Each class interval is represented by its class mark (midpoint), and frequency indicates how many observations fall in that interval.
Chapter 13 · CBSE Class X
📘 Definition
🎨 SVG Diagram
Concept Visualization
📌 Note
Methods to Find Mean
- Direct Method
- Assumed Mean Method
- Step Deviation Method
📌 Note
Direct Method
\[
\overline{x} = \frac{\sum f_i x_i}{\sum f_i}
\]
This method directly uses class marks and frequencies. It is best suited when values are small and calculations are manageable.
- Find class marks \(x_i\)
- Compute \(f_i x_i\)
- Apply formula
📌 Note
Assumed Mean Method
\[
\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i}, \quad d_i = x_i - a
\]
A suitable assumed mean \(a\) is taken to simplify calculations by reducing large numbers.
- Choose \(a\) near central class
- Compute deviations \(d_i\)
- Multiply with frequencies
📌 Note
Step Deviation Method
\[
\begin{aligned}
\overline{x} &= a + h \left(\frac{\sum f_i u_i}{\sum f_i}\right), \\\\ u_i &= \frac{x_i - a}{h}
\end{aligned}
\]
This method is highly efficient when class intervals are equal and values are large.
- Find class width \(h\)
- Compute \(u_i\)
- Apply simplified formula
✏️ Example
Find the mean of the following grouped data.
| Class Interval | Frequency |
|---|---|
| 10–20 | 5 |
| 20–30 | 9 |
| 30–40 | 14 |
| 40–50 | 8 |
Use Direct Method
Step by step Solution
- \[x_i = 15, 25, 35, 45\]
- \[\begin{aligned} \sum f_i x_i &= 5(15) + 9(25) + 14(35) + 8(45) \\&= 1150 \end{aligned}\]
- \[\begin{aligned}\sum f_i &=5+9+14+8\\&= 36\end{aligned}\]
- \[\begin{aligned} \overline{x} &= \frac{1150}{36} \\&\approx 31.94 \end{aligned}\]
📐 Derivation
The grouped mean formula originates from the basic definition:
\[ \overline{x} = \frac{\text{Sum of observations}}{\text{Number of observations}} \]Since exact values are unknown, each class is represented by its midpoint, leading to:
\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A survey records marks of students in grouped form. The school wants to analyze overall performance efficiently.
Question: Which method is most efficient and why?
Answer Insight:
Step deviation method is most efficient because it simplifies calculations by scaling deviations, reducing computational complexity in large datasets.
📘 Definition
💡 Concept
🎨 SVG Diagram
eometric Interpretation
The class mark lies exactly at the centre of the interval, balancing the distribution.
✏️ Example
Find the class marks of the following intervals.
| Class Interval | Class Mark |
|---|---|
| 10–20 | \[(10+20)/2 = 15\] |
| 20–30 | \[(20+30)/2 = 25\] |
| 30–40 | \[(30+40)/2 = 35\] |
Insight: These values are used in mean calculation formulas.
📐 Derivation
If values are uniformly distributed between lower limit \(L\) and upper limit \(U\), then the average value within that interval is:
\[ \text{Average} = \frac{L + U}{2} \]Hence, the class mark represents the expected central value of that class.
🔢 Formula
Where Class Mark is Used
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A dataset has unequal distribution within class intervals. Is class mark still reliable?
Answer Insight:
Class mark assumes uniform distribution. If data is skewed within intervals, the mean calculated may be slightly inaccurate.
📘 Definition
📘 Definition
Modal Class
🎨 SVG Diagram
Graphical Insight (Histogram Concept)
Mode corresponds to the peak of the distribution. In a histogram, it is located near the tallest bar.
🔢 Formula
📐 Derivation
The formula is derived using the histogram approach, where two lines are drawn from the top corners of adjacent rectangles to locate the peak more accurately within the modal class.
The ratio \((f_1 - f_0)\) and \((f_1 - f_2)\) determines how far the mode lies within the modal class.
