Cumulative frequency — ogive, median from graph: Worked examples

The scenario matrix

Every median-from-CF problem is one (or a blend) of these cells:

Cell	What makes it special	Trap it sets
A	$N /2$ lands exactly on a class boundary (CF equals $N /2$ )	Off-by-one median class
B	$N /2$ lands mid-class — real interpolation	Rounding / fraction slips
C	Discontinuous classes ( $0 - 9, 10 - 19$ )	Forget $\pm 0.5$ boundary fix
D	Twin-ogive — read median off the crossing	Reading wrong axis
E	Unequal class widths ( $h$ differs per class)	Use wrong $h$
F	Missing frequency given the median	Solve two equations together
G	Degenerate: median in first / last class, open-ended interval, all data in one class	$C F_{b} = 0$ ; open-end boundaries
H	Real-world word problem (waiting times, incomes)	Extract the table first

We now hit each cell at least once across 8 worked examples (Cell G is split into G(i) and G(ii)).

Example 1 — Cell A · $N /2$ on a boundary

Worked example Cell A

Marks of 40 students. Find the median.

Marks	$f$	CF
0–10	4	4
10–20	6	10
20–30	10	20
30–40	12	32
40–50	8	40

Forecast: $N /2 = 20$ . CF hits exactly 20 at "less than 30". Is the median class 20–30 or 30–40? Guess before reading on.

$N /2 = 40/2 = 20$ . Why this step? For grouped data we split by height $N /2$ , never $(N + 1) /2$ — the ogive is a smooth curve, not a discrete list.
Find the median class. CF goes from $10$ (below 20) up to $20$ within the 20–30 class. So median class $= 20 - 30$ . Why this step? The median class is where CF crosses or reaches $N /2$ from below — CF was $10 < 20$ at $x = 20$ , then climbs to $20$ across this class. It is NOT 30–40.
Extract: $L = 20, C F_{b} = 10, f = 10, h = 10$ . Why this step? $C F_{b}$ is the CF of the class before 20–30 (that's the 10–20 row, CF $= 10$ ).
Interpolate: $Median = 20 + (\frac{20 - 10}{10}) \times 10 = 20 + 1 \times 10 = 30.$ Why this step? We needed $20 - 10 = 10$ more counts; the class holds exactly $f = 10$ , so we travel the whole width — landing at the far edge, $x = 30$ .

Figure — Cumulative frequency — ogive, median from graph

Example 2 — Cell B · interpolation genuinely kicks in

$N /2 = 20$ . Why? Same halving rule.
Median class. CF jumps $13 \to 25$ inside 20–30, so it contains the 20th count. Median class $= 20 - 30$ . Why? $13 < 20 \leq 25$ — the crossing happens inside this class.
Extract: $L = 20, C F_{b} = 13, f = 12, h = 10$ .
Interpolate: $Median = 20 + (\frac{20 - 13}{12}) \times 10 = 20 + \frac{7}{12} \times 10 = 20 + 5.8 \overline{3} = 25.83.$ Why? We are $7$ counts into a class of $12$ ; assuming the 12 values are spread evenly across width 10, we go $\frac{7}{12}$ of the way in. Per our convention we round the final line to $25.83$ .

Example 3 — Cell C · discontinuous classes

Worked example Cell C

Ages (whole years) of 50 people. Classes are discontinuous.

Age	$f$
10–19	6
20–29	14
30–39	18
40–49	8
50–59	4

Forecast: The gap between 19 and 20 hides half a year. If you don't fix it, your median is off by $0.5$ . Guess the corrected boundaries first.

Convert to continuous boundaries. Subtract $0.5$ from each lower, add $0.5$ to each upper: $9.5 - 19.5, 19.5 - 29.5, 29.5 - 39.5, \dots$ Why this step? Real ages between 19 and 20 exist; the true boundary is the midpoint $19.5$ . Now $h = 10$ cleanly and no gaps.
CF table: $6, 20, 38, 46, 50$ .
$N /2 = 25$ . CF crosses $25$ going $20 \to 38$ in the $29.5 - 39.5$ class. Median class $= 29.5 - 39.5$ .
Extract: $L = 29.5, C F_{b} = 20, f = 18, h = 10$ .
Interpolate: $Median = 29.5 + (\frac{25 - 20}{18}) \times 10 = 29.5 + \frac{5}{18} \times 10 = 29.5 + 2.78 = 32.28.$ Why? $5$ counts into a class of $18$ , width 10; rounded to two decimals.

