---
title: "NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1"
url: https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-1
dateModified: 2026-10-07T15:56:30+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1

Chapter 13: Statistics. Every question from Exercise 13.1, with full working and the final answer.

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## EXERCISE 13.1

### Question 1

*3 marks · Short answer*

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?

**Solution**

1. Find the class mark $x_i$ for each interval using $x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$.
2. The class marks for $0-2, 2-4, 4-6, 6-8, 8-10, 10-12, 12-14$ are $1, 3, 5, 7, 9, 11, 13$ respectively.
3. Calculate $f_i x_i$ for each class: $1, 6, 5, 35, 54, 22, 39$.
4. Find the sum $\sum f_i = 20$ and $\sum f_i x_i = 162$.
5. Compute the mean using the direct method: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1$ plants.
6. The direct method was used because the numerical values of $x_i$ and $f_i$ are small.

**Answer:** 8.1 plants

> Common mistake: Arithmetic errors in finding class marks or products.

### Question 2

*3 marks · Short answer*

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.

**Solution**

1. Given class intervals and frequencies: 500-520 (12), 520-540 (14), 540-560 (8), 560-580 (6), 580-600 (10), with total frequency $\sum f_i = 50$.
2. Choose class marks $x_i$ for each interval: $510, 530, 550, 570, 590$, and take assumed mean $a = 550$ with class size $h = 20$.
3. Calculate $u_i = \frac{x_i - a}{h}$ to get $-2, -1, 0, 1, 2$, then find $f_i u_i$ products: $-24, -14, 0, 6, 20$ giving $\sum f_i u_i = -12$.
4. Apply the step-deviation formula $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 550 + \left(\frac{-12}{50}\right) \times 20 = 550 - 4.8 = 545.2$.

**Answer:** ₹545.20

> Common mistake: Arithmetic errors while computing deviation products or forgetting to multiply by the class size h.

### Question 3

*3 marks · Short answer*

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

**Solution**

1. Find the class marks $x_i$ for the intervals: $12, 14, 16, 18, 20, 22, 24$.
2. List the frequencies $f_i$: $7, 6, 9, 13, f, 5, 4$.
3. Calculate $f_i x_i$ for each class: $84, 84, 144, 234, 20f, 110, 96$.
4. Find the sum of frequencies $\sum f_i = 44 + f$ and sum of products $\sum f_i x_i = 752 + 20f$.
5. Use the given mean $\bar{x} = 18$ in the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
6. Substitute the values: $18 = \frac{752 + 20f}{44 + f}$, giving $18(44 + f) = 752 + 20f$.
7. Solve for $f$: $792 + 18f = 752 + 20f$, which gives $2f = 40$, so $f = 20$.

**Answer:** 20

> Common mistake: Algebraic errors while solving linear equation for $f$.

### Question 4

*3 marks · Short answer*

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

**Solution**

1. Find the class marks $x_i$ for $65-68, 68-71, 71-74, 74-77, 77-80, 80-83, 83-86$, which are $66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5$.
2. Choose assumed mean $a = 75.5$ and class size $h = 3$.
3. Calculate $u_i = \frac{x_i - a}{h}$, giving $-3, -2, -1, 0, 1, 2, 3$.
4. Multiply by frequencies to get $f_i u_i$: $-6, -8, -3, 0, 7, 8, 6$, whose sum $\sum f_i u_i = 4$.
5. Find total frequency $\sum f_i = 30$.
6. Compute mean: $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 75.5 + \left(\frac{4}{30}\right) \times 3 = 75.5 + 0.4 = 75.9$.

**Answer:** 75.9 heartbeats per minute

> Common mistake: Errors in calculating class marks with decimal values.

### Question 5

*3 marks · Short answer*

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?

