NCERT Solutions For Class 10 Maths Chapter 13 Statistics 2026-27

NCERT Solutions for Class 10 Maths Chapter 13 Statistics 2026-27 help students understand how to collect, organize, study, and interpret data in an easy way. Statistics is an important chapter in Class 10 Maths because it teaches students how to find useful information from numbers. In this chapter, students learn about grouped data, mean, median, and mode. They also understand how these concepts are used in real-life situations, such as comparing marks, studying population data, checking sales records, and analysing survey results.

The NCERT Solutions for Class 10 Maths Chapter 13 Statistics 2026-27 are prepared according to the latest NCERT syllabus and exam pattern. Each question is explained step by step so that students can understand the method clearly and avoid common mistakes. These solutions also help learners revise formulas, improve calculation skills, and prepare better for school exams and board exams.

Infinity Learn provides clear and student-friendly NCERT solutions that make difficult topics easier to understand. With the help of Infinity Learn, students can practice all important questions from Chapter 13 and build confidence in solving problems based on mean, median, and mode. The explanations are written in simple language, making them useful for quick learning and revision.

Using NCERT Solutions for Class 10 Maths Chapter 13 Statistics 2026-27 regularly can help students strengthen their basic concepts and improve accuracy. These solutions are also helpful for homework, class tests, assignments, and exam preparation. By studying the solved examples and practising similar questions, students can understand the chapter in a better way.

Download NCERT Solutions For Class 10 Maths Chapter 13 Statistics PDF

Students can download NCERT Solutions for Class 10 Maths Chapter 13 Statistics PDF to study the chapter easily anytime and anywhere. The PDF contains clear, step-by-step solutions to all the NCERT exercise questions based on mean, median, and mode. It is useful for completing homework, revising important formulas, and preparing for school and board exams. 

Access Class 10 Maths Chapter 13 Statistics NCERT Solutions

Exercise 13.1

Question 1: A group of students recorded the number of plants grown in 20 houses as part of an environmental awareness programme. Find the mean number of plants per house.

Solution:

The class marks and frequencies are small, so the direct method is convenient.

Class mark, x_i = (Lower class limit + Upper class limit)/2

Mean = (Σf_ix_i)/(Σf_i)

Mean = 162/20

Mean = 8.1

Answer: The mean number of plants per house is 8.1. The direct method is used because the class marks and frequencies are easy to multiply.

Question 2: The following distribution represents the daily wages of 50 factory workers. Find their mean daily wage using a suitable method.

Solution:

Since all class intervals have the same width, the step-deviation method is suitable.

Take the assumed mean, a = 150, and class size, h = 20.

u_i = (x_i - a)/h

Mean = a + [(Σf_iu_i)/(Σf_i)] × h

Mean = 150 + [(-12)/50] × 20

= 150 - 4.8

= 145.2

Answer: The mean daily wage is Rs. 145.20.

Question 3: The following distribution gives the daily pocket allowance of children in a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Solution:

Take the assumed mean a = 18.

Using the assumed-mean formula:

Mean = a + (Σf_id_i)/(Σf_i)

The given mean is 18.

18 = 18 + (2f - 40)/(44 + f)

Therefore, (2f - 40)/(44 + f) = 0

2f - 40 = 0

2f = 40

f = 20

Answer: The missing frequency is 20.

Question 4: The heartbeats per minute of 30 women were recorded in a hospital. Find the mean number of heartbeats per minute.

Solution:

Use the step-deviation method with assumed mean a = 75.5 and class size h = 3.

Mean = 75.5 + (4/30) × 3

= 75.5 + 0.4

= 75.9

Answer: The mean is 75.9 heartbeats per minute.

Question 5: Mangoes are packed in boxes in the following manner. Find the mean number of mangoes in a box. Also state the method used.

Solution:

The intervals are inclusive. They may be written as continuous intervals by subtracting 0.5 from every lower limit and adding 0.5 to every upper limit.

Take a = 57 and h = 3.

Mean = 57 + (25/400) × 3

= 57 + 0.1875

= 57.1875

Answer: The mean number of mangoes per box is 57.1875. The step-deviation method is convenient because the frequencies are large and the class marks have a common difference.

Question 6:The daily expenditure on food of 25 households is shown below. Find the mean daily expenditure.

Solution:

Take assumed mean a = 225 and class size h = 50.

Mean = 225 + [(-7)/25] × 50

= 225 - 14

= 211

Answer: The mean daily expenditure on food is Rs. 211.

Question 7: The concentration of SO₂ in the air was measured in 30 localities. Find the mean concentration.

Solution:

Take a = 0.14 and h = 0.04.

