CBSE Class 10 Mathematics Statistics Notes
1. Introduction to Statistics
Statistics is the branch of mathematics that deals with the collection, organisation, presentation, analysis, and interpretation of numerical data. In our everyday life, data surrounds us — from weather forecasts and sports scores to election results and economic surveys. The ability to read, summarise, and draw meaningful conclusions from data is one of the most practically useful skills you will develop in Class 10 Mathematics.
In earlier classes, you learned how to represent data using bar graphs, histograms, and frequency polygons. In Class 10, we go deeper — focusing on numerical measures that summarise a dataset into a single representative value. These are called measures of central tendency. CBSE Chapter 14 covers three such measures for grouped data: the Mean, the Median, and the Mode. Each has multiple methods for calculation, and the board exam frequently tests your ability to choose and apply the right method.
What You Will Learn in This Chapter • Mean of Grouped Data — Direct Method, Assumed Mean Method, Step Deviation Method • Mode of Grouped Data — Modal class, formula, and interpretation • Median of Grouped Data — Cumulative frequency, formula, and interpretation • Empirical Relationship between Mean, Median, and Mode • Ogive (Cumulative Frequency Curve) — Less Than and More Than types • How to find Median graphically using Ogive |
Why is this chapter important? Statistics is a guaranteed source of marks in every CBSE board exam. A 3-mark or 5-mark question on Mean, Median, or Mode appears almost without exception. The Ogive (graphical median) has appeared in several recent board papers as a 4-mark question. Students who master all three methods for Mean and understand the Median and Mode formula deeply can consistently score full marks in this chapter.
2. Basic Terms and Concepts
Before jumping into calculations, it is important to understand the key vocabulary used throughout this chapter. These terms will appear repeatedly in every question.
2.1 Key Terminology
Term | Definition | Example |
Data | A collection of facts or figures collected for a specific purpose | Marks of 30 students in a test |
Raw Data | Data in its original, unorganised form | 50, 45, 62, 38, 55, 70, ... |
Class Interval | A range into which data is grouped (e.g., 10–20) | 10–20, 20–30, 30–40 |
Class Width (h) | Difference between upper and lower class limits | 20 − 10 = 10 |
Frequency (f) | Number of data values falling in a class interval | 15 students scored 40–50 |
Cumulative Frequency | Running total of frequencies up to a class | 10, 25, 40, 52, 60 |
Class Mark (xᵢ) | Midpoint of a class interval: (Upper + Lower) / 2 | (10+20)/2 = 15 |
Modal Class | Class interval with the highest frequency | The class containing Mode |
Median Class | Class where cumulative frequency ≥ n/2 first | The class containing Median |
Assumed Mean (a) | A convenient value chosen to simplify Mean calculation | Middle class mark |
2.2 Types of Data Representation
Grouped data is usually presented in one of two table formats. Understanding which format you have is the first step before applying any formula.
Type | Format | Used For |
Frequency Distribution Table | Class interval | Frequency | Mean, Mode, Median |
Cumulative Frequency Table (Less Than) | Upper limit | Cumulative Frequency | Median, Less Than Ogive |
Cumulative Frequency Table (More Than) | Lower limit | Cumulative Frequency | More Than Ogive |
3. Mean of Grouped Data
The Mean (also called the arithmetic mean or average) is the most commonly used measure of central tendency. For grouped data, we cannot find the exact mean because we do not know individual values — only the class intervals. So we use the class mark (midpoint) as a representative value for each class.
CBSE Class 10 requires you to know three methods for finding the mean of grouped data. Each method gives the same answer, but different methods are easier depending on the numbers involved. The Assumed Mean and Step Deviation methods are specifically designed to reduce arithmetic effort when class marks are large numbers.
3.1 Direct Method
When to use: When class marks (xᵢ) are small, manageable numbers. This is the most straightforward method.