✏️ Example
Find the mode of the following data.
table class="table table-bordered text-center text-light">
Class Interval
Frequency
10–20 5 20–30 9 30–40 14 40–50 8
Modal class = 30–40
Step by step Solution
- \[ \begin{aligned} l &= 30,\; h \\&= 10,\; f_1 \\&= 14,\; f_0 \\&= 9,\; f_2 \\&= 8\end{aligned}\]
- \[\begin{aligned}\text{Mode} &= 30 + \left(\frac{14-9}{2(14)-9-8}\right)\times 10\\ &= 30 + \frac{5}{11}\times 10 \\&\approx 34.55 \end{aligned} \]
🧠 Remember
Special Cases
📌 Note
Uses of Mode
⚡ Exam Tip
⚠️ Warning
📋 Case Study
In a dataset, the modal class has very small difference between neighboring frequencies. What does it imply?
Answer Insight:
The distribution is relatively flat near the peak, so the mode is less sharply defined.
📘 Definition
Median
💡 Concept
📘 Definition
Cumulative Frequency (CF)
🎨 SVG Diagram
Graphical Interpretation (Ogive Concept)
The median can be visualized using an ogive (cumulative frequency curve). The x-coordinate corresponding to \(\frac{N}{2}\) gives the median.
📌 Note
Step-by-Step Procedure
🔢 Formula
📐 Derivation
The median lies at position \(\frac{N}{2}\). Within the median class, we assume uniform distribution of data.
The fraction \(\frac{N}{2} - cf\) gives how far the median lies inside the class, and dividing by \(f\) normalizes this position.
Multiplying by class width \(h\) converts this proportion into actual value distance.
✏️ Example
Find the median.
| Class | f | CF |
|---|---|---|
| 10–20 | 5 | 5 |
| 20–30 | 9 | 14 |
| 30–40 | 14 | 28 |
| 40–50 | 8 | 36 |
- \[ N = 36,\quad \frac{N}{2} = 18 \]
Median class = 30–40
- \[ \begin{aligned} l&=30,\\ cf&=14,\\ f&=14,\\ h&=10 \end{aligned} \]
- \[\text{Median} = 30 + \left(\frac{18-14}{14}\right)\times 10 \approx 32.86\]
🔎 Key Fact
Key Characteristics
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
In a highly skewed dataset, which measure is better: mean or median?
Answer Insight:
Median is better because it is not influenced by extreme values.
📘 Definition
Empirical Relation
💡 Concept
🎨 SVG Diagram
The relative positions of mean, median, and mode can be visualized in skewed distributions.
📐 Derivation
The relation is derived empirically by observing real datasets where:
- Mean shifts towards extreme values
- Median stays near central position
- Mode represents peak concentration
Based on these tendencies, the approximate linear relation:
\[ \text{Mode} \approx 3 \times \text{Median} - 2 \times \text{Mean} \]was established through statistical analysis.
📌 Note
Applications
✏️ Example
If Mean = 20 and Median = 18, find Mode.
\[
\begin{aligned}
\text{Mode} &= 3\;\text{Median}-2\;\text{Mode}\\&=3(18) - 2(20) \\&= 54 - 40 \\&= 14
\end{aligned}
\]
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
If Mean > Median > Mode, what is the nature of distribution?
Answer Insight:
The distribution is positively skewed.
❓ Question
The marks obtained by 30 students in a Mathematics test (out of 100) are given below. Find the mean marks.
