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
View or download the dedicated Chapter 13 Statistics revision resource below. Engaging with these structured concept notes daily ensures continuous academic progress and mastery of the 2026-27 curriculum for Mathematics.
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
\[ \bar{x} = \frac{f_1x_1 + f_2x_2 + \cdots + f_nx_n}{f_1 + f_2 + \cdots + f_n} = \frac{\sum f_ix_i}{\sum f_i} \]| Variable | Frequency | \( f_ix_i \) |
|---|---|---|
| \( x_1 \) | \( f_1 \) | \( f_1x_1 \) |
| \( x_2 \) | \( f_2 \) | \( f_2x_2 \) |
| \( \vdots \) | \( \vdots \) | \( \vdots \) |
| \( x_k \) | \( f_k \) | \( f_kx_k \) |
| Total | \( N = \sum f_i \) | \( \sum f_ix_i \) |
| Variable | 5 | 6 | 7 | 8 | 9 |
| Frequency | 4 | 10 | 15 | 11 | 3 |
Answer:
| \( x_i \) | \( f_i \) | \( f_ix_i \) |
|---|---|---|
| 5 | 4 | 20 |
| 6 | 10 | 60 |
| 7 | 15 | 105 |
| 8 | 11 | 88 |
| 9 | 3 | 27 |
| 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 \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} \).
| Class Interval | 10 - 30 | 30 - 50 | 50 - 70 | 70 - 90 | 90 - 110 |
| Frequency | 90 | 20 | 30 | 20 | 40 |
Answer:
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \).
| Class | \( f_i \) | \( x_i \) | \( f_ix_i \) |
|---|---|---|---|
| 10 - 30 | 90 | 20 | 1800 |
| 30 - 50 | 20 | 40 | 800 |
| 50 - 70 | 30 | 60 | 1800 |
| 70 - 90 | 20 | 80 | 1600 |
| 90 - 110 | 40 | 100 | 4000 |
| Total | \( \sum f_i = 200 \) | \( \sum f_ix_i = 10000 \) |
Mean \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{10000}{200} = 50 \).
| Class Interval | 1 - 10 | 11 - 20 | 21 - 30 | 31 - 40 | 41 - 50 |
| Frequency | 3 | 4 | 2 | 5 | 6 |
Answer:
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \).
| Class | \( f_i \) | \( x_i \) | \( f_ix_i \) |
|---|---|---|---|
| 1 - 10 | 3 | 5.5 | 16.5 |
| 11 - 20 | 4 | 15.5 | 62 |
| 21 - 30 | 2 | 25.5 | 51 |
| 31 - 40 | 5 | 35.5 | 177.5 |
| 41 - 50 | 6 | 45.5 | 273 |
| 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 \( \bar{x} = a + \frac{\sum f_id_i}{\sum f_i} \).
| Class | 25-29 | 30-34 | 35-39 | 40-44 | 45-49 | 50-54 | 55-59 |
| Frequency | 14 | 22 | 16 | 6 | 5 | 3 | 4 |
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 - 29.5 | 14 | 27 | -15 | -210 |
| 29.5 - 34.5 | 22 | 32 | -10 | -220 |
| 34.5 - 39.5 | 16 | 37 | -5 | -80 |
| 39.5 - 44.5 | 6 | 42 | 0 | 0 |
| 44.5 - 49.5 | 5 | 47 | 5 | 25 |
| 49.5 - 54.5 | 3 | 52 | 10 | 30 |
| 54.5 - 59.5 | 4 | 57 | 15 | 60 |
| Total | \( \sum f_i = 70 \) | \( \sum f_id_i = -395 \) |
Mean \( \bar{x} = a + \frac{\sum f_id_i}{\sum f_i} = 42 + \frac{-395}{70} = 36.36 \).
| x | 4 | 6 | 7 | p + 4 | 12 | 14 |
| f | 5 | 6 | 4 | 10 | 8 | 7 |
Answer:
Use the direct method.
| \( x_i \) | \( f_i \) | \( f_ix_i \) |
|---|---|---|
| 4 | 5 | 20 |
| 6 | 6 | 36 |
| 7 | 4 | 28 |
| p + 4 | 10 | \( 10p + 40 \) |
| 12 | 8 | 96 |
| 14 | 7 | 98 |
| 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 \( \bar{x} = a + \frac{\sum f_iu_i}{\sum f_i} \times h \).
| Class | 50 - 70 | 70 - 90 | 90 - 110 | 110 - 130 | 130 - 150 | 150 - 170 |
| Frequency | 18 | 12 | 13 | 27 | 8 | 22 |
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 - 70 | 18 | 60 | -3 | -54 |
| 70 - 90 | 12 | 80 | -2 | -24 |
| 90 - 110 | 13 | 100 | -1 | -13 |
| 110 - 130 | 27 | 120 | 0 | 0 |
| 130 - 150 | 8 | 140 | 1 | 8 |
| 150 - 170 | 22 | 160 | 2 | 44 |
| Total | \( \sum f_i = 100 \) | \( \sum f_iu_i = -39 \) |
Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 120 + 20 \times \frac{-39}{100} = 112.2 \).
| Marks | Below 10 | Below 20 | Below 30 | Below 40 | Below 50 | Below 60 | Below 70 | Below 80 | Below 90 | Below 100 |
| Number of students | 5 | 9 | 17 | 29 | 45 | 60 | 70 | 78 | 83 | 85 |
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 - 10 | 5 | 5 | -5 | -25 |
| 10 - 20 | 4 | 15 | -4 | -16 |
| 20 - 30 | 8 | 25 | -3 | -24 |
| 30 - 40 | 12 | 35 | -2 | -24 |
| 40 - 50 | 16 | 45 | -1 | -16 |
| 50 - 60 | 15 | 55 | 0 | 0 |
| 60 - 70 | 10 | 65 | 1 | 10 |
| 70 - 80 | 8 | 75 | 2 | 16 |
| 80 - 90 | 5 | 85 | 3 | 15 |
| 90 - 100 | 2 | 95 | 4 | 8 |
| 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 \( \bar{x} - a \).
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Chapter 13 Statistics Concepts and Summary for Class 10 Mathematics
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