CBSE Class 10 Mathematics Statistics Notes Set 03

Revision Notes for Class 10 Mathematics: Chapter 13 Statistics

Review targeted revision notes for Class 10 Mathematics with the CBSE Class 10 Mathematics Statistics Notes Set 03. Built according to official educational guidelines for the 2026-27 academic year, these downloadable summaries for Chapter 13 Statistics support daily study and last-minute exam readiness.

Review Chapter 13 Statistics for Class 10 Mathematics

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Introduction

Statistics is the study of the collection, organisation, analysis, interpretation and presentation of data. In earlier classes we learnt how to organise data, represent it graphically and find the mean, median and mode of ungrouped data. In this chapter we learn to calculate the mean, median and mode of grouped data.

Mean

The arithmetic mean is the sum of the values of all the observations divided by the total number of observations. Let \( x_1, x_2, x_3, \ldots, x_n \) be n observations. Then the mean is

\[ \bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \frac{1}{n}\sum_{i=1}^{n} x_i \]

Here the symbol \( \sum_{i=1}^{n} x_i \) denotes the sum \( x_1 + x_2 + x_3 + \cdots + x_n \).

Mean of Grouped Data

Let \( x_1, x_2, \ldots, x_n \) be observations with respective frequencies \( f_1, f_2, \ldots, f_n \). Then the mean \( \bar{x} \) is defined by

\[ \bar{x} = \frac{f_1x_1 + f_2x_2 + \cdots + f_nx_n}{f_1 + f_2 + \cdots + f_n} = \frac{\sum f_ix_i}{\sum f_i} \]
VariableFrequency\( f_ix_i \)
\( x_1 \)\( f_1 \)\( f_1x_1 \)
\( x_2 \)\( f_2 \)\( f_2x_2 \)
\( \vdots \)\( \vdots \)\( \vdots \)
\( x_k \)\( f_k \)\( f_kx_k \)
Total\( N = \sum f_i \)\( \sum f_ix_i \)
Example 1. Calculate the mean for the following distribution:
Variable56789
Frequency41015113

Answer:
\( x_i \)\( f_i \)\( f_ix_i \)
5420
61060
715105
81188
9327
Total\( N = \sum f_i = 43 \)\( \sum f_ix_i = 300 \)

Mean \( = \frac{\sum f_ix_i}{\sum f_i} = \frac{300}{43} = 6.976 \).

In real-life situations, data is usually so large that to make a meaningful study we need to convert the given ungrouped data into grouped data and devise some method to find its mean. The arithmetic mean of grouped data can be calculated by any one of the following methods: (i) direct method, (ii) assumed mean method, (iii) step-deviation method.

Direct Method

Steps for finding the arithmetic mean of grouped data by the direct method.
Step 1: Prepare a table with columns for the class interval, the mid-point or class mark \( x_i \) and the corresponding frequency \( f_i \). The class mark of each class is \( x_i = \frac{\text{lower class limit} + \text{upper class limit}}{2} \).
Step 2: Calculate \( f_ix_i \) for each row.
Step 3: Use the formula \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} \).

Example 2. Find the mean of the following frequency distribution by the direct method.
Class Interval10 - 3030 - 5050 - 7070 - 9090 - 110
Frequency9020302040

Answer:
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \).
Class\( f_i \)\( x_i \)\( f_ix_i \)
10 - 3090201800
30 - 502040800
50 - 7030601800
70 - 9020801600
90 - 110401004000
Total\( \sum f_i = 200 \)\( \sum f_ix_i = 10000 \)

Mean \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{10000}{200} = 50 \).
Example 3. Find the mean of the following distribution by the direct method.
Class Interval1 - 1011 - 2021 - 3031 - 4041 - 50
Frequency34256

Answer:
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \).
Class\( f_i \)\( x_i \)\( f_ix_i \)
1 - 1035.516.5
11 - 20415.562
21 - 30225.551
31 - 40535.5177.5
41 - 50645.5273
Total\( \sum f_i = 20 \)\( \sum f_ix_i = 580 \)

Mean \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{580}{20} = 29 \).

Assumed Mean Method

In some cases, especially when the class marks are large, finding the mean by the direct method can be quite lengthy. In such cases we can use the assumed mean method. We choose an arbitrary constant a (the assumed mean) and subtract it from each class mark \( x_i \). The reduced value \( d_i = x_i - a \) is called the deviation.
Note. The assumed mean should be one of the central values of \( x_i \) when they are arranged in ascending or descending order.

Steps for finding the arithmetic mean by the assumed mean method.
Step 1: Prepare a table with columns for the class interval, class mark \( x_i \) and frequency \( f_i \).
Step 2: Choose a number a (preferably among the central values of the class marks) as the assumed mean.
Step 3: Obtain the deviations \( d_i = x_i - a \) and write them against the corresponding frequencies.
Step 4: Multiply each frequency by its deviation to get \( f_id_i \).
Step 5: Find \( \sum f_id_i \) and \( \sum f_i \).
Step 6: Use the formula \( \bar{x} = a + \frac{\sum f_id_i}{\sum f_i} \).