\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline x_i & 10&20&36&40&50&56&60&70&72&80&88&92&95\\ \hline f_i & 1&1&3&4&3&2&4&4&1&1&2&3&1\\ \hline \end{array} \]
🧩 Solution
Method 1: Direct Method (Ungrouped Data)
\[
\overline{x} = \frac{\sum f_i x_i}{\sum f_i}
\]
\[
\begin{aligned}
\text{Mean }(\overline{x}) &= \dfrac{\sum f_ix_i}{\sum f_i}\\\\&=\frac{1779}{30} \\\\&= 59.3
\end{aligned}
\]
| \(x_i\) | \(f_i\) | \(f_i x_i\) |
|---|---|---|
| 10 | 1 | 10 |
| 20 | 1 | 20 |
| 36 | 3 | 108 |
| 40 | 4 | 160 |
| 50 | 3 | 150 |
| 56 | 2 | 112 |
| 60 | 4 | 240 |
| 70 | 4 | 280 |
| 72 | 1 | 72 |
| 80 | 1 | 80 |
| 88 | 2 | 176 |
| 92 | 3 | 276 |
| 95 | 1 | 95 |
| Total | \(\sum f_i = 30\) | \(\sum f_i x_i = 1779\) |
💡 Concept
🧩 Solution
Method 2: Direct Method (Grouped Data)
Class width = 15
\[ \begin{array}{|c|c|} \hline \text{Class Interval} & f_i\\ \hline 10-25 & 2\\ 25-40 & 3\\ 40-55 & 7\\ 55-70 & 6\\ 70-85 & 6\\ 85-100 & 6\\ \hline \end{array} \]| Class | \(f_i\) | \(x_i\) | \(f_i x_i\) |
|---|---|---|---|
| 10-25 | 2 | 17.5 | 35 |
| 25-40 | 3 | 32.5 | 97.5 |
| 40-55 | 7 | 47.5 | 332.5 |
| 55-70 | 6 | 62.5 | 375 |
| 70-85 | 6 | 77.5 | 465 |
| 85-100 | 6 | 92.5 | 555 |
| Total | \(\sum f_i =30\) | \(\sum f_ix_i = 1860\) |
🧩 Solution
Method 3: Assumed Mean Method
Let assumed mean \((a) = 47.5\)
\[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]| Class Interval | \(f_i\) | \(x_i\) | \(d_i = x_i - a\) | \(f_i d_i\) |
|---|---|---|---|---|
| 10–25 | 2 | 17.5 | -30 | -60 |
| 25–40 | 3 | 32.5 | -15 | -45 |
| 40–55 | 7 | 47.5 | 0 | 0 |
| 55–70 | 6 | 62.5 | 15 | 90 |
| 70–85 | 6 | 77.5 | 30 | 180 |
| 85–100 | 6 | 92.5 | 45 | 270 |
| Total | \(\sum f_i = 30\) | \(\sum f_i d_i = 435\) |
Substituting Values
\[ \begin{aligned} \overline{x} &= 47.5 + \frac{435}{30} &\\= 47.5 + 14.5 &\\= 62 \end{aligned} \]
🧩 Solution
Method 4: Step Deviation Method
Let \(h = 15\), \(a = 47.5\)
\[ \overline{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right) \]| Class Interval | \(f_i\) | \(x_i\) | \(u_i = \frac{x_i - a}{h}\) | \(f_i u_i\) |
|---|---|---|---|---|
| 10–25 | 2 | 17.5 | -2 | -4 |
| 25–40 | 3 | 32.5 | -1 | -3 |
| 40–55 | 7 | 47.5 | 0 | 0 |
| 55–70 | 6 | 62.5 | 1 | 6 |
| 70–85 | 6 | 77.5 | 2 | 12 |
| 85–100 | 6 | 92.5 | 3 | 18 |
| Total | \(\sum f_i = 30\) | \(\sum f_i u_i = 29\) |
Substituting Values
\[ \begin{aligned} \overline{x} &= 47.5 + 15\left(\frac{29}{30}\right) \\&= 47.5 + 14.5 \\&= 62 \end{aligned} \]
🎨 SVG Diagram
Why Means Differ
Grouping causes approximation because all values in a class are replaced by midpoint.
