STATISTICS-Exercise 13.2

Statistics is one of the most practical and scoring chapters in the Class X Mathematics syllabus, as it bridges numerical computation with real-world interpretation of data. The textbook exercise solutions for Chapter 13, Statistics, are designed to help learners systematically master the calculation and analysis of data presented in grouped frequency distributions. These solutions focus not only on obtaining correct numerical answers but also on developing a clear understanding of statistical reasoning and methodology. The exercises guide students through essential concepts such as mean, median, and mode of grouped data, along with their respective formulae and stepwise application. Special emphasis is placed on structured presentation, correct identification of class intervals, cumulative frequencies, and the selection of appropriate methods such as the direct method, assumed mean method, and step-deviation method. Each solution is presented in a logical, exam-ready format to help students avoid common errors and gain confidence in problem-solving. These textbook exercise solutions serve as a reliable reference for board examination preparation, classroom revision, and self-study. By following these worked examples, learners can strengthen conceptual clarity, improve calculation accuracy, and develop the ability to interpret statistical results meaningfully—skills that are vital not only for examinations but also for practical data analysis in everyday contexts.

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December 19, 2025  |  By Academia Aeternum

STATISTICS-Exercise 13.2

Maths - Exercise

Q1. 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

Lets us put the data in table
Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_id_i\)
5-15 6 10 -20 -120
15-25 11 20 -10 -110
25-35 21 \((f_0)\) 30 0 0
35-45 23 \((f_1)\) 40 10 230
45-55 14 \((f_2)\) 50 20 280
55-65 5 60 30 150
Total \(\sum{f_i=80}\) \(\sum{f_id_i=430}\)

Modal Class 35-45 (class with highest frequency)
class interval (h) = Upper class limit-lower class limit = 15-5- 10

$$\begin{aligned}l=35\\ f_{1}=23\\ f_{0}=21\\ f_{2}=14\\ h=10\\ \text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) h\\\\ &=35+\left( \dfrac{23-21}{2\times 23-21-14}\right) 10\\\\ &=35+\left( \dfrac{2}{46-35}\right) \times 10\\\\ &=35+\dfrac{2}{11}\times 10\\\\ &=35+1.8\\\\ &=36.8\end{aligned}$$

Assumed mean = 30
Substitute the values

$$\begin{aligned}\overline{x}&=a+\left( \dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\right) \\\\ &=30+\left( \dfrac{430}{80}\right) \\\\ &=30+\dfrac{43}{8}\\\\ &=35.37\end{aligned}$$ Mode = 36.8 Mean = 35.37
Maximum Number of patients admitted in the hospital are of age 36.8, while on an average the age of patient admitted to the hospital is 35.37 years

Q2. 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
\(\tiny\text{(Modal Class)}\)		 80-100 100-120
Frequency 10 35 52 \((f_0)\) 61 \((f_1)\) 38 \((f_2)\) 29
Determine the modal lifetimes of the components.

Solution

Modal Class = 60-80 (class having Maxium frequency)
class interval (h) = Upper class limit-lower class limit = 20-0 = 20

\(\begin{aligned}l=60\\ h=20\\ f_{0}=52\\ f_{1}=61\\ f_{2}=38\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) h\\\\ &=60+\left( \dfrac{61-52}{2\times 61-52-38}\right) \times 20\\\\ &=60+\left( \dfrac{9}{122-90}\right) \times 20\\\\ &=60+\left( \dfrac{9}{32}\right) \times 20\\\\ &=65.625\end{aligned}$$
Modal lifetime of components = 65.625 hours

Q3. 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

Solution

Lets us put the data in table
Class Interval \(f_i\) \(x_i\) \(d_i=x_i-a\) \(u_i=\frac{d_i}{h}\) \(f_iu_i\)
1000-1500 24 \((f_0)\) 1250 -1500 -3 -72
1500-2000 40 \((f_1)\) 1750 -1000 -2 -80
2000-2500 33 2250 -500 -1 -33
2500-3000 28 2750 0 0 0
3000-3500 30 3250 500 1 30
3500-4000 22 3750 1000 2 44
4000-4500 16 3750 1500 3 48
4500-5000 7 4250 2000 4 28
Total \(\sum{f_i}=(200)\) \(\sum{f_iu_i}=(-35)\)