Step 1: Find class mark (midpoint): xᵢ = (Upper limit + Lower limit) / 2 Step 2: Multiply each class mark by its frequency: fᵢxᵢ Step 3: Sum all fᵢxᵢ and sum all fᵢ
Mean (x̄) = Σ(fᵢxᵢ) / Σfᵢ
where fᵢ = frequency of iᵗʰ class, xᵢ = class mark of iᵗʰ class |
Worked Example 3.1 — Direct Method Problem: Find the mean of the following distribution: Class: 0–10 10–20 20–30 30–40 40–50 Freq: 5 10 15 8 2
Step 1: Class marks (xᵢ): 5, 15, 25, 35, 45 Step 2: fᵢxᵢ: 5×5=25, 10×15=150, 15×25=375, 8×35=280, 2×45=90 Step 3: Σfᵢxᵢ = 25+150+375+280+90 = 920 Σfᵢ = 5+10+15+8+2 = 40
Mean = 920 / 40 = 23 |
3.2 Assumed Mean Method
When to use: When class marks are large numbers (e.g., 500, 750, 1000). Subtracting an assumed mean 'a' makes the deviations (dᵢ) small and easy to work with.
Step 1: Choose assumed mean 'a' — preferably the middle class mark Step 2: Calculate deviation: dᵢ = xᵢ − a Step 3: Calculate: fᵢdᵢ for each class
Mean (x̄) = a + [ Σ(fᵢdᵢ) / Σfᵢ ]
where a = assumed mean, dᵢ = xᵢ − a |
Worked Example 3.2 — Assumed Mean Method Problem: Find the mean daily wages of workers: Wages (₹): 500–520 520–540 540–560 560–580 580–600 Workers: 12 14 8 6 10
xᵢ: 510 530 550 570 590 Choose a = 550 (middle class mark) dᵢ: −40 −20 0 +20 +40 fᵢdᵢ: −480 −280 0 +120 +400
Σfᵢdᵢ = −480 − 280 + 0 + 120 + 400 = −240 Σfᵢ = 12 + 14 + 8 + 6 + 10 = 50
Mean = 550 + (−240/50) = 550 − 4.8 = ₹545.20 |
3.3 Step Deviation Method
When to use: When all class widths are equal. You divide dᵢ by the class width h to get even smaller numbers called uᵢ (step deviations). This is the most calculation-efficient method.
Step 1: Choose assumed mean 'a' and note class width 'h' Step 2: Calculate step deviation: uᵢ = (xᵢ − a) / h Step 3: Calculate: fᵢuᵢ for each class
Mean (x̄) = a + h × [ Σ(fᵢuᵢ) / Σfᵢ ]
where h = class width (must be equal for all classes) |
Worked Example 3.3 — Step Deviation Method Problem: Find the mean of marks scored by students: Marks: 10–25 25–40 40–55 55–70 70–85 85–100 Students: 2 3 7 6 6 6
xᵢ: 17.5 32.5 47.5 62.5 77.5 92.5 Choose a = 47.5, h = 15 uᵢ = (xᵢ−47.5)/15: −2 −1 0 +1 +2 +3 fᵢuᵢ: −4 −3 0 +6 +12 +18
Σfᵢuᵢ = −4 − 3 + 0 + 6 + 12 + 18 = 29 Σfᵢ = 2+3+7+6+6+6 = 30
Mean = 47.5 + 15 × (29/30) = 47.5 + 14.5 = 62 |
3.4 Comparison of All Three Mean Methods
Method | Formula | Best Used When | Effort Level |
Direct Method | x̄ = Σfᵢxᵢ / Σfᵢ | Class marks are small numbers | High |
Assumed Mean | x̄ = a + Σfᵢdᵢ / Σfᵢ | Class marks are large numbers | Medium |
Step Deviation | x̄ = a + h × Σfᵢuᵢ / Σfᵢ | Equal class widths | Low (easiest) |
Important: All Three Methods Give the Same Answer Direct Method = Assumed Mean Method = Step Deviation Method The choice of method only affects how easy the calculation is — not the result. In the board exam, use whichever method the question specifies. If no method is specified, Step Deviation is usually fastest when class width is equal. |
4. Mode of Grouped Data
The Mode of a dataset is the value that appears most frequently. For grouped data, we cannot find the exact mode directly — instead, we first identify the modal class (the class with the highest frequency) and then apply a formula to estimate the mode within that class.