⚡ Exam Tip
⚠️ Warning
Common Mistakes
❓ Question
Example-2: Mode of Grouped Data (Family Size)
A survey conducted on 20 households gives the following distribution of family size. Find the mode.
\[ \begin{array}{|c|c|} \hline \text{Family Size} & 1\text{-}3 & 3\text{-}5 & 5\text{-}7 & 7\text{-}9 & 9\text{-}11\\ \hline \text{Number of Families} & 7 & 8 & 2 & 2 & 1\\ \hline \end{array} \]
🧩 Solution
Mode represents the most frequent value. For grouped data, we locate it using the
modal class and refine its position using interpolation.
Step by step Solution
- Identify Modal Class
Highest frequency = 8
⇒ Modal class = 3–5 - \(\begin{aligned} l &= 3\\ h &= 2\\ f_1 &= 8\\ f_0 &= 7\\ f_2 &= 2\\ \end{aligned}\)
- Apply Formula \[ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right)h \]
- Calculation \[ \begin{aligned} \text{Mode} &= 3 + \left(\frac{8-7}{2\times 8 - 7 - 2}\right)\times 2 \\\\ &= 3 + \left(\frac{1}{16 - 9}\right)\times 2 \\\\ &= 3 + \frac{1}{7}\times 2 \\\\ &= 3 + \frac{2}{7} \\\\ &= 3.286 \ (\text{approx}) \end{aligned} \]
Final Answer: Mode ≈ 3.29
🎨 SVG Diagram
Graphical Insight
Mode lies inside the modal class and is shifted depending on neighboring frequencies.
📝 Summary
Interpretation
⚡ Exam Tip
⚠️ Warning
Common Mistake
📋 Case Study
If \(f_1\) becomes equal to \(f_0\), what happens to mode?
Answer Insight:
Mode shifts toward the lower boundary of the modal class.
📄 Question
Median from Less Than Type Data
Heights (in cm) of 51 girls are given in less than cumulative form. Find the median height.
| Height (cm) | Cumulative Frequency |
|---|---|
| < 140 | 4 |
| < 145 | 11 |
| < 150 | 29 |
| < 155 | 40 |
| < 160 | 46 |
| < 165 | 51 |
💡 Concept
🧩 Solution
Convert to Class Intervals
| Class Interval | Frequency \(f\) | Cumulative Frequency (CF) |
|---|---|---|
| 0–140 | 4 | 4 |
| 140–145 | 7 | 11 |
| 145–150 | 18 | 29 |
| 150–155 | 11 | 40 |
| 155–160 | 6 | 46 |
| 160–165 | 5 | 51 |
Locate Median Class
\[
N = 51,\quad \frac{N}{2} = 25.5
\]
The cumulative frequency just greater than 25.5 is 29, so the median class is 145–150.
- \(l = 145\)
- \(cf = 11\)
- \(f = 18\)
- \(h = 5\)
Apply Formula
\[
\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right)h
\]
Calculation
\[ \begin{aligned}
\text{Median} &= 145 + \left(\frac{25.5 - 11}{18}\right)\times 5 \\
&= 145 + \left(\frac{14.5}{18}\right)\times 5 \\
&= 145 + \frac{72.5}{18} \\
&= 145 + 4.03 \\
&= 149.03
\end{aligned} \]
Final Answer: Median ≈ 149.03 cm
🎨 SVG Diagram
Graphical Insight (Ogive Method)
The median corresponds to the x-value at \(\frac{N}{2}\) on the ogive curve.