Modal Class = 1500-2000 (Class having Maximum frequency)
Class interval (h) = Upper class limit- lower class limit = 1500-1000 = 500

\(\begin{aligned}l&=1500\\ h&=500\\ f_{0}&=24\\ f_{1}&=40\\ f_{2}&=33\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) h\\\\ &=1500+\left( \dfrac{40-24}{2\times 40-24-33}\right) \times 500\\\\ &=1500+\left( \dfrac{16}{80-57}\right) 500\\\\ &=1500+\dfrac{16\times 500}{23}\\\\ &=1500+347.83\\\\ &=1847.83\end{aligned}$$

Assumed mean = 2750
Class interval (h) = 500
Substituting values in Mean formula

$$\begin{aligned}\overline{x}&=a+\left( \dfrac{\sum f_{i}u_{i}}{\sum f_{i}}\right) \times h\\\\ &=2750+\left( \dfrac{-35}{200}\right) 500\\\\ &=2750-\dfrac{35\times 5}{2}\\\\ &=2750-87.5\\\\ &=2662.5\end{aligned}$$
Modal Month y Expense = 1847.83
Mean Monthly Expense = 2662.50

Q4. 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 Number of states / U.T.
15-20 3
20-25 8
25-30 9
30-35 10
35-40 3
40-45 0
45-50 0
50-55 2

Solution

Lets put data in table
Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_id_i\)
10-20 3 17.5 -15 -45
20-25 8 22.5 -10 -80
25-30 9 \((f_0)\) 27.5 -5 -45
30-35 10 \((f_1)\) 32.5 0 0
35-40 3 \((f_2)\) 37.5 5 15
40-45 0 42.5 10 0
45-50 0 47.5 15 0
50-55 2 52.5 20 40
Total \(\sum{f_i}=35\) \(\sum{f_id_i}=-115\)

Modal Class - 30-35 (Class having maximum frequency)
class interval h = Upper Class limit-lower class limit = 20-15 = 5

\(\begin{aligned}l&=30\\ h&=5\\ f_{0}&=9\\ f_{1}&=10\\ f_{2}&=3\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) \times h\\\\ &=30+\left( \dfrac{10-9}{2\times 10-9-3}\right) 5\\\\ &=30+\left( \dfrac{1}{20-12}\right) 5\\\\ &=30+\dfrac{5}{8}\\\\ &=30.6\end{aligned}$$

Assumed mean = 32.5
Substitute values in formula

$$\begin{aligned}\overline{x}&=a+\left( \dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\right) \\\\ &=32.5+\left( \dfrac{-115}{35}\right) \\\\ &=32.5-\dfrac{115}{35}\\\\ &=32.5-\dfrac{23}{7}\\\\ &=29.2\end{aligned}$$
Mode=30.6
Mean = 29.2

Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs score Number of batsmen
3000-4000 4 \((f_0)\)
4000-5000 18 \((f_1)\)
5000-6000 9 \((f_2)\)
6000-7000 7
7000-8000 6
8000-9000 3
9000-10000 2
10000-11000 1
Find the mode of the data.

