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

# NCERT Solutions for Class 10 Maths Chapter 13 Statistics

This chapter's questions cover calculations and concepts related to measures of central tendency for grouped and ungrouped data, including mean, median, and mode.

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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.

## EXERCISE 13.2

### Question 1

*3 marks · Short answer*

The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years): 5-15, 15-25, 25-35, 35-45, 45-55, 55-65
Number of patients: 6, 11, 21, 23, 14, 5
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

**Solution**

1. The maximum class frequency is 23, so the modal class is $35 - 45$, with $l = 35$, $h = 10$, $f_1 = 23$, $f_0 = 21$, and $f_2 = 14$.
2. Using the mode formula $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$, we get $\text{Mode} = 35 + \left(\frac{23 - 21}{2(23) - 21 - 14}\right) \times 10 = 35 + \frac{20}{11} = 36.8$ years.
3. For the mean, the class marks $x_i$ for $5-15, 15-25, 25-35, 35-45, 45-55, 55-65$ are $10, 20, 30, 40, 50, 60$.
4. Using the assumed mean $a = 40$ and $h = 10$, $u_i = \frac{x_i - 40}{10}$ gives $u_i$ values $-3, -2, -1, 0, 1, 2$, and $f_i u_i$ values $-18, -22, -21, 0, 14, 10$, with $\sum f_i u_i = -37$ and $\sum f_i = 80$.
5. Using the step-deviation method $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 40 + \left(\frac{-37}{80}\right) \times 10 = 35.37$ years.
6. Interpretation: The maximum number of patients admitted are of age $36.8$ years, whereas the average age of the patients admitted is $35.37$ years.

**Answer:** Modal age = 36.8 years, Mean age = 35.37 years

> Common mistake: Taking incorrect lower limit $l$ or confusing preceding and succeeding frequencies in the mode formula.

### Question 2

*3 marks · Short answer*

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :
Lifetimes (in hours): 0-20, 20-40, 40-60, 60-80, 80-100, 100-120
Frequency: 10, 35, 52, 61, 38, 29
Determine the modal lifetimes of the components.

**Solution**

1. The maximum class frequency is 61, corresponding to the class interval 60 - 80, so the modal class is 60 - 80.
2. Identify the formula parameters: lower limit $l = 60$, class size $h = 20$, frequency of modal class $f_1 = 61$, frequency of preceding class $f_0 = 52$, and frequency of succeeding class $f_2 = 38$.
3. Substitute the values into the mode formula: $\text{Mode} = 60 + \left(\frac{61 - 52}{2(61) - 52 - 38}\right) \times 20$.
4. Calculate the result: $\text{Mode} = 60 + \frac{9}{122 - 90} \times 20 = 60 + \frac{180}{32} = 60 + 5.625 = 65.625$ hours.

**Answer:** 65.625 hours

> Common mistake: Incorrect identification of $f_0$ as the frequency after the modal class instead of before.

### Question 3

*3 marks · Short answer*

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :
Expenditure (in ₹): 1000-1500, 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000
Number of families: 24, 40, 33, 28, 30, 22, 16, 7

**Part (i)**

1. The maximum frequency is 40, which corresponds to the class interval 1500 - 2000, so the modal class is 1500 - 2000.
2. Substitute $l = 1500$, $h = 500$, $f_1 = 40$, $f_0 = 24$, and $f_2 = 33$ into the mode formula.
3. $\text{Mode} = 1500 + \left(\frac{40 - 24}{2(40) - 24 - 33}\right) \times 500 = 1500 + \frac{8000}{23} = 1847.83$.

Answer (i): Rs 1847.83

**Part (ii)**

1. Using the step-deviation method with assumed mean $a = 2750$ and class size $h = 500$, calculate $u_i = \frac{x_i - 2750}{500}$ to get $\sum f_i = 200$ and $\sum f_i u_i = -15$.
2. Substitute the values into the mean formula $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$.
3. $\bar{x} = 2750 + \left(\frac{-15}{200}\right) \times 500 = 2750 - 37.5 = 2662.50$.

