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CBSE • Class 9 • Mathematics • Ch 12
Estimated Time: 45 Mins
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Statistics

In Class 9 Mathematics, Chapter 12 "Statistics" delivers an authoritative, comprehensive exploration of graphical data representation aligned with the latest 2026–27 NCERT rationalized curriculum. This master resource equips students with rigorous conceptual understanding of continuous grouped frequency distributions, the geometry of histograms with equal and unequal class widths, the construction of frequency polygons, and the critical analytical skills needed for high-stakes school and board examinations.

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📊 Have You Ever Wondered?

How did a simple pencil-drawn map of London in 1854 save thousands of lives from a deadly cholera outbreak?

When cholera ravaged London, doctor John Snow did not just look at lists of death certificates. He marked each death as a bar on a street map. Instantly, a massive cluster emerged around the Broad Street water pump! By removing the pump handle, the epidemic was halted. That was the dawn of graphical statistics.

A raw spreadsheet containing 10,000 patient temperatures or exam marks is overwhelming noise to the human brain. But when transformed into a Histogram or a Frequency Polygon, the underlying distribution—peaks, spreads, symmetries, and outliers—leaps into vivid clarity. In this chapter, you will master the exact geometric tools used by epidemiologists, data scientists, and economists to turn raw numbers into actionable truth.

Why This Chapter Matters

In Class 9 Mathematics, Chapter 12 "Statistics" delivers an authoritative, comprehensive exploration of graphical data representation aligned with the latest 2026–27 NCERT rationalized curriculum. This master resource equips students with rigorous conceptual understanding of continuous grouped frequency distributions, the geometry of histograms with equal and unequal class widths, the construction of frequency polygons, and the critical analytical skills needed for high-stakes school and board examinations.

Before You Begin (Prerequisites)

  • Understanding raw vs tabulated data and tally marks from Class 7 & 8.
  • Basic concepts of Central Tendency: Mean, Median, and Mode.
  • Constructing single and double bar graphs on Cartesian coordinate axes.
  • Calculating fractions and proportions for scaling data.

What You Will Learn (Core Objectives)

  • Convert discontinuous (inclusive) frequency distributions into continuous (exclusive) class intervals.
  • Calculate Class Limits, Class Width ($h = \text{Upper} - \text{Lower}$), and Class Marks ($x_i = \frac{\text{Upper} + \text{Lower}}{2}$).
  • Construct precise Histograms for continuous grouped data with equal class widths and proper scale selection (including axis kinks).
  • Compute Adjusted Frequencies (Frequency Density) to construct accurate Histograms for distributions with unequal class widths.
  • Construct Frequency Polygons with and without histograms, including proper closure at both ends using zero-frequency reference midpoints.
  • Compare two independent statistical distributions on the same coordinate axes using dual frequency polygons.

Chapter Roadmap & Progression

1 1. Continuous Grouped Frequency Dis...
2 2. Histograms for Continuous Data (...
3 3. Histograms with Varying (Unequal...
4 4. Frequency Polygons (Construction...

Complete Concept Guide (100% Curriculum Coverage)

1. Continuous Grouped Frequency Distributions & Class Intervals

1. The Intuition

Suppose 40 students take a 100-mark mathematics test. Listing individual scores like 34, 78, 56, 92... gives zero immediate insight into overall performance. To extract meaning, we condense raw observations into manageable chunks called Class Intervals (e.g., $0-10, 10-20, 20-30\dots$).

2. The Formal Concept & Mathematical Definitions

A continuous class interval $[a, b)$ is characterized by three fundamental properties:

  • Lower Class Limit ($a$): The smallest value that can fall within the interval.
  • Upper Class Limit ($b$): The greatest boundary value of the interval.
  • Class Width / Size ($h$): The difference between the upper and lower limits: $$h = \text{Upper Limit} - \text{Lower Limit} = b - a$$
  • Class Mark / Midpoint ($x_i$): The central value representing the entire class: $$x_i = \frac{\text{Upper Class Limit} + \text{Lower Class Limit}}{2} = \frac{a + b}{2}$$

Exclusive (Continuous) vs Inclusive (Discontinuous) Series:

Feature Exclusive / Continuous Series Inclusive / Discontinuous Series
Interval Pattern $10-20, 20-30, 30-40$ $11-20, 21-30, 31-40$
Continuity Upper limit of one class = Lower limit of next class. A gap of 1 unit exists between successive classes.
Graphical Usability Directly usable for Histograms. Must be converted by subtracting $0.5$ from lower limits and adding $0.5$ to upper limits!
3. Concrete Worked Example: Conversion to Continuous Form

Problem: The lengths of 40 leaves of a plant are measured in millimeters: $118-126, 127-135, 136-144, 145-153$. Convert these into continuous intervals suitable for a histogram.