Solution

Modal class = 4000-5000 (having maximum frequency)
h = 5000-4000 = 1000

\(\begin{aligned}l&=4000\\ h&=1000\\ f_{0}&=4\\ f_{1}&=18\\ f_{2}&=9\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) \times h\\\\ &=4000t\left( \dfrac{18-4}{2\times 18-4-9}\right) \times 1000\\\\ &=4000+\dfrac{14\times 1000}{36-13}\\\\ &=4000+\dfrac{14\times 1000}{23}\\\\ &=4000+608.7\\\\ &=4608.7\end{aligned}$$
Mode = 4608.7

Q6. 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 \((f_0)\) 20 \((f_1)\) 11 \((f_2)\) 15 8

Solution

Modal class = 40-50 (class having maximum frequency)
h = upper Class limit-lower Class limit = 50-40 = 10

\(\begin{aligned}l&=40\\ h&=10\\ f_{0}&=12\\ f_{1}&=20\\ f_{2}&=11\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) \times 10\\\\ &=40+\left( \dfrac{20-12}{2\times 20-12-11}\right) \times 10\\\\ &=40+\left( \dfrac{8}{40-23}\right) \times 10\\\\ &=40+\dfrac{8\times 10}{17}\\\\ &=40+4.7\\\\ &=44.7\end{aligned}$$
Mode = 44.7

Frequently Asked Questions

Statistics is the branch of mathematics that deals with the collection, organisation, presentation, analysis, and interpretation of numerical data.

Statistics helps in understanding trends, making comparisons, predicting outcomes, and taking data-based decisions in real-life situations.

Data refers to numerical information collected from observations, surveys, or experiments for analysis.

Raw data is unorganised data collected directly from a source without any classification or arrangement.

Grouped data is data organised into class intervals to simplify analysis when observations are large in number.

It is a table that shows how often values occur within defined class intervals.

Class intervals are divisions of data into fixed ranges used to group observations.

Class width is the difference between the upper and lower limits of a class interval.

The class mark is the midpoint of a class interval, calculated as \((\text{upper limit} + \text{lower limit})/2\).

Measures of central tendency describe a central or typical value of data, such as mean, median, and mode.

Mean is the average value of grouped data calculated using class marks and frequencies.

\(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\), where \(x_i\) are class marks and \(f_i\) are frequencies.

It is a method to calculate mean by assuming a convenient value as the mean to simplify calculations.

A short-cut method of finding mean using deviations divided by class width to reduce computation.

It is preferred when class intervals are equal and numbers are large.

Median is the value that divides the data into two equal parts when arranged in order.

The class interval that contains the median value.

\(\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right)h\)

\(l\): lower limit, \(N\): total frequency, \(cf\): cumulative frequency before median class, \(f\): frequency, \(h\): class width

Mode is the value that occurs most frequently in a data set.

The class interval with the highest frequency.

\(\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right)h\)

\(f_1\) is modal class frequency, \(f_0\) preceding frequency, \(f_2\) succeeding frequency.

It is the running total of frequencies in a distribution.

A graphical representation of cumulative frequencies, also called an ogive.

An ogive formed using cumulative frequencies less than the upper class limits.

An ogive drawn using cumulative frequencies greater than the lower class limits.

By plotting both ogives and locating the x-coordinate of their intersection.

They summarise large data sets using a single representative value.

Mean is best for symmetrical distributions.

Median is preferred for skewed distributions.

When identifying the most common value, such as shoe size or popular choice.

Economics, science, medicine, education, population studies, and business analysis.

To analyse results, performance trends, and assessment outcomes.

Numerical problems, formula-based questions, graphical interpretation, and case-study questions.

Errors in tables lead to incorrect calculations and wrong conclusions.

Wrong class marks, incorrect cumulative frequencies, and formula substitution errors.

All major steps with formulas must be clearly shown for full marks.

Yes, for a perfectly symmetrical distribution.

Changing the origin and scale using assumed mean and step deviation.

Drawing conclusions and inferences from analysed data.

Analytical thinking, logical reasoning, and numerical accuracy.

Yes, due to formula-based questions and structured solutions.

Memorise formulas, practice numericals, and avoid calculation mistakes.

It carries significant weightage in the Class X Mathematics examination.

Proper scale, labeling, and accuracy are essential for full marks.

Frequency per unit class width, used in unequal class intervals.

Yes, they help compare distributions visually.

Displaying data using tables, graphs, and curves.

It teaches how data can be analysed logically to draw meaningful conclusions.

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