Answer (ii): Rs 2662.50

**Answer:** Modal expenditure = Rs 1847.83, Mean expenditure = Rs 2662.50

> Common mistake: Wrong choice of assumed mean leading to tedious arithmetic.

### Question 4

*3 marks · Short answer*

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
Number of students per teacher: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55
Number of states/U.T.: 3, 8, 9, 10, 3, 0, 0, 2

**Part (i)**

1. The maximum frequency is 10, corresponding to the class interval $30 - 35$. So, the modal class is $30 - 35$.
2. Here $l = 30$, $h = 5$, $f_1 = 10$, $f_0 = 9$, and $f_2 = 3$.
3. Using the formula, $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h = 30 + \left(\frac{10 - 9}{2(10) - 9 - 3}\right) \times 5$
4. $\text{Mode} = 30 + \frac{1}{8} \times 5 = 30 + 0.625 = 30.6$ (approx).

Answer (i): 30.6

**Part (ii)**

1. Let us choose the assumed mean $a = 32.5$ and class size $h = 5$.
2. Constructing the table with class marks $x_i$: for $15-20$, $x_i = 17.5$, $u_i = -3$, $f_i u_i = -9$; for $20-25$, $x_i = 22.5$, $u_i = -2$, $f_i u_i = -16$; for $25-30$, $x_i = 27.5$, $u_i = -1$, $f_i u_i = -9$; for $30-35$, $x_i = 32.5$, $u_i = 0$, $f_i u_i = 0$; for $35-40$, $x_i = 37.5$, $u_i = 1$, $f_i u_i = 3$; for $40-45$, $x_i = 42.5$, $u_i = 2$, $f_i u_i = 0$; for $45-50$, $x_i = 47.5$, $u_i = 3$, $f_i u_i = 0$; for $50-55$, $x_i = 52.5$, $u_i = 4$, $f_i u_i = 8$.
3. Total frequency $\sum f_i = 3 + 8 + 9 + 10 + 3 + 0 + 0 + 2 = 35$, and $\sum f_i u_i = -9 - 16 - 9 + 0 + 3 + 0 + 0 + 8 = -23$.
4. Using the step-deviation method, $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 32.5 + \left(\frac{-23}{35}\right) \times 5 = 32.5 - \frac{23}{7} = 32.5 - 3.29 = 29.21$.

Answer (ii): 29.2

**Answer:** The mode is 30.6 and the mean is 29.2.

> Common mistake: Taking the wrong frequency for $f_0$ or $f_2$ while calculating the mode.

### Question 5

*3 marks · Short answer*

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
Runs scored: 3000-4000, 4000-5000, 5000-6000, 6000-7000, 7000-8000, 8000-9000, 9000-10000, 10000-11000
Number of batsmen: 4, 18, 9, 7, 6, 3, 1, 1
Find the mode of the data.

**Solution**

1. Identify the maximum class frequency as 18, which corresponds to the class interval $4000-5000$.
2. Set the parameters for the mode formula: lower limit $l = 4000$, class size $h = 1000$, $f_1 = 18$, $f_0 = 4$, and $f_2 = 9$.
3. Substitute these values into the mode formula: $\text{Mode} = 4000 + \frac{18-4}{2(18)-4-9} \times 1000 = 4000 + \frac{14}{23} \times 1000 = 4608.7$.

**Answer:** 4608.7 runs

> Common mistake: Forgetting to add the lower limit $l$ after calculating the fraction part.

### Question 6

*3 marks · Short answer*

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :
Number of cars: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80
Frequency: 7, 14, 13, 12, 20, 11, 15, 8

**Solution**

1. Identify the maximum frequency as 20, which corresponds to the class interval $40-50$.
2. Set the parameters for the mode formula: lower limit $l = 40$, class size $h = 10$, $f_1 = 20$, $f_0 = 12$, and $f_2 = 11$.
3. Substitute these values into the mode formula: $\text{Mode} = 40 + \frac{20-12}{2(20)-12-11} \times 10 = 40 + \frac{8}{17} \times 10 = 44.7$.