Example 4 — Cell D · twin-ogive (read the crossing)

Forecast: Both curves cross at $y = N /2 = 20$ . Guess: the $x$ there should match $25.83$ from Example 2.

Less-than CF: $5, 13, 25, 35, 40$ plotted at upper boundaries $10, 20, 30, 40, 50$ . Why upper? Less-than CF counts everything below the upper boundary.
More-than CF: $40, 35, 27, 15, 5$ plotted at lower boundaries $0, 10, 20, 30, 40$ . Why lower? More-than CF at a lower boundary $=$ everything from that boundary rightward.
They cross where "number below" $=$ "number above" $= N /2 = 20$ . Drop a perpendicular → $x \approx 25.8$ . Why this step? Both halves equal $N /2$ only at the median — by definition of the median.

Example 5 — Cell E · unequal class widths

Worked example Cell E

$N = 60$ . Widths differ — read $h$ per class.

Class	width $h$	$f$	CF
0–10	10	8	8
10–30	20	20	28
30–40	10	22	50
40–70	30	10	60

Forecast: $N /2 = 30$ . Which class contains it, and what is its own width?

$N /2 = 30$ . CF crosses $30$ going $28 \to 50$ in the 30–40 class.
Extract using THIS class's width: $L = 30, C F_{b} = 28, f = 22, h = 10$ . Why this step? The formula's $h$ is the width of the median class only — here $10$ , even though the class before it had width $20$ . Using a neighbour's width is the Cell-E trap.
Interpolate: $Median = 30 + (\frac{30 - 28}{22}) \times 10 = 30 + \frac{2}{22} \times 10 = 30 + 0.909 = 30.91.$ Why? $2$ counts into a class of $22$ , width 10; rounded to two decimals.

Example 6 — Cell F · solve for a missing frequency (the honest algebra)

Worked example Cell F

$N = 100$ and the median is known to be $35$ . The frequency $x$ of the 30–40 class is missing. Find $x$ .

Class	$f$
0–10	8
10–20	12
20–30	20
30–40	$x$
40–50	18
50–60	$y$

Forecast: Two unknowns ( $x$ and $y$ ), so we need two equations: the total $= 100$ and the median formula $= 35$ . Solve them together. Guess whether $x$ comes out a whole number.

Equation 1 — the total constraint. $8 + 12 + 20 + x + 18 + y = 100 \Rightarrow x + y = 42.$ Why this step? All frequencies must sum to $N$ ; with two blanks this is one equation in two unknowns, so we cannot finish on it alone — we need the median equation too.
Locate the median class from the given median. The stated median $35 \in (30, 40)$ , so the median class is 30–40. Why this step? The median always lies inside its class; being told it equals $35$ pins the class to the one straddling $35$ .
Extract the five numbers (median class 30–40). $L = 30, C F_{b} = 8 + 12 + 20 = 40, f = x, h = 10, N = 100.$ Why this step? $C F_{b}$ is the running total of the three classes before 30–40; $f$ is the unknown frequency of the median class itself. Consistency check: for the median to sit inside 30–40 we need $C F_{b} < N /2 = 50$ ; here $40 < 50$ ✓, so the problem does have a solution — no "misprint" surgery required.
Equation 2 — the median formula. Substitute into the one formula and set it equal to $35$ : $35 = 30 + (\frac{\frac{100}{2} - 40}{x}) \times 10 \Rightarrow 5 = \frac{50 - 40}{x} \times 10 = \frac{100}{x} .$ Why this step? Everything except $x$ is known, so the median equation becomes a single equation in $x$ .
Solve for $x$ . $5 = \frac{100}{x} \Rightarrow x = \frac{100}{5} = 20.$ Why this step? Multiply both sides by $x$ , then divide by $5$ — pure algebra, clean whole number.
Use Equation 1 to finish. $x + y = 42$ and $x = 20$ , so $y = 42 - 20 = 22.$ Why this step? The total constraint delivers the second unknown once the median has pinned the first.