🔎 Key Fact
Interpretation
⚡ Exam Tip
⚠️ Warning
Common Mistake
❓ Question
Finding Unknown Frequencies using Median
The median of the following data is 525. Find the values of \(x\) and \(y\), given total frequency is 100.
| Class Interval | Frequency |
|---|---|
| 0–100 | 2 |
| 100–200 | 5 |
| 200–300 | \(x\) |
| 300–400 | 12 |
| 400–500 | 17 |
| 500–600 | 20 |
| 600–700 | \(y\) |
| 700–800 | 9 |
| 800–900 | 7 |
| 900–1000 | 4 |
💡 Concept
🧩 Solution
Cumulative Frequency Table
| Class | \(f\) | CF |
|---|---|---|
| 0–100 | 2 | 2 |
| 100–200 | 5 | 7 |
| 200–300 | \(x\) | \(7+x\) |
| 300–400 | 12 | \(19+x\) |
| 400–500 | 17 | \(36+x\) |
| 500–600 | 20 | \(56+x\) |
| 600–700 | \(y\) | \(56+x+y\) |
| 700–800 | 9 | \(65+x+y\) |
| 800–900 | 7 | \(72+x+y\) |
| 900–1000 | 4 | \(76+x+y\) |
Use Total Frequency
\[ 76 + x + y = 100 \Rightarrow x + y = 24 \tag{1} \]
Identify Median Class
\[ \frac{N}{2} = \frac{100}{2} = 50 \]
CF just greater than 50 is \(56 + x\), corresponding to class 500–600.
- \(l = 500\)
- \(cf = 36 + x\)
- \(f = 20\)
- \(h = 100\)
Apply Median Formula
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right)h \]
Solve for \(x\)
\[ \begin{aligned}
525 &= 500 + \left(\frac{50 - (36 + x)}{20}\right)\times 100 \\
25 &= \left(\frac{14 - x}{20}\right)\times 100 \\
25 &= 5(14 - x) \\
25 &= 70 - 5x \\
5x &= 45 \\
x &= 9
\end{aligned} \]
Solve for \(y\)
\[ \begin{aligned}
x + y &= 24\\
9 + y &= 24 \\\Rightarrow y &= 15
\end{aligned} \]
\[
\boxed{x = 9,\quad y = 15}
\]
🎨 SVG Diagram
Median lies inside the class 500–600 and divides the distribution into two equal halves.
⚡ Exam Tip
⚠️ Warning
Common Mistakes
NCERT Class X · Chapter 13
Statistics Engine
Master measures of central tendency through formulas, step-by-step solutions, interactive drills, and concept-building exercises.
📐 Mean
📊 Median
🎯 Mode
📈 Ogive
⚡ Step Solver
🧩 Quizzes
📐 Core Formulas at a Glance
All key formulas for Chapter 13 Statistics — Mean, Median, Mode, and Cumulative Frequency — with symbol legends and usage notes.
① Arithmetic Mean (Grouped Data)
Direct Method
Mean — Direct Method
x̄ = Σfᵢxᵢ / Σfᵢ
xᵢ = class mark (midpoint) = (lower + upper) / 2
fᵢ = frequency of that class
Use when: values and frequencies are small & manageable.
fᵢ = frequency of that class
Use when: values and frequencies are small & manageable.
Assumed Mean Method
Mean — Short-cut / Assumed Mean
x̄ = A + (Σfᵢdᵢ / Σfᵢ)
dᵢ = xᵢ − A
dᵢ = xᵢ − A
A = assumed mean (any convenient class mark)
dᵢ = deviation of xᵢ from A
Use when: values are large; reduces arithmetic load.
dᵢ = deviation of xᵢ from A
Use when: values are large; reduces arithmetic load.
Step-Deviation Method
Mean — Step-Deviation Method
x̄ = A + (Σfᵢuᵢ / Σfᵢ) × h
uᵢ = (xᵢ − A) / h
uᵢ = (xᵢ − A) / h
h = class width (assumed equal)
uᵢ = step deviation
Use when: class widths are equal — most efficient method.
uᵢ = step deviation
Use when: class widths are equal — most efficient method.