Step 1 (Find the Gap): Gap between upper limit of 1st class ($126$) and lower limit of 2nd class ($127$) is $127 - 126 = 1\text{ mm}$.

Step 2 (Calculate Half-Correction Factor): $\frac{1}{2} = 0.5\text{ mm}$.

Step 3 (Apply to Limits): Subtract $0.5$ from each lower limit, add $0.5$ to each upper limit:

  • $118 - 0.5$ to $126 + 0.5$ → $117.5 - 126.5$
  • $127 - 0.5$ to $135 + 0.5$ → $126.5 - 135.5$
  • $136 - 0.5$ to $144 + 0.5$ → $135.5 - 144.5$
  • $145 - 0.5$ to $153 + 0.5$ → $144.5 - 153.5$
4. Pitfall & Examiner Trap
⚠️ Common Trap: The Boundary Value Ambiguity
In a continuous distribution with intervals $10-20$ and $20-30$, which interval does an observation of exactly $20$ belong to?
Rule: By mathematical convention in exclusive series, an observation equal to the upper limit is strictly excluded from that interval and belongs to the next interval as its lower limit. Therefore, $20$ belongs strictly to $20-30$, never to $10-20$!
5. Real-World Relevance

In modern sports analytics, cricket broadcasters group bowler delivery speeds into continuous speed bands (e.g. $135.0-139.9\text{ km/h}, 140.0-144.9\text{ km/h}$) to track fast-bowler stamina throughout a 5-day Test match.

2. Histograms for Continuous Data (Equal Class Widths)

1. The Intuition

Why can we not simply use a standard bar graph for continuous variables like age, height, or time? In a bar graph, bars are separated by empty spaces, which suggests discrete categories (like Apples, Oranges, Bananas). But height flows seamlessly without gaps—between $150\text{ cm}$ and $151\text{ cm}$ lie infinite decimal values. A Histogram glues adjacent rectangular columns together to visually honor that continuity.

2. The Formal Anatomy of a Histogram

A Histogram is a graphical display consisting of adjacent rectangles erected over continuous class intervals:

  • Horizontal Axis ($X$-axis): Represents the continuous variable divided into class intervals.
  • Vertical Axis ($Y$-axis): Represents the frequencies (or frequency densities) on a uniform scale.
  • Rectangle Base: Corresponds to the class width ($h = b - a$).
  • Rectangle Height: Proportional to the frequency when class widths are equal.
  • Total Enclosed Area: The fundamental law of histograms states that the area of each rectangle is directly proportional to the frequency of that class: $$\text{Area of Rectangle} \propto \text{Frequency}$$
  • The Axis Kink / Zigzag Line: If the lowest class interval does not begin at $0$ (for instance, starting at $110\text{ cm}$), a broken zigzag line (called a kink) must be drawn along the $X$-axis near the origin to indicate that the scale is compressed between $0$ and $110$.
3. Concrete Worked Example: Constructing a Histogram

Problem: Draw a histogram for the following distribution of weights of 36 students:

Weight (kg)30.5 – 35.535.5 – 40.540.5 – 45.545.5 – 50.550.5 – 55.5
No. of Students961533

Step 1 (Analyze Intervals): All class intervals are continuous and possess equal width: $35.5 - 30.5 = 5\text{ kg}$.

Step 2 (Set Axes & Scales):

  • $X$-axis: $1\text{ cm} = 5\text{ kg}$, starting from $30.5$ (draw a kink between $0$ and $30.5$).
  • $Y$-axis: $1\text{ cm} = 2\text{ students}$, ranging from $0$ to $16$.

Step 3 (Erect Rectangles): Erect rectangles of heights $9, 6, 15, 3, 3$ directly over each respective 5-kg base with no gaps between them.

4. Pitfall & Examiner Trap
⚠️ Fatal Mistake: Leaving Spacing Between Rectangles
Students frequently leave gaps between bars like in Class 7 bar graphs. In a histogram for continuous data, rectangles must touch each other without any space! Drawing gaps will cost full marks on the question.
5. Real-World Relevance

Digital cameras and image editing software (like Photoshop and Lightroom) display a real-time luminosity histogram showing tonal distribution from pure shadows ($0$) to blown highlights ($255$) to help photographers expose images properly.

3. Histograms with Varying (Unequal) Class Widths (Area Principle)

1. The Intuition

What happens if one class interval is $10$ units wide, while another is $50$ units wide? If you plot both bars at a height of 20, the second bar will look five times larger to the human eye simply because its base is wider! Because the human eye judges proportion by total area, plotting raw frequencies with unequal widths creates visual deception.