**Answer:** 44.7 cars

> Common mistake: Incorrectly identifying $f_0$ and $f_2$ from the frequency list.

## EXERCISE 13.3

### Question 1

*6 marks · Long answer*

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption (in units): 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205
Number of consumers: 4, 5, 13, 20, 14, 8, 4

**Part (i)**

1. Construct the frequency distribution table with class intervals, frequency ($f_i$), class mark ($x_i$), $f_i x_i$, cumulative frequency ($cf$), and step-deviation ($u_i$).
2. For class 125-145, $cf = 22$, $f = 20$, $l = 125$, and $h = 20$.
3. Calculate median using Median = $l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h = 125 + \left(\frac{34 - 22}{20}\right) \times 20 = 137$ units.

Answer (i): Median = 137 units

**Part (ii)**

1. Using the direct method or step-deviation method with assumed mean $a = 135$ and $h = 20$, compute $\sum f_i x_i = 9320$ and $\sum f_i = 68$.
2. Calculate mean using $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{9320}{68} = 137.06$ units.

Answer (ii): Mean = 137.06 units

**Part (iii)**

1. Identify the modal class with maximum frequency 20, which is 125-145.
2. Use the mode formula: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h = 125 + \left(\frac{20 - 13}{2(20) - 13 - 14}\right) \times 20 = 135.77$ units.
3. Comparing the three measures, we observe that they are approximately equal, indicating that the data is moderately skewed and representative of a normal distribution.

Answer (iii): Mode = 135.77 units, and the three measures are approximately equal.

**Answer:** Median = 137 units, Mean = 137.06 units, Mode = 135.77 units

> Common mistake: Taking incorrect cumulative frequency or wrong lower limit for the median or modal class.

### Question 2

*3 marks · Short answer*

If the median of the distribution given below is 28.5, find the values of $x$ and $y$.
Class interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60
Frequency: 5, $x$, 20, 15, $y$, 5 (Total = 60)

**Solution**

1. Set up the cumulative frequency table: class intervals with frequencies $5, x, 20, 15, y, 5$ and cumulative frequencies $5, 5+x, 25+x, 40+x, 40+x+y, 45+x+y$.
2. Use the given total frequency $60$ to write $45 + x + y = 60$, which simplifies to $x + y = 15$.
3. Use the given median $28.5$ to identify the median class as $20 - 30$ with $l = 20$, $f = 20$, $cf = 5 + x$, and $h = 10$.
4. Substitute into the median formula: $28.5 = 20 + \left(\frac{30 - (5 + x)}{20}\right) \times 10$.
5. Solve the equation to find $x = 8$, and substitute $x = 8$ into $x + y = 15$ to get $y = 7$.

**Answer:** x = 8 and y = 7

> Common mistake: Taking the wrong cumulative frequency for the median class.

### Question 3

*3 marks · Short answer*

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
Age (in years): Below 20, Below 25, Below 30, Below 35, Below 40, Below 45, Below 50, Below 55, Below 60
Number of policy holders: 2, 6, 24, 45, 78, 89, 92, 98, 100

