NDA Maths · Statistics

Frequency Distributions and Graphical Representation

How to organise raw data into class intervals + frequencies, and which graph (histogram, polygon, ogive, pie chart) tells the story best.

All Statistics notes NDA Maths strategy

Why this matters

14 PYQs across 2017–2025 — small subtopic but reliable scoring territory. Three shapes dominate: picking the right graph for given data, computing a histogram class's relative height when widths are unequal, and reading values straight off a frequency table (mode, cumulative count, median).

Concept 1 of 3

Histograms, Frequency Polygons & Ogives

Intuition

A histogram is bars whose AREA represents frequency — so when class widths are unequal, the bar height must be frequency divided by class width (frequency density), not raw frequency. Connecting the midpoints of histogram tops gives a frequency polygon. Plotting cumulative frequency against class boundaries gives an ogive — used to read the median directly.

Definition

Histogram bars have width = class width and height = $\text{frequency density}$ . For equal class widths, density is proportional to raw frequency. An ogive is a cumulative-frequency curve; the value of $x$ at which the ogive equals $n/2$ is the median.

Frequency Density (for unequal class widths)

\text{Density} = \dfrac{\text{Frequency}}{\text{Class width}}

Class widthupper bound − lower bound of the class

Visualization · change the bin width, watch the shape change

Bin width:

Same 30 data points each time — only the bin width changes. At width 50 the shape looks almost uniform; at width 5 it looks jagged. Choosing bin width is part of the analysis, not the data.

Worked example

Three classes have widths 5, 10, 5 with frequencies 30, 50, 25. Find the height of each histogram bar (relative).

Density = frequency / class width.
Class 1: $30 / 5 = 6$ .
Class 2: $50 / 10 = 5$ .
Class 3: $25 / 5 = 5$ .
Heights are in ratio 6 : 5 : 5 — note class 2 has the highest frequency but NOT the tallest bar.

Answer:Heights = 6, 5, 5 (in density units)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1StatisticsEASY

For a histogram based on a frequency distribution with unequal class intervals, the frequency of a class should be proportional to

[Q103 · Apr · 2021]

Bar height $\neq$ frequency when class widths differ

If two classes have the same frequency but different widths, the wider class has the SHORTER bar — because density (height) divides frequency by width. Students draw bars of equal height for equal frequencies; correct histograms make AREAS equal, not heights.

Drill 4 more on histograms, frequency polygons & ogives

Concept 2 of 3

Pie Charts

Intuition

A pie chart shows how a whole is split into parts. Each part's sector angle is proportional to its share of the total — and all the sector angles together must add to

360^\circ

Definition

For a category with frequency $f_i$ and total frequency $\sum f_i = N$ , the sector angle is $\theta_i = \dfrac{f_i}{N} \times 360^\circ$ . Equivalently, the angle is proportional to the frequency, with proportionality constant $360/N$ .

Sector Angle in a Pie Chart

\theta_i = \dfrac{f_i}{N} \times 360^\circ \qquad \sum_i \theta_i = 360^\circ

$f_i$ frequency / count of category $i$
$N$ total frequency $\sum f_i$

Diagram · pie sector = (f / total) × 360°

Walk8/20 → 144° (40%)
Cycle6/20 → 108° (30%)
Bus4/20 → 72° (20%)
Car2/20 → 36° (10%)

Each slice's central angle is its share of 360°: Walk = 8/20 × 360° = 144°. The four angles add to 360° and the frequencies to the total — the check most pie-chart questions turn on.

Worked example

A company has 30 Science, 70 Arts and 50 Commerce graduates. Find the pie-chart angle for Science.

Total $N = 30 + 70 + 50 = 150$ .
Apply the formula: $\theta_{\text{Science}} = \dfrac{30}{150} \times 360^\circ$ .
Simplify: $\theta_{\text{Science}} = \dfrac{1}{5} \times 360^\circ = 72^\circ$ .