2. The Adjusted Frequency Formula

To preserve the cardinal law of histograms—that Area must be proportional to Frequency—we must adjust the height (length of rectangle) using the Minimum Class Width ($h_{\text{min}}$):

$$\mathbf{\text{Adjusted Length of Rectangle} = \frac{\text{Minimum Class Width}}{\text{Class Width of this Interval}} \times \text{Frequency of this Class}}$$

$$\text{Adjusted Height} = \frac{h_{\text{min}}}{h_i} \times f_i$$

3. Concrete Worked Example: Complete Adjustment Table

Problem: A teacher examined the marks scored by 100 students in an exam with unequal grading brackets. Prepare the adjusted frequency table to draw an accurate histogram.

Marks (Class Interval) Frequency ($f_i$) Class Width ($h_i$) Adjustment Calculation $\left(\frac{h_{\text{min}}}{h_i} \times f_i\right)$ Adjusted Length (Height)
$0 - 20$ $7$ $20$ $\frac{10}{20} \times 7$ $3.5$
$20 - 30$ $10$ $10$ $\frac{10}{10} \times 10$ $10$
$30 - 40$ $10$ $10$ $\frac{10}{10} \times 10$ $10$
$40 - 50$ $20$ $10$ $\frac{10}{10} \times 20$ $20$
$50 - 60$ $20$ $10$ $\frac{10}{10} \times 20$ $20$
$60 - 70$ $15$ $10$ $\frac{10}{10} \times 15$ $15$
$70 - 100$ $8$ $30$ $\frac{10}{30} \times 8$ $2.67$

Notice: For the broad $70-100$ bracket (width 30), the height is compressed to $2.67$, ensuring its area ($30 \times 2.67 \approx 80$) remains strictly proportional to the raw frequency ($8$).

4. Pitfall & Examiner Trap
⚠️ Highest Mark-Loss Trap in Class 9:
Whenever an examination problem provides class intervals of varying sizes (e.g. some width 10, some width 20 or 30), DO NOT plot the given frequencies directly on the $Y$-axis! You must construct the adjustment column first. Plotting raw frequencies for unequal widths receives zero credit.
5. Real-World Relevance

National income and tax statistics always use unequal intervals (e.g. ₹0–₹5 Lakh, ₹5–₹10 Lakh, ₹10–₹25 Lakh, ₹25 Lakh–₹1 Crore). Economists rely strictly on adjusted frequency densities to avoid distorting income inequality graphs.

4. Frequency Polygons (Construction With & Without Histograms)

1. The Intuition

Suppose you want to compare the performance of Class 9 Section A and Section B on the same test. If you draw two histograms on top of each other, the overlapping solid rectangles will obscure each other and create visual chaos. A Frequency Polygon solves this by replacing blocky rectangles with clean, elegant line graphs that can easily be superimposed.

2. Construction Method & The Closing Principle

A Frequency Polygon can be constructed using two standard techniques:

  • Method A (Using a Histogram):
    1. Draw the histogram for the given continuous frequency distribution.
    2. Mark the midpoint of the top horizontal edge of each rectangular bar (these are the Class Marks).
    3. Join these consecutive midpoints with straight line segments.
  • Method B (Without Drawing a Histogram):
    1. Calculate the Class Mark ($x_i$) for each interval: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.
    2. Plot the coordinate points $(x_i, f_i)$ on graph paper where the $X$-coordinate is the class mark and $Y$-coordinate is the frequency.
    3. Connect adjacent plotted points with straight line segments.
  • The Golden Rule of Polygon Closure: A polygon is by definition a closed figure! To complete the frequency polygon:
    • Create an imaginary preceding class interval before the first class with frequency 0, and find its midpoint.
    • Create an imaginary succeeding class interval after the last class with frequency 0, and find its midpoint.
    • Connect the first and last plotted points down to these two zero-frequency midpoints on the horizontal axis.
    • Fundamental Theorem: The total area enclosed under the closed frequency polygon is exactly equal to the total area of the histogram!
3. Concrete Worked Example: Constructing Without a Histogram

Problem: Draw a frequency polygon directly from the following data without drawing a histogram:

Class Interval Imaginary / Real Class Mark ($x_i$) Frequency ($f_i$) Coordinates $(x, y)$
$130 - 140$ Imaginary Preceding $135$ $0$ $(135, 0)$
$140 - 150$ Actual 1st Class $145$ $5$ $(145, 5)$
$150 - 160$ Actual 2nd Class $155$ $10$ $(155, 10)$
$160 - 170$ Actual 3rd Class $165$ $20$ $(165, 20)$
$170 - 180$ Actual 4th Class $175$ $9$ $(175, 9)$
$180 - 190$ Imaginary Succeeding $185$ $0$ $(185, 0)$

Plotting Instructions: Plot points $(135, 0), (145, 5), (155, 10), (165, 20), (175, 9), (185, 0)$ and join them in sequence with straight line segments to produce the completed closed polygon.