Mean = 0.14 + [(-31)/30] × 0.04

= 0.14 - 0.04133

= 0.09867 ppm

Rounded to three decimal places, the mean is 0.099 ppm.

Answer: The mean concentration of SO₂ is approximately 0.099 ppm.

Question 8: The following table records the number of days for which 40 students were absent during a term. Find the mean number of absent days.

Solution:

The class widths are not equal, so use the assumed-mean method. Take a = 17.

Mean = 17 + (-181/40)

= 17 - 4.525

= 12.475

Answer: The mean number of absent days is approximately 12.48 days.

Question 9: The following table gives the literacy rates of 35 cities. Find the mean literacy rate.

Solution:

Take a = 70 and h = 10.

Mean = 70 + (-2/35) × 10

= 70 - 20/35

= 69.4286

Answer: The mean literacy rate is approximately 69.43%.

Exercise 13.2

Question 1: The ages of patients admitted to a hospital during one year are given below. Find the mean and mode. Compare and interpret the results.

Mean:

Take a = 30 and d_i = x_i - 30.

Mean = 30 + 430/80

= 30 + 5.375

= 35.375 years

Therefore, mean age = 35.38 years, approximately.

Mode:

The highest frequency is 23, so the modal class is 35-45.

Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h

Here, l = 35, f₁ = 23, f₀ = 21, f₂ = 14 and h = 10.

Mode = 35 + [(23 - 21)/(46 - 21 - 14)] × 10

= 35 + (2/11) × 10

= 36.82 years, approximately.

Answer: The mean age is approximately 35.38 years, while the modal age is approximately 36.82 years. Thus, the average patient was about 35 years old, while the largest concentration of patients was around 37 years old.

Question 2: The lifetimes of 225 electrical components are given below. Determine their modal lifetime.

Solution:

The highest frequency is 61, so the modal class is 60-80.

Here, l = 60, h = 20, f₁ = 61, f₀ = 52 and f₂ = 38.

Mode = 60 + [(61 - 52)/(2 × 61 - 52 - 38)] × 20

= 60 + (9/32) × 20

= 60 + 5.625

= 65.625 hours

Answer: The modal lifetime of the components is 65.625 hours.

Question 3: The monthly expenditure of 200 families is given below. Find the modal and mean monthly expenditure.

Mode:

The highest frequency is 40, so the modal class is 1500-2000.

Here, l = 1500, f₁ = 40, f₀ = 24, f₂ = 33 and h = 500.

Mode = 1500 + [(40 - 24)/(80 - 24 - 33)] × 500

= 1500 + (16/23) × 500

= 1500 + 347.826

= Rs. 1847.83, approximately.

Mean:

Take a = 2750 and h = 500.

Mean = 2750 + (-35/200) × 500

= 2750 - 87.5

= Rs. 2662.50

Answer: The modal monthly expenditure is approximately Rs. 1847.83, and the mean monthly expenditure is Rs. 2662.50.

Question 4: The state-wise teacher-student ratios in higher secondary schools are shown below. Find the mean and mode, and interpret the results.

Mode:

The modal class is 30-35 because it has the highest frequency, 10.

Here, l = 30, f₁ = 10, f₀ = 9, f₂ = 3 and h = 5.

Mode = 30 + [(10 - 9)/(20 - 9 - 3)] × 5

= 30 + (1/8) × 5

= 30.625

Mode ≈ 30.63 students per teacher.

Mean:

Take a = 32.5 and h = 5.

Mean = 32.5 + (-23/35) × 5

= 29.2143

Mean ≈ 29.21 students per teacher.

Answer: The mean ratio is approximately 29.21 students per teacher, while the modal ratio is approximately 30.63 students per teacher. Thus, the overall average is about 29 students per teacher, while the most common ratio is about 31 students per teacher.

Question 5: The following distribution shows the runs scored by some leading one-day international batsmen. Find the mode.

Solution:

The modal class is 4000-5000 because its frequency, 18, is the highest.

Here, l = 4000, f₁ = 18, f₀ = 4, f₂ = 9 and h = 1000.

Mode = 4000 + [(18 - 4)/(36 - 4 - 9)] × 1000

= 4000 + (14/23) × 1000

= 4000 + 608.6957

= 4608.7 runs, approximately.

Answer: The mode is approximately 4608.7 runs.

Question 6: A student counted the number of cars passing a point on a road during 100 periods of three minutes each. Find the mode.

Solution:

The highest frequency is 20, so the modal class is 40-50.

Here, l = 40, f₁ = 20, f₀ = 12, f₂ = 11 and h = 10.