Answer:

\theta_{\text{Science}} = 72^\circ

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2StatisticsMODERATE

The central angles

p

q

r

and

s

(in degrees) of four sectors in a Pie Chart satisfy the relation

9p=3q=2r=6s

. What is the value of

4p-q

[Q107 · Apr · 2023]

All angles MUST sum to $360^\circ$

If your computed angles don't add to 360, you have an arithmetic error. PYQs that give angle relations ("

9p = 3q = 2r = 6s

") use

p + q + r + s = 360

as the closing equation — without that, the system is underdetermined.

Drill 4 more on pie charts

Concept 3 of 3

Reading Frequency Tables — Mode, Cumulative, Median

Intuition

A frequency table compresses a lot of data into one grid. Three things you read directly off it: the MODAL class (highest frequency), the CUMULATIVE frequency up to any class (running total), and the MEDIAN (the value where cumulative frequency crosses

n/2

Definition

Modal class: the class with the highest frequency. Cumulative frequency at class $k$ : $\sum_{i \leq k} f_i$ . Median for grouped data: $M = L + \dfrac{n/2 - F}{f}\,h$ , where $L$ = lower bound of the median class, $F$ = cumulative frequency before it, $f$ = frequency of median class, $h$ = class width.

Median from a Grouped Frequency Distribution

M = L + \dfrac{\tfrac{n}{2} - F}{f}\,h

$L$ lower bound of the median class
$F$ cumulative frequency BEFORE the median class
$f$ frequency of the median class
$h$ class width

Worked example

Heights (cm) of 20 students:

150{-}155 \to 4,\ 155{-}160 \to 6,\ 160{-}165 \to 7,\ 165{-}170 \to 3

. Find the median height.

Total $n = 4 + 6 + 7 + 3 = 20$ , so $n/2 = 10$ .
Cumulative frequencies: 4, 10, 17, 20. The 10th observation falls at the END of the $155{-}160$ class — but $n/2 = 10$ is reached at the boundary, so by convention the median class is $160{-}165$ .
Identify: $L = 160,\ F = 10,\ f = 7,\ h = 5$ .
Apply: $M = 160 + \dfrac{10 - 10}{7} \times 5 = 160 + 0 = 160$ .

Answer:

M = 160

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3StatisticsMODERATE

The frequency distribution of height of students of a class is given below: Height (cm): 160–162, 162–164, 164–166, 166–168 Number of Students: 12, 15, 24, 13

What is the median height of the class?

[Q102 · Apr · 2025]

Cumulative frequency is RUNNING total, not class total

The cumulative frequency at class

k

is the sum of frequencies from class 1 through class

k

— not the frequency of class

k

alone. Tripping on this turns every median-from-grouped-data question into nonsense.

Identify the median class FIRST, then plug into the formula

The median class is the class where the cumulative frequency first reaches or exceeds

n/2

. Don't pick the class with the highest frequency (that's the modal class) or the middle row of the table.

Drill 3 more on reading frequency tables — mode, cumulative, median

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Histograms, Frequency Polygons & Ogives
Frequency Density (for unequal class widths)
$\text{Density} = \dfrac{\text{Frequency}}{\text{Class width}}$
Pie Charts
Sector Angle in a Pie Chart
$\theta_i = \dfrac{f_i}{N} \times 360^\circ \qquad \sum_i \theta_i = 360^\circ$
Reading Frequency Tables — Mode, Cumulative, Median
Median from a Grouped Frequency Distribution
$M = L + \dfrac{\tfrac{n}{2} - F}{f}\,h$

Watch out for (4)

Bar height $\neq$ frequency when class widths differ
→ Histograms, Frequency Polygons & Ogives
All angles MUST sum to $360^\circ$
→ Pie Charts
Cumulative frequency is RUNNING total, not class total
→ Reading Frequency Tables — Mode, Cumulative, Median
Identify the median class FIRST, then plug into the formula
→ Reading Frequency Tables — Mode, Cumulative, Median

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Related notes

Measures of Central Tendency — grouped-data median and mode use these tables →