Example 7 — Cell G · degenerate cases

Worked example Cell G(i) — median in the FIRST class

$N = 20$ . Data piled at the low end.

Class	$f$	CF
0–5	12	12
5–10	5	17
10–15	3	20

Forecast: $N /2 = 10$ . CF reaches $10$ inside the very first class. What is $C F_{b}$ when there is no class before?

$N /2 = 10$ . CF crosses $10$ inside 0–5 (CF goes $0 \to 12$ ). Median class $= 0 - 5$ .
Extract: $L = 0, C F_{b} = 0, f = 12, h = 5$ . Why $C F_{b} = 0$ ? There is no class before the first, so nothing has accumulated — the running total starts at $0$ . This is the degenerate boundary case.
Interpolate: $Median = 0 + (\frac{10 - 0}{12}) \times 5 = \frac{10}{12} \times 5 = 4.17.$ Why this step? We need $10$ counts out of the $f = 12$ living in this class; those $12$ are spread evenly across width $5$ , so we travel $\frac{10}{12}$ of the width. With $L = 0$ the whole median is that step.

Worked example Cell G(ii) — an OPEN-ENDED interval ("60+") and the LAST class

Monthly incomes (₹ thousand) of 50 people. The top class has no upper bound.

Income	$f$	CF
0–20	8	8
20–40	20	28
40–60	14	42
60+	8	50

Forecast: The "60+" has no upper edge — does that wreck the median? Guess: does the open end even matter here?

$N /2 = 25$ . CF crosses $25$ going $8 \to 28$ in the 20–40 class. Median class $= 20 - 40$ . Why this step? $8 < 25 \leq 28$ — the crossing is inside 20–40, nowhere near the open top class.
Extract: $L = 20, C F_{b} = 8, f = 20, h = 20$ . Why this step? Every one of the five numbers lives in a closed class (20–40 and its predecessor), so the missing upper edge of "60+" never enters the formula.
Interpolate: $Median = 20 + (\frac{25 - 8}{20}) \times 20 = 20 + \frac{17}{20} \times 20 = 20 + 17 = 37.$ Why this step? $17$ counts into a class of $20$ , width $20$ .

Example 8 — Cell H · real-world word problem

Worked example Cell H

A café records how long 80 customers wait (minutes). "Half of customers waited less than how long?" — that is the median.

Wait (min)	$f$
0–2	10
2–4	22
4–6	28
6–8	14
8–10	6

Forecast: "Half waited less than…" screams median. $N /2 = 40$ . Guess which 2-minute band.

Translate the words. "The time below which half the customers fall" $=$ the median. Why this step? Word problems hide the table; the phrase "half fall below" is the median's definition.
CF: $10, 32, 60, 74, 80$ . $N /2 = 40$ .
Median class: CF crosses $40$ going $32 \to 60$ in 4–6. Extract $L = 4, C F_{b} = 32, f = 28, h = 2$ .
Interpolate: $Median = 4 + (\frac{40 - 32}{28}) \times 2 = 4 + \frac{8}{28} \times 2 = 4 + 0.571 = 4.57 min .$ Why? $8$ counts into a class of $28$ , width $2$ ; rounded to two decimals.

First class median ::: $C F_{b} = 0$ (nothing accumulated yet) Discontinuous classes ::: subtract/add $0.5$ before finding $L, h$ $N /2$ on a boundary ::: median class is the one CF crosses from below, not the next one Unequal widths ::: use the median class's own $h$ Missing frequency + given median ::: write the total equation AND the median equation, solve together Missing frequency has no solution when ::: $C F_{b} \geq N /2$ (classes before already reach halfway) Open-ended interval ::: harmless if the median class is closed; else assume neighbour's $h$ Twin ogive ::: median is the $x$ where the two curves cross at height $N /2$

Worked examples — Cumulative frequency — ogive, median from graph