STATISTICS-Exercise 13.1

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 18, 2025  |  By Academia Aeternum

STATISTICS-Exercise 13.1

Maths - Exercise

Q1. 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

Lets us put the data in table
Class Interval \(f_i\) \(x_i\) \(f_ix_i\)
0-2 1 1 1
2-4 2 3 6
4-6 1 5 5
6-8 5 7 35
8-10 6 9 54
10-12 2 11 22
12-14 3 13 39
Total \(\sum{f_i=20}\) \(\sum{f_ix_i=162}\)
Substituting Values in Mean Formula $$\begin{aligned}\sum f_{i}&=20\\\\ \sum f_{i}x_{i}&=162\\\\ \overline{x}&=\dfrac{\sum f_{i}x_{i}}{\sum f_{i}}\\\\ \overline{x}&=\dfrac{162}{20}\\\\ &=\dfrac{162}{20}\\\\ &=8.1\end{aligned}$$
Mean=8.1, We will use the direct method as the data is of low values

Q2. 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

Lets us put the data in table
Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_id_i\)
500-520 12 510 -40 -480
520-540 14 530 -20 -280
540-560 8 550 0 0
560-580 6 570 20 120
580-600 10 590 40 400
Total \(\sum{f_i}=50\) \(\sum{f_id_i}=-240\)

Let Assumed Mean=550
Substitute value in Mean Formula

$$\begin{aligned}\overline{x}&=a+\dfrac{\sum{f_{i}d_{i}}}{\sum{f_{i}}}\\\\ &=550+\dfrac{\left( -240\right) }{50}\\\\ &=550-\dfrac{24}{5}\\\\ &=550-4.8\\\\ &=545.20\end{aligned}$$
Mean daily wages of the workers is ₹ 545.20

Q3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 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

Lets us put the data in table
Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_id_i\)
11-13 7 12 -6 -42
13-15 6 14 -4 -24
15-17 9 16 -2 -18
17-19 13 18 0 0
19-21 \(f\) 20 2 \(2f\)
21-23 5 22 4 20
23-25 4 24 6 24
Total \(\sum{f_i}=(44+f)\) \(\sum{f_id_i}=(2f-40)\)

Let Assumed Mean \(a\)= 18
Let substitute value in Mean Formula

$$\begin{aligned}\overline{x}=a+\dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\\\\ 18=18+\dfrac{2f-40}{44+f}\\\\ 18-18=\dfrac{2t-40}{44+f}\\\\ \Rightarrow 2f-40=0\\\\ 2f=40\\\\ f=\dfrac{40}{2}\\\\ =20\end{aligned}$$
Value of missing frequency is 20

Q4. 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

Lets put data in table
Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_id_i\)
65-68 2 66.5 -9 -18
68-71 4 69.5 -6 -24
71-74 3 72.5 -3 -9
74-77 8 75.5 0 0
77-80 7 78.5 3 21
80-83 4 81.5 6 24
83-86 2 84.5 9 18
Total \(\sum{f_i}=30\) \(\sum{f_id_i}=12\)

Let Assumed Mean \(a\)= 75.5
Subtitute the values in Mean Formula

$$\begin{aligned}\overline{x}&=a+\dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\\\\ \overline{x}&=75.5+\dfrac{12}{30}\\\\ &=75.5+0.4\\\\ &=75.9\end{aligned}$$
Mean heartbeats per minute for these women is 75.9

Q5. 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 55-58 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

Class is not continuous, we have to make class continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit. Let put data in table
Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_id_i\)
49.5-52.5 15 51 -6 -90
52.5-55.5 110 54 -3 -330
55.5-58.5 135 57 0 0
58.5-61.5 115 59 3 345
61.5-63.5 25 61 6 150
Total \(\sum{f_i}=400\) \(\sum{f_id_i}=75\)

Let Assumed Mean \(a\) = 57
Substitute the values in Mean Formula

$$\begin{aligned}\overline{x}&=a+\dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\\\\ &=57+\dfrac{75}{400}\\\\ &=57.187\\\\ &\approx 57.19\end{aligned}$$
Mean number of mangoes kept in a packing box is 57.19

Q6. 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

Lets Put data in a table
Class Interval \(f_i\) \(x_i\) \(d_i\) \(u_i=\frac{d_i}{50}\) \(f_iu_i\)
100-150 4 125 -100 -2 -8
150-200 5 175 -50 -1 -5
200-250 12 225 0 0 0
250-300 2 275 50 1 2
300-350 2 325 100 2 4
Total \(\sum{f_i}=25\) \(\sum{f_iu_i}=-7\)

Let Assumed Mean \(a\)=225
Class Interval \(h\)=50
Substitute values in Mean Formula

$$\begin{aligned}\overline{x}&=a+\left( \dfrac{\sum f_{i}u_{i}}{\sum f_{i}}\right) \times h\\\\ &=225+\left( \dfrac{-7}{25}\right) 50\\\\ &=225+\left( -14\right) \\\\ &=225-14\\\\ &=211\end{aligned}$$

Q7. To find out the concentration of SO2 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 SO2 Frequency
0.00-0.04 4
0.04-0.08 9
0.08-0.12 9
0.12-0.16 2
0.16-0.20 4
0.20-0.24 2

Find the mean concentration of \(SO_2\) in the air.

Solution

Lets put data in table
Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_id_i\)
0.00-0.04 4 0.02 -0.08 -0.32
0.04-0.08 9 0.06 -0.04 -0.36
0.08-0.12 9 0.10 0 0
0.12-0.16 2 0.14 0.04 0.08
0.16-0.20 4 0.18 0.08 0.32
0.20-0.24 2 0.22 0.12 0.24
Total \(\sum{f_i}=30\) \(\sum{f_id_i}=-0.04\)

Let Assumed Mean \(a\)=0.10
Substitute values in Mean Fromula

$$\begin{aligned}\overline{x}&=a+\dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\\\\ &=0.10+\dfrac{\left( -0.04\right) }{30}\\\\ &=0.10-\dfrac{0.04}{30}\\\\ &=0.0986\ldots \\\\ &=0.099\end{aligned}$$
Mean concentration of \(SO_2\) in the air is 0.099 ppm

Q8. 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 10 11 7 4 4 3 1

Solution

Let put this data in table
Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_id_i\)
0-6 11 3 -14 -154
6-10 10 8 -9 -90
10-14 7 12 -5 -35
14-20 4 17 0 0
20-28 4 24 7 28
28-38 3 33 16 48
38-40 1 39 22 22
Total \(\sum{f_i}\) \(\sum{f_id_i}=-181\)

Let Assumed Mean \(a\)=17
Substitute values in Mean Formula

$$\begin{aligned}\overline{x}&=a+\dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\\\\ &=17-\dfrac{181}{40}\\\\ &=12.475\\\\ &\approx 12.48\end{aligned}$$
Mean number of days a student was absent is 12.48

Q9.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

Lets put data in table
Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_id_i\)
45-55 3 50 -20 -60
55-65 10 60 -10 -100
65-75 11 70 0 0
75-85 8 80 10 80
85-95 3 90 20 60
Total \(\sum{f_i}=35\) \(\sum{f_id_i}=-20\)

Let Assumed mean \(a\) is =70
Substitute values in Mean Formula

$$\begin{aligned}\overline{x}&=a+\dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\\\\ &=70+\dfrac{\left( -20\right) }{35}\\\\ &=70-\dfrac{20}{35}\\\\ &=69.428\\\\ &\approx 69.43\% \end{aligned}$$
Mean literacy rate is 69.43%

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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