4. Pitfall & Examiner Trap
⚠️ Common Trap: Dropping Vertically to the Origin
Students often end a frequency polygon by dropping a vertical line down to the lower limit or to the origin $(0, 0)$. This is mathematically incorrect!
Rule: The ends must be joined to the class marks of the imaginary classes having frequency zero. In the example above, anchor to $(135, 0)$ and $(185, 0)$, not to $(140, 0)$ or $(180, 0)$.
5. Real-World Relevance

Meteorologists plot daily temperatures over 12 months using dual frequency polygons on the exact same axis to clearly contrast historical 50-year climate averages against current global warming spikes.

Frequency Polygon & Histogram Architectural Matrix

Histogram & Frequency Polygon Dual Integration Showing continuous bars, midpoint connection, and zero-frequency anchor closures 0 5 10 15 20 Frequency (f) 140 150 160 170 180 Class Intervals (Continuous Scale) (165, 20) Anchor (135, 0) Anchor (185, 0) Histogram (Continuous) Frequency Polygon (Closed)

Chapter Summary & 10 Key Takeaways

Takeaway 1
Histogram Definition: A 2-dimensional graphical representation of continuous grouped data where adjacent rectangular bars touch without any spacing.
Takeaway 2
Area Law of Histograms: The fundamental rule dictates that the area of each rectangular bar is directly proportional to its class frequency.
Takeaway 3
The Kink Convention: A jagged break or zigzag kink along the horizontal axis is mandatory when intervals start at a non-zero value, indicating scale compression from origin $0$.
Takeaway 4
Unequal Class Widths: When widths vary, rectangular heights must be adjusted as $\text{Height} = \frac{\text{Minimum Width}}{\text{Class Width}} \times \text{Frequency}$ to prevent visual distortion.
Takeaway 5
Class Mark Formulation: Midpoint of any class interval is given by $x_i = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}$, acting as the single representative coordinate of that class.
Takeaway 6
Closed Polygon Property: A frequency polygon must be anchored to the horizontal axis by joining its endpoints to the midpoints of imaginary preceding and succeeding zero-frequency classes.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
What is the class mark of the interval $130 - 150$? How is it derived?
Reveal Answer & Explanation
Answer: The class mark is the arithmetic mean of the upper and lower limits: $$\text{Class Mark} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} = \frac{130 + 150}{2} = \frac{280}{2} = 140$$ It represents the central focal point of that class interval.
Class Mark is 140 (midpoint of interval).
2
Why are adjacent rectangles in a histogram drawn touching each other without any gaps?
Reveal Answer & Explanation
Answer: Because a histogram represents continuous grouped numerical data where the upper limit of one class interval is identical to the lower limit of the next class interval. Gaps are used exclusively in discrete bar graphs to denote distinct categorical entities.
Reflects continuity of class boundaries.
3
In a frequency distribution, one class interval is $10 - 20$ with frequency 6, and another is $70 - 100$ with frequency 9. If the minimum class width is 10, calculate the adjusted height of the rectangle for the interval $70 - 100$.
Reveal Answer & Explanation
Answer: Minimum class width $h_{\text{min}} = 10$. For the class $70 - 100$, width $h_i = 100 - 70 = 30$, and frequency $f_i = 9$. Using the formula: $$\text{Adjusted Height} = \frac{h_{\text{min}}}{h_i} \times f_i = \frac{10}{30} \times 9 = \frac{1}{3} \times 9 = 3$$ The bar will be drawn at a height of 3 units.
Adjusted height is 3 units.
4
What are the coordinates plotted to construct a frequency polygon without drawing a histogram?
Reveal Answer & Explanation
Answer: For each class interval, the plotted coordinates are $(x_i, f_i)$, where the $X$-coordinate is the Class Mark (midpoint $\frac{\text{Upper} + \text{Lower}}{2}$) and the $Y$-coordinate is the class frequency $f_i$.
Coordinates are (Class Mark, Frequency).
5
How is a frequency polygon closed at both ends, and why is this closure mathematically necessary?
Reveal Answer & Explanation
Answer: It is closed by extending line segments to the class marks of two imaginary classes—one immediately preceding the first class and one immediately succeeding the last class—both assigned frequency 0. This is mathematically necessary because the area under a properly closed frequency polygon is strictly equal to the total area of the corresponding histogram.
Anchors to midpoints of imaginary zero-frequency classes.
6
The class marks of a distribution are $47, 52, 57, 62, 67$. Determine the class size and find the actual class limits of the first class.
Reveal Answer & Explanation
Answer: Class size $h = 52 - 47 = 5$. Half the class size is $\frac{5}{2} = 2.5$. For the first class mark ($47$): Lower Limit $= 47 - 2.5 = 44.5$; Upper Limit $= 47 + 2.5 = 49.5$. The first class interval is $44.5 - 49.5$.
Class size is 5; first class is 44.5 – 49.5.
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