Example 4. Find the arithmetic mean of the following frequency distribution by the assumed mean method.
Class25-2930-3435-3940-4445-4950-5455-59
Frequency1422166534

Answer:
Since the upper limit of a class is not equal to the lower limit of the next class, the classes are not continuous. Add 0.5 to each upper limit and subtract 0.5 from each lower limit (the gap between two classes is 1).
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take the assumed mean \( a = 42 \) and \( d_i = x_i - a \).
Class\( f_i \)\( x_i \)\( d_i = x_i - a \)\( f_id_i \)
24.5 - 29.51427-15-210
29.5 - 34.52232-10-220
34.5 - 39.51637-5-80
39.5 - 44.564200
44.5 - 49.5547525
49.5 - 54.53521030
54.5 - 59.54571560
Total\( \sum f_i = 70 \)\( \sum f_id_i = -395 \)

Mean \( \bar{x} = a + \frac{\sum f_id_i}{\sum f_i} = 42 + \frac{-395}{70} = 36.36 \).
Example 5. If the mean of the distribution given below is 9.2, find the value of p.
x467p + 41214
f5641087

Answer:
Use the direct method.
\( x_i \)\( f_i \)\( f_ix_i \)
4520
6636
7428
p + 410\( 10p + 40 \)
12896
14798
Total\( \sum f_i = 40 \)\( \sum f_ix_i = 318 + 10p \)

Mean \( = \frac{318 + 10p}{40} = 9.2 \Rightarrow 318 + 10p = 368 \Rightarrow 10p = 50 \Rightarrow p = 5 \).

Step Deviation Method

When the values of both \( x_i \) and \( f_i \) are large, the calculation by the above methods becomes quite lengthy. In that case we use the step-deviation method, also known as the shortcut method. Here the deviations \( x_i - a \) are divided by a constant h, the class size (the length of a class interval):

\[ u_i = \frac{x_i - a}{h}, \qquad h = b_{n+1} - b_n \]

Steps for finding the arithmetic mean by the step-deviation method.
Step 1: Prepare a table with columns for the class interval, class mark \( x_i \) and frequency \( f_i \).
Step 2: Choose a number a from the central values of \( x_i \) as the assumed mean.
Step 3: Find the class size \( h = b_{n+1} - b_n \).
Step 4: Obtain \( u_i = \frac{x_i - a}{h} \).
Step 5: Multiply each frequency \( f_i \) by \( u_i \) to get \( f_iu_i \).
Step 6: Find \( \sum f_iu_i \).
Step 7: Use the formula \( \bar{x} = a + \frac{\sum f_iu_i}{\sum f_i} \times h \).

Example 6. Find the mean of the following distribution by the step-deviation method.
Class50 - 7070 - 9090 - 110110 - 130130 - 150150 - 170
Frequency18121327822

Answer:
The values of the class marks are large, so take \( a = 120 \) and \( h = 20 \).
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take \( a = 120 \), class size \( h = 20 \) and \( u_i = \frac{x_i - a}{h} \).
Class\( f_i \)\( x_i \)\( u_i = \frac{x_i - a}{h} \)\( f_iu_i \)
50 - 701860-3-54
70 - 901280-2-24
90 - 11013100-1-13
110 - 1302712000
130 - 150814018
150 - 17022160244
Total\( \sum f_i = 100 \)\( \sum f_iu_i = -39 \)

Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 120 + 20 \times \frac{-39}{100} = 112.2 \).
Example 7. Find the mean marks from the following data by the step-deviation method.
MarksBelow 10Below 20Below 30Below 40Below 50Below 60Below 70Below 80Below 90Below 100
Number of students591729456070788385

Answer:
To convert to a frequency distribution, write below 10 as 0 - 10, below 20 as 10 - 20 and so on. The frequencies are found by subtracting the previous cumulative number from the current one.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take \( a = 55 \), class size \( h = 10 \) and \( u_i = \frac{x_i - a}{h} \).
Class\( f_i \)\( x_i \)\( u_i = \frac{x_i - a}{h} \)\( f_iu_i \)
0 - 1055-5-25
10 - 20415-4-16
20 - 30825-3-24
30 - 401235-2-24
40 - 501645-1-16
50 - 60155500
60 - 701065110
70 - 80875216
80 - 90585315
90 - 10029548
Total\( \sum f_i = 85 \)\( \sum f_iu_i = -56 \)

Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 55 + 10 \times \frac{-56}{85} = 48.41 \).

Knowledge Hub

1. If \( \bar{x} \) is the mean of n observations \( x_1, x_2, \ldots, x_n \), then the mean of \( x_1 + a, x_2 + a, \ldots, x_n + a \) is \( \bar{x} + a \). If each observation is increased by a, the mean is also increased by a.
2. The mean of \( ax_1, ax_2, \ldots, ax_n \) is \( a\bar{x} \) (a is a non-zero number): if each observation is multiplied by a non-zero number a, the mean is also multiplied by a.
3. The mean of \( \frac{x_1}{a}, \frac{x_2}{a}, \ldots, \frac{x_n}{a} \) is \( \frac{\bar{x}}{a} \), where a is any non-zero number.
4. The mean of \( x_1 - a, x_2 - a, \ldots, x_n - a \) is \( \bar{x} - a \).

Chapter 13 Statistics Concepts and Summary for Class 10 Mathematics

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