② Median (Grouped Data)
Median Formula
Median of Grouped Data
M = l + [(n/2 − cf) / f] × h
l = lower boundary of median class
n = total frequency (Σfᵢ)
cf = cumulative frequency of class before median class
f = frequency of median class
h = class width
Median class: the class where cumulative frequency ≥ n/2 for the first time.
n = total frequency (Σfᵢ)
cf = cumulative frequency of class before median class
f = frequency of median class
h = class width
Median class: the class where cumulative frequency ≥ n/2 for the first time.
Locating Median
Steps to Find Median Class
1. Find n = Σfᵢ
2. Compute n/2
3. Find cf ≥ n/2 (first time)
4. That row's class = Median Class
2. Compute n/2
3. Find cf ≥ n/2 (first time)
4. That row's class = Median Class
Build a cumulative frequency column. The row where cf first reaches or exceeds n/2 is your median class.
③ Mode (Grouped Data)
Mode Formula
Mode of Grouped Data
Mo = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
l = lower boundary of modal class
f₁ = frequency of modal class
f₀ = frequency of class before modal class
f₂ = frequency of class after modal class
h = class width
Modal class: class with the highest frequency.
f₁ = frequency of modal class
f₀ = frequency of class before modal class
f₂ = frequency of class after modal class
h = class width
Modal class: class with the highest frequency.
Empirical Relation
Relation Between Mean, Median & Mode
Mode = 3 × Median − 2 × Mean
This empirical (approximate) relationship holds for moderately skewed distributions. Useful to find one measure when the other two are known. Note: This is not exact — just a rule of thumb.
④ Cumulative Frequency & Ogive
Less Than Ogive
Less-Than Cumulative Frequency
cf_less = f₁ + f₂ + … + fₙ
Plot: (upper boundary, cf)
Plot: (upper boundary, cf)
Start from 0 at the lower boundary of the first class. Add each frequency successively. Points are plotted at upper class boundaries.
More Than Ogive
More-Than Cumulative Frequency
cf_more = n − (f₁ + f₂ + … + fₖ₋₁)
Plot: (lower boundary, cf)
Plot: (lower boundary, cf)
Start from n at the lower boundary of the first class. Subtract frequencies successively. Points plotted at lower class boundaries. Intersection of both ogives → Median.
⚡ Step-by-Step AI Solver
Choose a method below, enter your data, and get a complete worked solution with every step explained.
Enter comma-separated class intervals and frequencies. Class intervals as 10-20,20-30,…
📝 Practice Questions
Concept-building questions with full step-by-step solutions. Filtered by topic — none are textbook repeats.
💡 Tips & Tricks
Smart strategies, shortcuts, and mental models to solve Statistics problems faster and more accurately.
⚠️ Common Mistakes to Avoid
The most frequent errors students make — with correct approaches and explanations.
🧩 Interactive Learning Modules
Choose a module to deepen your understanding through interactive drills, concept builders, and explorations.
Concept Quiz
10 MCQs with instant feedback and detailed explanations. Tests deep understanding.
CF Table Builder
Enter data and watch the cumulative frequency table build in real time.
Method Selector
Given a dataset scenario, decide which method of finding mean to use and why.
Guess the Median Class
Random frequency distributions — identify the correct median class before the answer reveals.
Formula Match
Match each symbol or term to its correct definition across all three formulas.
Flash Drill
Rapid-fire numeric inputs — compute class marks, deviations, and uᵢ values instantly.
🎯 Concept Quiz
📊 Cumulative Frequency Table Builder
Enter class intervals and frequencies below. The less-than and more-than CF tables are built automatically.
🔀 Method Selector Challenge
🎲 Guess the Median Class
🔗 Formula Symbol Match
⚡ Flash Drill
📚
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Statistics plays a crucial role in transforming raw numerical information into meaningful conclusions. In Class X Mathematics, Chapter 12 introduces students to the systematic study of data collected from real-life situations and demonstrates how such data can be organised, analysed, and interpreted logically. This chapter strengthens the learner’s ability to deal with large sets of information, which is essential not only for academic assessments but also for informed decision-making in daily…
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