory.** | Daily wages (in ₹) | 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.** #### **A2. Solution:** We use the **Assumed Mean Method** to simplify calculations. Let Assumed Mean ($a$) = 550 (mid-value of 540-560). Class size ($h$) = 20. | Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $d_i = x_i - a$ | $f_i d_i$ | | :--- | :---: | :---: | :---: | :---: | | 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** | **$\sum f_i = 50$** | | | **$\sum f_i d_i = -240$** | Mean ($\bar{x}$) = $a + \frac{\sum f_i d_i}{\sum f_i}$ $= 550 + \frac{-240}{50}$ $= 550 - 4.8 = 545.2$ **Mean daily wages of the workers is ₹ 545.20.** #### **Q3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency $f$.** | Daily pocket allowance (in ₹) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of children** | 7 | 6 | 9 | 13 | $f$ | 5 | 4 | #### **A3. Solution:** Given: Mean ($\bar{x}$) = 18. We use the **Direct Method** (or Assumed Mean Method). Let's use Direct Method here for clarity. | Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $f_i x_i$ | | :--- | :---: | :---: | :---: | | 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** | **$\sum f_i = 44 + f$** | | **$\sum f_i x_i = 752 + 20f$** | Mean ($\bar{x}$) = $\frac{\sum f_i x_i}{\sum f_i}$ $18 = \frac{752 + 20f}{44 + f}$ $18(44 + f) = 752 + 20f$ $792 + 18f = 752 + 20f$ $792 - 752 = 20f - 18f$ $40 = 2f$ $f = 20$ **The missing frequency $f$ is 20.** #### **Q4. 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 heartbeats 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 | #### **A4. Solution:** We use the **Assumed Mean Method**. Let Assumed Mean ($a$) = 75.5. Class size ($h$) = 3. | Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $d_i = x_i - a$ | $f_i d_i$ | | :--- | :---: | :---: | :---: | :---: | | 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 ($a$) | 0 | 0 | | 77 - 80 | 7 | 78.5 | 3 | 21 | | 80 - 83 | 4 | 81.5 | 6 | 24 | | 83 - 86 | 2 | 84.5 | 9 | 18 | | **Total** | **$\sum f_i = 30$** | | | **$\sum f_i d_i = 12$** | Mean ($\bar{x}$) = $a + \frac{\sum f_i d_i}{\sum f_i}$ $= 75.5 + \frac{12}{30}$ $= 75.5 + 0.4 = 75.9$ **Mean heartbeats per minute is 75.9.** #### **Q5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers 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?** #### **A5. Solution:** The class intervals are not continuous. We make them continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit (though for finding class marks $x_i$, the result remains the same). We use the **Step Deviation Method**. Let Assumed Mean ($a$) = 57. Class size ($h$) = 3. | Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $u_i = \frac{x_i - a}{h}$ | $f_i u_i$ | | :--- | :---: | :---: | :---: | :---: | | 49.5 - 52.5 | 15 | 51 | -2 | -30 | | 52.5 - 55.5 | 110 | 54 | -1 | -110 | | 55.5 - 58.5 | 135 | 57 ($a$) | 0 | 0 | | 58.5 - 61.5 | 115 | 60 | 1 | 115 | | 61.5 - 64.5 | 25 | 63 | 2 | 50 | | **Total** | **$\sum f_i = 400$** | | | **$\sum f_i u_i = 25$** | Mean ($\bar{x}$) = $a + (\frac{\sum f_i u_i}{\sum f_i}) \times h$ $= 57 + (\frac{25}{400}) \times 3$ $= 57 + \frac{3}{16} = 57 + 0.1875$ $= 57.19$ (approx). **Mean number of mangoes is 57.19.