**Solution**

1. Given cumulative frequency distribution: Below 20: 2, Below 25: 6, Below 30: 24, Below 35: 45, Below 40: 78, Below 45: 89, Below 50: 92, Below 55: 98, Below 60: 100.
2. Convert the given less-than type cumulative frequency distribution into class intervals with frequencies: 20-25: 4, 25-30: 18, 30-35: 21, 35-40: 33, 40-45: 11, 45-50: 3, 50-55: 6, 55-60: 2.
3. Here $n = 100$, so $\frac{n}{2} = 50$, which lies in the median class $35-40$ with lower limit $l = 35$, cumulative frequency of preceding class $\text{cf} = 45$, frequency $f = 33$, and class size $h = 5$.
4. Using the formula $\text{Median} = l + \left(\frac{\frac{n}{2} - \text{cf}}{f}\right) \times h$, we substitute the values to get $\text{Median} = 35 + \left(\frac{50 - 45}{33}\right) \times 5$.
5. Simplifying the expression gives $\text{Median} = 35 + \frac{25}{33} = 35 + 0.76 = 35.76\text{ years}$.

**Answer:** 35.76 years

> Common mistake: Subtracting cumulative frequencies incorrectly while finding the class frequencies.

### Question 4

*3 marks · Short answer*

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
Length (in mm): 118-126, 127-135, 136-144, 145-153, 154-162, 163-171, 172-180
Number of leaves: 3, 5, 9, 12, 5, 4, 2
Find the median length of the leaves.
(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5-126.5, 126.5-135.5, ..., 171.5-180.5.)

**Solution**

1. Convert the discontinuous class intervals into continuous class intervals by subtracting $0.5$ from the lower limits and adding $0.5$ to the upper limits: $117.5 - 126.5, 126.5 - 135.5, \dots, 171.5 - 180.5$.
2. Write the frequencies and compute the cumulative frequencies: $3, 8, 17, 29, 34, 38, 40$.
3. Find $n = 40$, so $\frac{n}{2} = 20$.
4. Identify the median class as $144.5 - 153.5$ with $l = 144.5$, $f = 12$, $cf = 17$, and $h = 9$.
5. Substitute the values in the median formula: $\text{Median} = 144.5 + \left(\frac{20 - 17}{12}\right) \times 9 = 144.5 + 2.25 = 146.75$ mm.

**Answer:** 146.75 mm

> Common mistake: Applying the median formula directly without converting the classes into continuous intervals.

### Question 5

*3 marks · Short answer*

The following table gives the distribution of the life time of 400 neon lamps :
Life time (in hours): 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000
Number of lamps: 14, 56, 60, 86, 74, 62, 48
Find the median life time of a lamp.

**Solution**

1. Prepare the cumulative frequency table for the given life times: frequencies are $14, 56, 60, 86, 74, 62, 48$ and cumulative frequencies are $14, 70, 130, 216, 290, 352, 400$.
2. Find $n = 400$, so $\frac{n}{2} = 200$.
3. Identify the median class as $3000 - 3500$ with lower limit $l = 3000$, frequency $f = 86$, cumulative frequency of preceding class $cf = 130$, and class size $h = 500$.
4. Substitute the values into the median formula: $\text{Median} = 3000 + \left(\frac{200 - 130}{86}\right) \times 500$.
5. Calculate the final value: $\text{Median} = 3000 + \frac{70 \times 500}{86} = 3000 + 406.98 = 3406.98$ hours.

**Answer:** 3406.98 hours

> Common mistake: Arithmetic errors in computing the fractional part of the median.

### Question 6

*3 marks · Short answer*

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters: 1-4, 4-7, 7-10, 10-13, 13-16, 16-19
Number of surnames: 6, 30, 40, 16, 4, 4
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

**Part (i)**

1. Find cumulative frequencies for the classes: $6, 36, 76, 92, 96, 100$.
2. Find $n = 100$, so $\frac{n}{2} = 50$.
3. Identify median class as $7 - 10$ with $l = 7$, $f = 40$, $cf = 36$, and $h = 3$, giving $\text{Median} = 7 + \left(\frac{50 - 36}{40}\right) \times 3 = 8.05$ letters.

Answer (i): 8.05 letters

**Part (ii)**

1. Find class marks $x_i$: $2.5, 5.5, 8.5, 11.5, 14.5, 17.5$.
2. Calculate $f_i x_i$: $15, 165, 340, 184, 58, 70$, giving $\sum f_i x_i = 832$.
3. Compute mean as $\bar{x} = \frac{832}{100} = 8.32$ letters.

