Class 10 Maths Chapter 13 Statistics Exercise 13.2 NCERT Solutions

Exercise 13.2 of Class 10 Maths Chapter 13, Statistics, focuses on a quicker and more convenient way to find the mean of grouped data using the Step Deviation Method.

This method is especially helpful when the class intervals are the same, as it reduces lengthy calculations and makes the process easier to handle. It’s one of the key methods you’ll study in this chapter as part of the  CBSE Class 10 Maths syllabus. 

With the help of NCERT Solutions for Class 10, each step is explained clearly. This will make it easier to understand the method and apply it correctly in exams. 

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2

1. The following table shows the ages of the patients admitted to a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution: To find out the modal class, let us consider the class interval with high frequency. Here, the greatest frequency = 23
so the modal class = 35 – 45
Lower limit of modal class = l = 35,
class width (h) = 10,
f m = 23
f 1 = 21 and f 2 = 14 

The formula to find the mode is Mode = l + [(f m – f 1 )/ (2f m – f 1 – f 2 )] × h
Substitute the values in the formula, we get Mode = 35+[(23-21)/(46-21-14)]×10
= 35 + (20/11)
= 35 + 1.8
= 36.8 years
So the mode of the given data = 36.8 years
Calculation of Mean: First find the midpoint using the formula, x i = (upper limit +lower limit)/2

2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.

Solution: From the given data the modal class is 60–80. Lower limit of modal class = l = 60, 

The frequencies are: f m = 61, f 1 = 52, f 2 = 38 and h = 20 

The formula to find the mode is Mode = l+ [(f m – f 1 )/(2f m – f 1 – f 2 )] × h
Substitute the values in the formula, we get Mode = 60 + [(61 – 52)/ (122 – 52 – 38)] × 20
Mode = 60 + [(9 × 20)/32]
Mode = 60 + (45/8) = 60 + 5.625
Therefore, modal lifetime of the components = 65.625 hours.

3. 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:

Solution: Given data: Modal class = 1500-2000, l = 1500, Frequencies: f m = 40 f 1 = 24, f 2 = 33 and h = 500
Mode formula: Mode = l + [(f m – f 1 )/ (2f m – f 1 – f 2 )] × h
Substitute the values in the formula, we get Mode = 1500 + [(40 – 24)/ (80 – 24 – 33)] × 500
Mode = 1500 + [(16 × 500)/23]
Mode = 1500 + (8000/23)
= 1500 + 347.83
Therefore, modal monthly expenditure of the families = Rupees 1847.83 

Calculation for mean: First find the midpoint using the formula, x i =(upper limit +lower limit)/2 Let us assume a mean, (a) be 2750.

The formula to calculate the mean,
Mean = x̄ = a +(∑f i u i /∑f i ) × h
Substitute the values in the given formula = 2750 + (-35/200) × 500
= 2750 – 87.50
= 2662.50
So, the mean monthly expenditure of the families = Rs. 2662.50

4. 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

Solution: Given data: Modal class = 30 – 35
l = 30
Class width (h) = 5
f m = 10
f 1 = 9 and
f 2 = 3
Mode Formula: Mode = l + [(f m – f 1 )/ (2f m – f 1 – f 2 )] × h
Substitute the values in the given formula Mode = 30 + [(10 – 9)/ (20 – 9 – 3)] × 5
= 30 + (5/8)
= 30 + 0.625
= 30.625
Therefore, the mode of the given data = 30.625
Calculation of mean: Find the midpoint using the formula, x i =(upper limit +lower limit)/2

Mean = x̄ = ∑f i x i /∑f i = 1022.5/35 = 29.2 (approx) Therefore, mean = 29.2

5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.

Find the mode of the data.

Solution: Given data: Modal class = 4000 – 5000
l = 4000
Class width (h) = 1000, f m = 18, f 1 = 4 and f 2 = 9
Mode Formula: Mode = l + [(f m – f 1 )/ (2f m – f 1 – f 2 )] × h
Substitute the values Mode = 4000 + [(18 – 4)/ (36 – 4 – 9)] × 1000
= 4000 + (14000/23)
= 4000 + 608.695
= 4608.695 = 4608.7 (approximately)
Thus, the mode of the given data is 4608.7 runs.

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

Solution: Given Data: Modal class = 40 – 50,
l = 40,
Class width (h) = 10,
f m = 20,
f 1 = 12
f 2 = 11
Mode = l + [(f m – f 1 )/(2f m – f 1 – f 2 )] × h
Substitute the values Mode = 40 + [(20 – 12)/ (40 – 12 – 11)] × 10
= 40 + (80/17)
= 40 + 4.7
= 44.7
Thus, the mode of the given data is 44.7 cars.

How to Score Better in Class 10 Maths Exam?

Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:

• Build Strong Concepts:

Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.

• Work on Weak Areas:

Focus on difficult topics from the Class 10 Maths syllabus instead of skipping them to avoid losing marks.

• Revise Formulas Daily:

Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.

• Practise Regularly:

Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.

• Solve Previous Year Papers:

Practising  class 10 Maths previous year questions (PYQs) helps you understand question patterns and important topics.