** #### **Q6. The table below shows the daily expenditure on food of 25 households in a locality.** | Daily expenditure (in ₹) | 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.** #### **A6. Solution:** We use the **Step Deviation Method**. Let Assumed Mean ($a$) = 225. Class size ($h$) = 50. | Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $u_i = \frac{x_i - a}{h}$ | $f_i u_i$ | | :--- | :---: | :---: | :---: | :---: | | 100 - 150 | 4 | 125 | -2 | -8 | | 150 - 200 | 5 | 175 | -1 | -5 | | 200 - 250 | 12 | 225 ($a$) | 0 | 0 | | 250 - 300 | 2 | 275 | 1 | 2 | | 300 - 350 | 2 | 325 | 2 | 4 | | **Total** | **$\sum f_i = 25$** | | | **$\sum f_i u_i = -7$** | Mean ($\bar{x}$) = $a + (\frac{\sum f_i u_i}{\sum f_i}) \times h$ $= 225 + (\frac{-7}{25}) \times 50$ $= 225 + (-7 \times 2)$ $= 225 - 14 = 211$. **Mean daily expenditure is ₹ 211.** #### **Q7. 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) | 0.00-0.04 | 0.04-0.08 | 0.08-0.12 | 0.12-0.16 | 0.16-0.20 | 0.20-0.24 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | | **Frequency** | 4 | 9 | 9 | 2 | 4 | 2 | **Find the mean concentration of SO₂ in the air.** #### **A7. Solution:** We use the **Step Deviation Method**. Let Assumed Mean ($a$) = 0.10. Class size ($h$) = 0.04. | Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $u_i = \frac{x_i - a}{h}$ | $f_i u_i$ | | :--- | :---: | :---: | :---: | :---: | | 0.00 - 0.04 | 4 | 0.02 | -2 | -8 | | 0.04 - 0.08 | 9 | 0.06 | -1 | -9 | | 0.08 - 0.12 | 9 | 0.10 ($a$) | 0 | 0 | | 0.12 - 0.16 | 2 | 0.14 | 1 | 2 | | 0.16 - 0.20 | 4 | 0.18 | 2 | 8 | | 0.20 - 0.24 | 2 | 0.22 | 3 | 6 | | **Total** | **$\sum f_i = 30$** | | | **$\sum f_i u_i = -1$** | Mean ($\bar{x}$) = $a + (\frac{\sum f_i u_i}{\sum f_i}) \times h$ $= 0.10 + (\frac{-1}{30}) \times 0.04$ $= 0.10 - 0.00133...$ $= 0.099$ ppm (approx). **Mean concentration of SO₂ is 0.099 ppm.** #### **Q8. 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 | #### **A8. Solution:** Here, the class sizes are unequal. We use the **Direct Method**. | Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $f_i x_i$ | | :--- | :---: | :---: | :---: | | 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** | **$\sum f_i = 40$** | | **$\sum f_i x_i = 499$** | Mean ($\bar{x}$) = $\frac{\sum f_i x_i}{\sum f_i}$ $= \frac{499}{40} = 12.475$. **Mean number of days absent is 12.48 days.** #### **Q9. 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 | 85-95 | | :--- | :---: | :---: | :---: | :---: | :---: | | **Number of cities** | 3 | 10 | 11 | 8 | 3 | #### **A9. Solution:** We use the **Step Deviation Method**. Let Assumed Mean ($a$) = 70. Class size ($h$) = 10. | Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $u_i = \frac{x_i - a}{h}$ | $f_i u_i$ | | :--- | :---: | :---: | :---: | :---: | | 45 - 55 | 3 | 50 | -2 | -6 | | 55 - 65 | 10 | 60 | -1 | -10 | | 65 - 75 | 11 | 70 ($a$) | 0 | 0 | | 75 - 85 | 8 | 80 | 1 | 8 | | 85 - 95 | 3 | 90 | 2 | 6 | | **Total** | **$\sum f_i = 35$** | | | **$\sum f_i u_i = -2$** | Mean ($\bar{x}$) = $a + (\frac{\sum f_i u_i}{\sum f_i}) \times h$ $= 70 + (\frac{-2}{35}) \times 10$ $= 70 - \frac{4}{7}$ $= 70 - 0.57 = 69.43$. **Mean literacy rate is 69.43%.**