Understanding the Mode formula: The formula accounts for the fact that the modal class has a higher frequency than both the class before it (f₀) and the class after it (f₂). The terms f₁ − f₀ and f₁ − f₂ measure how dominant the modal class is compared to its neighbours.
4.1 Mode Formula
Step 1: Identify the Modal Class — the class with the highest frequency (f₁) Step 2: Note f₀ = frequency of class before modal class Note f₂ = frequency of class after modal class Note l = lower limit of modal class Note h = class width
Mode = l + [ (f₁ − f₀) / (2f₁ − f₀ − f₂) ] × h
where l = lower boundary of modal class, h = class width |
4.2 Worked Example — Mode
Worked Example 4.1 — Mode of Grouped Data Problem: Find the mode of the following data: Class: 0–20 20–40 40–60 60–80 80–100 Frequency: 6 8 12 9 5
Step 1: Identify Modal Class Highest frequency = 12, corresponding to class 40–60 ∴ Modal class = 40–60
Step 2: Identify values l = 40, h = 20, f₁ = 12, f₀ = 8 (class before), f₂ = 9 (class after)
Step 3: Apply formula Mode = 40 + [(12−8) / (2×12 − 8 − 9)] × 20 = 40 + [4 / (24 − 17)] × 20 = 40 + [4/7] × 20 = 40 + 11.43 = 51.43 |
4.3 Key Points About Mode
• Modal class is always the class with the highest frequency — not the highest class mark.
• If two classes have the same highest frequency, the data is bimodal — this rarely appears in CBSE but is good to know.
• The denominator (2f₁ − f₀ − f₂) will always be positive if f₁ is truly the highest frequency.
• Mode is the easiest measure to calculate — it only needs the modal class and its two neighbours.
• Mode is not affected by extreme values (outliers) in the data.
5. Median of Grouped Data
The Median is the middle value of a dataset when arranged in order. For grouped data, the Median divides the total frequency into two equal halves — 50% of the data lies below the Median and 50% lies above it. To find it, we first build a cumulative frequency table, then locate the median class (the class where cumulative frequency first reaches or exceeds n/2), and finally apply the Median formula.
Why is Median useful? Unlike the Mean, the Median is not affected by extreme values (outliers). For example, in a dataset of incomes where a few people earn very high salaries, the Median gives a more realistic picture of the 'typical' income than the Mean would.
5.1 Median Formula
Step 1: Build cumulative frequency table Step 2: Find n/2 (where n = Σfᵢ = total frequency) Step 3: Locate Median Class — first class where cumulative frequency ≥ n/2 Step 4: Note: l = lower limit of median class cf = cumulative frequency of class BEFORE median class f = frequency of median class h = class width
Median = l + [ (n/2 − cf) / f ] × h |
5.2 Worked Example — Median
Worked Example 5.1 — Median of Grouped Data Problem: Find the median of the following frequency distribution: Class: 0–10 10–20 20–30 30–40 40–50 Frequency: 5 8 20 15 7
Step 1: Cumulative Frequency Table 0–10: cf = 5 10–20: cf = 5 + 8 = 13 20–30: cf = 13 + 20 = 33 ← first cf ≥ n/2 30–40: cf = 33 + 15 = 48 40–50: cf = 48 + 7 = 55
Step 2: n = 55, n/2 = 27.5 Median class = 20–30 (cf = 33 ≥ 27.5)
Step 3: l = 20, cf = 13 (cf before median class), f = 20, h = 10 Median = 20 + [(27.5 − 13) / 20] × 10 = 20 + [14.5 / 20] × 10 = 20 + 7.25 = 27.25 |
5.3 Key Points About Median
• cf in the formula stands for the cumulative frequency of the class JUST BEFORE the median class, not of the median class itself.