Answer (ii): 8.32 letters

**Part (iii)**

1. Identify modal class as $7 - 10$ with $l = 7$, $f_1 = 40$, $f_0 = 30$, $f_2 = 16$, and $h = 3$.
2. Substitute into the mode formula: $\text{Mode} = 7 + \left(\frac{40 - 30}{2(40) - 30 - 16}\right) \times 3$.
3. Simplify to get $\text{Mode} = 7 + \frac{10}{34} \times 3 = 7.88$ letters.

Answer (iii): 7.88 letters

**Answer:** Median = 8.05 letters, Mean = 8.32 letters, Mode = 7.88 letters

> Common mistake: Using incorrect class marks for intervals with decimal mid-points.

### Question 7

*3 marks · Short answer*

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight (in kg): 40-45, 45-50, 50-55, 55-60, 60-65, 65-70, 70-75
Number of students: 2, 3, 8, 6, 6, 3, 2

**Solution**

1. Prepare the cumulative frequency table for the given weight distribution: classes 40-45 ($f=2, cf=2$), 45-50 ($f=3, cf=5$), 50-55 ($f=8, cf=13$), 55-60 ($f=6, cf=19$), 60-65 ($f=6, cf=25$), 65-70 ($f=3, cf=28$), 70-75 ($f=2, cf=30$).
2. Total frequency $n = 30$, so $\frac{n}{2} = 15$. The cumulative frequency just greater than 15 is 19, corresponding to the median class 55-60.
3. Here $l = 55$, $cf = 13$, $f = 6$, and $h = 5$.
4. Substitute the values into the median formula: $\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h = 55 + \left(\frac{15 - 13}{6}\right) \times 5 = 55 + \frac{10}{6} = 56.67$ kg.

**Answer:** Median weight = 56.67 kg

> Common mistake: Using the wrong cumulative frequency of the preceding class or incorrect class size $h$.

## Frequently asked questions

### How many exercises and questions are there in NCERT Solutions for Class 10 Maths Chapter 13 Statistics?

This chapter has a total of 3 exercises with 22 questions in all, including 9 questions in Exercise 13.1, 6 questions in Exercise 13.2, and 7 questions in Exercise 13.3. You can find step-by-step solutions for all these questions in SwaVid's free PDF available on this page only.

### What main topics and concepts are covered in the exercises of Class 10 Chapter 13 Statistics?

The exercises cover important concepts such as finding the mean using direct, assumed mean, and step-deviation methods, calculating the mode of grouped data, and finding the median of grouped data. Additionally, you will learn how to handle missing frequencies and discontinuous class intervals across these topics.

### Which are the hardest question types in this Statistics chapter and how should I approach them?

Questions involving missing frequencies for mean, median, or combined central tendencies are often considered the trickiest. To approach them, clearly set up the given formulas like the median formula $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$ and substitute the known values carefully to solve for the unknowns.

### How should I write my answers in the board exams to score full marks in Class 10 Statistics?

To secure full marks, always write down the given values clearly, state the formula you are using before applying it, and show your calculation steps in a neat table. SwaVid's free PDF on this page only provides well-formatted solutions that demonstrate the exact presentation style required for board exams.

### Is a free PDF for Class 10 Maths Chapter 13 Statistics solutions available?

Yes, complete and reliable solutions aligned with the latest NCERT textbook for the 2026-27 session are available on this page. You can easily access SwaVid's free PDF to practice all textbook questions and improve your exam preparation.

## Related pages

- [Exercise 13.1 solutions](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-1)
- [Exercise 13.2 solutions](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-2)
- [Exercise 13.3 solutions](https://www.swavid.com/maths/class/10/chapter/statistics/ncert-solutions/exercise-13-3)
- [Statistics: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/statistics)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

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