Mode = 40 + [(20 - 12)/(40 - 12 - 11)] × 10

= 40 + (8/17) × 10

= 44.7059

Answer: The mode is approximately 44.7 cars.

Exercise 13.3

Question 1: The monthly electricity consumption of 68 consumers is given below. Find the mean, median and mode, and compare the three values.

Mean:

Take a = 135 and h = 20.

Mean = 135 + (7/68) × 20

= 137.0588

Mean ≈ 137.06 units.

Mode:

The modal class is 125-145.

Here, l = 125, f₁ = 20, f₀ = 13, f₂ = 14 and h = 20.

Mode = 125 + [(20 - 13)/(40 - 13 - 14)] × 20

= 125 + (7/13) × 20

= 135.77 units, approximately.

Median:

Median = l + [((n/2) - cf)/f] × h

n = 68, so n/2 = 34.

The first cumulative frequency greater than 34 is 42. Therefore, the median class is 125-145.

Here, l = 125, cf = 22, f = 20 and h = 20.

Median = 125 + [(34 - 22)/20] × 20

= 125 + 12

= 137 units

Answer: Mean = 137.06 units, median = 137 units and mode = 135.77 units. The three values are close to one another.

Question 2: The median of the following distribution is 28.5. Find x and y.

Solution:

From the total frequency:

5 + x + 20 + 15 + y + 5 = 60

x + y + 45 = 60

x + y = 15 ...(1)

The given median, 28.5, lies in the class 20-30.

Therefore, l = 20, n = 60, cf = 5 + x, f = 20 and h = 10.

Median = l + [((n/2) - cf)/f] × h

28.5 = 20 + [(30 - 5 - x)/20] × 10

8.5 = (25 - x)/2

17 = 25 - x

x = 8

Using equation (1):

8 + y = 15

y = 7

Answer:x = 8 and y = 7.

Question 3: An insurance agent collected the following cumulative age data for 100 policyholders. Policies are issued only to people aged 18 years or more but below 60 years. Find the median age.

Solution:

Convert the less-than cumulative frequencies into ordinary frequencies by subtracting each cumulative frequency from the next one.

n = 100, so n/2 = 50.

The first cumulative frequency greater than 50 is 78. Hence, the median class is 35-40.

Here, l = 35, cf = 45, f = 33 and h = 5.

Median = 35 + [(50 - 45)/33] × 5

= 35 + 25/33

= 35.7576

Answer: The median age is approximately 35.76 years.

Question 4: The lengths of 40 leaves were measured to the nearest millimetre. Find the median length.

Solution:

Because the measurements are correct to the nearest millimetre, convert the classes into continuous intervals by subtracting 0.5 from every lower limit and adding 0.5 to every upper limit.

n = 40, so n/2 = 20.

The first cumulative frequency greater than 20 is 29. Thus, the median class is 144.5-153.5.

Here, l = 144.5, cf = 17, f = 12 and h = 9.

Median = 144.5 + [(20 - 17)/12] × 9

= 144.5 + 2.25

= 146.75 mm

Answer: The median length of the leaves is 146.75 mm.

Solution:

n = 400, so n/2 = 200.

The first cumulative frequency greater than 200 is 216. Therefore, the median class is 3000-3500.

Here, l = 3000, cf = 130, f = 86 and h = 500.

Median = 3000 + [(200 - 130)/86] × 500

= 3000 + (70/86) × 500

= 3406.9767 hours

Answer: The median lifetime is approximately 3406.98 hours.

Question 6: One hundred surnames were selected from a telephone directory. Their distribution according to the number of letters is given below. Find the median, mean and mode.

Median:

n = 100, so n/2 = 50.

The median class is 7-10.

Here, l = 7, cf = 36, f = 40 and h = 3.

Median = 7 + [(50 - 36)/40] × 3

= 7 + 1.05

= 8.05 letters

Mean:

Mean = 832/100

= 8.32 letters

Mode:

The modal class is 7-10.

Here, l = 7, f₁ = 40, f₀ = 30, f₂ = 16 and h = 3.

Mode = 7 + [(40 - 30)/(80 - 30 - 16)] × 3

= 7 + (10/34) × 3

= 7.8824 letters

Answer: Median = 8.05 letters, mean = 8.32 letters and mode ≈ 7.88 letters.

Question 7: The following table gives the weights of 30 students. Find the median weight.

Solution:

n = 30, so n/2 = 15.

The first cumulative frequency greater than 15 is 19. Therefore, the median class is 55-60.

Here, l = 55, cf = 13, f = 6 and h = 5.

Median = 55 + [(15 - 13)/6] × 5

= 55 + 10/6

= 56.6667 kg

Answer: The median weight is approximately 56.67 kg.

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