• Always check: cumulative frequency of the median class must be ≥ n/2, not just > n/2.
• The Median is not affected by extreme values — it is robust against outliers.
• Median divides the area of a histogram into two equal halves.
• If n is odd, n/2 gives a decimal — use it as-is (do not round).
6. Empirical Relationship Between Mean, Median and Mode
In practice, for a moderately skewed frequency distribution, there exists an approximate relationship between the three measures of central tendency. This relationship was established empirically (through observation of real datasets) and is known as the Empirical Formula or Karl Pearson's Empirical Relation.
Mode = 3 × Median − 2 × Mean
Equivalently: Median = (Mode + 2 × Mean) / 3 Mean = (3 × Median − Mode) / 2 |
When to Use the Empirical Relationship Use this formula when two of the three values are given and you need to find the third. Example: 'Mean = 54 and Mode = 60. Find the Median.' Median = (Mode + 2 × Mean) / 3 = (60 + 108) / 3 = 168 / 3 = 56
This is a common 1-mark or 2-mark question in CBSE board exams. Note: This formula is approximate — it holds for moderately skewed distributions only. |
6.1 Worked Examples — Empirical Relationship
Worked Example 6.1 Problem: The mean and median of a distribution are 14 and 15 respectively. Find the mode.
Mode = 3 × Median − 2 × Mean = 3 × 15 − 2 × 14 = 45 − 28 = 17 |
Worked Example 6.2 Problem: For a distribution, Mode = 9.3 and Mean = 7.88. Find the Median.
Median = (Mode + 2 × Mean) / 3 = (9.3 + 2 × 7.88) / 3 = (9.3 + 15.76) / 3 = 25.06 / 3 = 8.35 |
7. Ogive — Cumulative Frequency Curves
An Ogive (pronounced oh-jive) is a smooth S-shaped curve drawn by plotting cumulative frequency against upper class limits (Less Than Ogive) or lower class limits (More Than Ogive). The Ogive is a graphical representation of the cumulative frequency distribution and is primarily used to find the Median graphically.
Why is the Ogive important for CBSE? The board exam regularly asks students to draw one or both types of Ogive and use them to find the median graphically. The 4-mark or 5-mark Ogive question is one of the most predictable question types in this chapter.
7.1 Less Than Ogive
• Plot points: (Upper class limit, cumulative frequency) for each class.
• Add a starting point at the lower boundary of the first class with cf = 0.
• Join all points with a smooth freehand curve.
• The curve rises from left to right (monotonically increasing).
7.2 More Than Ogive
• Plot points: (Lower class limit, more-than cumulative frequency) for each class.
• More-than cf for a class = total frequency (n) minus less-than cf of the previous class.
• Add a final point at the upper boundary of the last class with cf = 0.
• The curve falls from left to right (monotonically decreasing).
7.3 Finding Median from Ogive
1. Method 1 (Using one Ogive): Draw the Less Than Ogive. Locate n/2 on the Y-axis. Draw a horizontal line from n/2 to meet the curve. From that point, draw a vertical line to the X-axis. The X-value is the Median.
2. Method 2 (Using both Ogives): Draw both Less Than and More Than Ogives on the same graph. The X-coordinate of their intersection point gives the Median directly.
Worked Example 7.1 — Building an Ogive Table Problem: Construct Less Than and More Than Ogive tables for: Class: 0–10 10–20 20–30 30–40 40–50 Frequency: 5 8 20 15 7 Total n = 55
Less Than Ogive Points: More Than Ogive Points: Upper limit | cf Lower limit | cf 10 | 5 0 | 55 20 | 13 10 | 50 30 | 33 20 | 42 40 | 48 30 | 22 50 | 55 40 | 7 50 | 0
To find Median graphically: locate n/2 = 27.5 on Y-axis of Less Than Ogive. Draw horizontal line → meets curve → drop vertical → X-axis reads ≈ 27.25 |
7.4 Comparison of Less Than and More Than Ogive
Feature | Less Than Ogive | More Than Ogive |
X-axis value plotted | Upper class limit | Lower class limit |
Y-axis value plotted | Less than cumulative frequency | More than cumulative frequency |
Starting point | (lower boundary of 1st class, 0) | (upper boundary of last class, 0) |
Shape of curve | Rising S-curve (left to right) | Falling S-curve (left to right) |
Intersection of both | X-coordinate = Median | X-coordinate = Median |
Used alone for Median? | Yes — using n/2 method | Yes — using n/2 method |
8. Mean vs Median vs Mode — Full Comparison
Understanding when to use which measure is just as important as knowing how to calculate it. This section gives you a complete side-by-side comparison.