**Solution**

1. Notice that the classes are inclusive ($50-52, 53-55$), so convert them to continuous classes by subtracting $0.5$ from lower limits and adding $0.5$ to upper limits: $49.5-52.5, 52.5-55.5, 55.5-58.5, 58.5-61.5, 61.5-64.5$.
2. Find the class marks $x_i$: $51, 54, 57, 60, 63$.
3. Choose assumed mean $a = 57$ and class size $h = 3$.
4. Calculate $u_i = \frac{x_i - 57}{3}$, giving $-2, -1, 0, 1, 2$.
5. Multiply by frequencies to get $f_i u_i$: $-30, -110, 0, 115, 50$, with sum $\sum f_i u_i = 25$.
6. Find total frequency $\sum f_i = 400$.
7. Compute mean: $\bar{x} = 57 + \left(\frac{25}{400}\right) \times 3 = 57 + 0.1875 = 57.19$ (approx).
8. The step-deviation method was chosen because frequencies and numbers are large.

**Answer:** 57.19 mangoes

> Common mistake: Failing to convert discontinuous class intervals into continuous ones.

### Question 6

*3 marks · Short answer*

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.

**Solution**

1. Find the class marks $x_i$ for $100-150, 150-200, 200-250, 250-300, 300-350$, which are $125, 175, 225, 275, 325$.
2. Choose assumed mean $a = 225$ and class size $h = 50$.
3. Calculate $u_i = \frac{x_i - 225}{50}$, giving $-2, -1, 0, 1, 2$.
4. Multiply by frequencies to get $f_i u_i$: $-8, -5, 0, 2, 4$, with sum $\sum f_i u_i = -7$.
5. Find total frequency $\sum f_i = 25$.
6. Compute mean: $\bar{x} = 225 + \left(\frac{-7}{25}\right) \times 50 = 225 - 14 = 211$.

**Answer:** ₹211

> Common mistake: Arithmetic errors while multiplying large class sizes.

### Question 7

*3 marks · Short answer*

To find out the concentration of $\text{SO}_2$ 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 $\text{SO}_2$ (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 $\text{SO}_2$ in the air.

**Solution**

1. Find the class mark $x_i$ for each interval as the average of upper and lower limits: $0.02, 0.06, 0.10, 0.14, 0.18, 0.22$.
2. Multiply each frequency $f_i$ by its corresponding class mark $x_i$ to get $f_i x_i$: $0.08, 0.54, 0.90, 0.28, 0.72, 0.44$.
3. Find the sum of frequencies $\sum f_i = 30$ and the sum of products $\sum f_i x_i = 2.96$.
4. Apply the direct method formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2.96}{30} = 0.099\text{ ppm}$.
5. State the final mean concentration of $\text{SO}_2$.

**Answer:** 0.099 ppm

> Common mistake: Arithmetic errors while calculating decimals for class marks and products.

### Question 8

*3 marks · Short answer*

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

**Solution**

1. Find the class mark $x_i$ for each interval: $3, 8, 12, 17, 24, 33, 39$.
2. Multiply each $f_i$ by $x_i$ to obtain $f_i x_i$: $33, 80, 84, 68, 96, 99, 39$.
3. Find the sum of frequencies $\sum f_i = 40$ and the sum of products $\sum f_i x_i = 499$.
4. Apply the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{499}{40} = 12.475\text{ days}$.
5. State the final mean number of days.

**Answer:** 12.48 days

> Common mistake: Incorrect class mark calculation for intervals of varying width.

### Question 9

*3 marks · Short answer*

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

**Solution**

1. Find the class mark $x_i$ for each interval: $50, 60, 70, 80, 90$.
2. Choose an assumed mean $a = 70$ and class size $h = 10$, then compute $u_i = \frac{x_i - 70}{10}$ as $-2, -1, 0, 1, 2$.
3. Multiply $f_i$ by $u_i$ to get $f_i u_i$: $-6, -10, 0, 8, 6$, and find their sum $\sum f_i u_i = -2$.
4. Find the sum of frequencies $\sum f_i = 35$.
5. Apply the step-deviation formula $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 70 + \left(\frac{-2}{35}\right) \times 10 = 69.43\%$.
6. State the final mean literacy rate.

**Answer:** 69.43%

> Common mistake: Forgetting to multiply by the class size $h$ in the step-deviation formula.

## Related pages

- [All Chapter 13 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions)
- [Exercise 13.2](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-2)
- [Exercise 13.3](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-3)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student