Feature | Mean | Median | Mode |
Definition | Sum of values ÷ total count | Middle value when data is ordered | Most frequently occurring value |
For grouped data | Uses class mark (xᵢ) | Uses cumulative frequency | Uses modal class |
Formula | Σfᵢxᵢ / Σfᵢ | l + [(n/2−cf)/f] × h | l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h |
Affected by outliers? | Yes — very sensitive | No — not affected | No — not affected |
Number of methods | 3 (Direct, AM, SD) | 1 formula | 1 formula |
When best used | Symmetric distributions | Skewed data / income data | Most common item / fashion |
Board exam weight | High (3–5 marks, 3 methods) | High (3–5 marks + Ogive) | Medium (1–3 marks) |
9. Common Mistakes to Avoid
The following mistakes are the most frequently made by students in board exams. Study each one carefully and make sure you know the correct approach.
Mistake | What Students Do Wrong | Correct Approach |
Wrong cf in Median | Use cf of the median class itself | Use cf of the class BEFORE the median class |
Wrong modal class | Choose class with highest mark, not frequency | Modal class = highest FREQUENCY class |
Not computing uᵢ correctly | Forget to divide by h in step deviation | uᵢ = (xᵢ − a) / h — always divide by h |
Assumed mean not a class mark | Choose a random number as assumed mean | Choose 'a' as a class mark (midpoint) |
Ogive plotted at class marks | Plot cumulative freq vs class marks | Plot vs upper limit (LT) or lower limit (MT) |
Rounding n/2 before median | Round 27.5 to 28 before using formula | Use n/2 as decimal — never round it |
f₀/f₂ identification in mode | Swap f₀ and f₂ | f₀ = class before modal class; f₂ = class after |
10. Complete Formula Summary
Concept | Formula | Notes |
Class Mark | xᵢ = (Upper + Lower) / 2 | Midpoint of class interval |
Deviation | dᵢ = xᵢ − a | Used in Assumed Mean method |
Step Deviation | uᵢ = (xᵢ − a) / h | Used in Step Deviation method |
Mean (Direct) | x̄ = Σfᵢxᵢ / Σfᵢ | Multiply marks by frequency |
Mean (Assumed) | x̄ = a + Σfᵢdᵢ / Σfᵢ | a = assumed mean |
Mean (Step Dev) | x̄ = a + h × Σfᵢuᵢ / Σfᵢ | Equal class widths only |
Median | l + [(n/2 − cf) / f] × h | cf = cf before median class |
Mode | l + [(f₁−f₀) / (2f₁−f₀−f₂)] × h | f₁ = modal frequency |
Empirical (Mode) | Mode = 3×Median − 2×Mean | Approximate relation |
Empirical (Median) | Median = (Mode + 2×Mean) / 3 | Approximate relation |
Empirical (Mean) | Mean = (3×Median − Mode) / 2 | Approximate relation |
11. Key Points to Remember
• Three methods for Mean: Direct, Assumed Mean, Step Deviation — all give the same answer.
• Step Deviation method is fastest when class widths are equal.
• Modal class = class with highest frequency (not highest class mark).
• Median class = first class where cumulative frequency ≥ n/2.
• 'cf' in Median formula = cumulative frequency of the class BEFORE the median class.
• Empirical relation: Mode = 3 Median − 2 Mean (approximate, for skewed data).
• Less Than Ogive: plot (upper limit, cf). More Than Ogive: plot (lower limit, more-than cf).
• Intersection of both Ogives gives the Median graphically.
• To find Median from one Ogive: locate n/2 on Y-axis → horizontal to curve → drop to X-axis.
• Never round n/2 before using it in the Median formula.
• The Median and Mode are not affected by outliers; the Mean is.
12. Practice Questions
The following questions are based on the CBSE board exam pattern. Attempt all categories. Show complete working — step marks are awarded in CBSE for each correct step.
12.1 — 1 Mark Questions (VSA)
The mean of a distribution is 20 and mode is 26. Find the Median using the empirical formula.
Write the formula for Median of grouped data and identify what each variable represents.
The modal class of a distribution is 30–40. The frequency of this class is 15, the class before has frequency 10, and the class after has frequency 8. Find the mode.
For a less than ogive, what values are plotted on the X-axis and Y-axis?
What is the class mark of the interval 25–35?
If Σfᵢuᵢ = 12, Σfᵢ = 40, a = 25, and h = 5, find the mean using the step deviation method.
12.2 — 3 Mark Questions (SA)
Find the mean of the following data using the Assumed Mean Method: Class: 10–30, 30–50, 50–70, 70–90, 90–110 | Frequency: 5, 9, 19, 8, 4
Find the mode of the following distribution: Class: 25–35, 35–45, 45–55, 55–65, 65–75 | Frequency: 6, 10, 22, 14, 8
Find the median of the following data: Class: 0–8, 8–16, 16–24, 24–32, 32–40 | Frequency: 6, 7, 11, 9, 7
The mean of the following frequency distribution is 57.6 and total frequency is 50. Find the missing frequencies f₁ and f₂: Class: 0–20, 20–40, 40–60, 60–80, 80–100 | Frequency: 7, f₁, 12, f₂, 8
For the following data, find the mean using the Direct Method and verify using the Step Deviation Method: Class: 0–10, 10–20, 20–30, 30–40, 40–50 | Frequency: 4, 8, 14, 10, 4
12.3 — 5 Mark Questions (LA)
The following table gives the daily income of 50 workers. Find mean, mode, and median. Also find the empirical relationship and verify. Class: 100–120, 120–140, 140–160, 160–180, 180–200 | Frequency: 12, 14, 8, 6, 10
The following data shows the ages of patients admitted to a hospital. Draw a Less Than Ogive and find the median graphically: Age: 5–15, 15–25, 25–35, 35–45, 45–55, 55–65 | Patients: 6, 11, 21, 23, 14, 5
Draw both Less Than and More Than Ogives for the following data on the same axes and find the Median from their intersection: Class: 0–10, 10–20, 20–30, 30–40, 40–50 | Frequency: 5, 8, 20, 15, 7
The mean of the following data is 50. Using the Step Deviation Method, find the missing frequency: Class: 0–20, 20–40, 40–60, 60–80, 80–100 | Frequency: 17, f₁, 32, f₂, 19 | Total = 120
The following distribution shows marks of 100 students. Find mean by step deviation method, mode by the mode formula, and verify the empirical relation Mode = 3 Median − 2 Mean. Class: 0–20, 20–40, 40–60, 60–80, 80–100 | Frequency: 10, 22, 46, 16, 6
Board Exam Strategy for Statistics 1. Memorise all three Mean formulas and know when to use each — this is non-negotiable. 2. When finding Median, always build the full cumulative frequency table first. 3. In Mode, clearly identify f₀, f₁, and f₂ before substituting — labelling prevents errors. 4. Empirical formula questions are easy 1–2 mark pickups — never skip them. 5. For Ogive: remember Less Than uses upper limits; More Than uses lower limits. 6. Show all intermediate columns (xᵢ, dᵢ or uᵢ, fᵢxᵢ etc.) in table form for step marks. 7. CBSE awards marks for correct method even if arithmetic has a small error. 8. Practise at least 3 full Mean + Median + Mode questions before your board exam. |
