Cumulative frequency — ogive, median from graph: Exercises

Before we start, three shared reminders in plain words, so no symbol and no idea is unearned.

Intuition WHY the formula looks like that (derive it, don't memorise it)

Start at the left edge $L$ of the median class. At that point exactly $C F_{b}$ observations sit below us — but we want to stand where $N /2$ observations are below us. So we still need to climb past $extra counts still needed \frac{N}{2} - C F_{b}$ more observations. Now the key assumption: inside the median class the $f$ observations are treated as spread evenly across the width $h$ (see the uniform-spread reminder below). Even spreading means each unit of x carries the same number of counts, namely $f / h$ counts per unit of x. To gain $(\frac{N}{2} - C F_{b})$ counts we must therefore walk a horizontal distance $Δ x = \frac{counts needed}{counts per unit x} = \frac{\frac{N}{2} - C F _{b}}{f / h} = (\frac{\frac{N}{2} - C F _{b}}{f}) \times h .$ Adding this walk $Δ x$ to our starting point $L$ gives $Median = L + Δ x$ — exactly the boxed formula. Every letter is now earned, not accepted on faith.

Figure — Cumulative frequency — ogive, median from graph

Level 1 — Recognition

(Can you name the pieces and read the graph?)

Exercise 1.1. A table has classes $0 - 10, 10 - 20, 20 - 30$ with frequencies $3, 7, 5$ . Write the less-than cumulative frequency table and state $N$ .

Recall Solution 1.1

Accumulate left to right — each CF is the previous CF plus the current frequency.

Less than	CF
10	3
20	$3 + 7 = 10$
30	$10 + 5 = 15$

The final CF is $15$ , so $N = 15$ . That last value must equal the sum of all frequencies ( $3 + 7 + 5 = 15$ ) — your sanity check. ✓

Exercise 1.2. For a less-than ogive, against which x-value do we plot the CF of the class $20 - 30$ : the midpoint $25$ or the upper boundary $30$ ? Explain in one sentence.

Recall Solution 1.2

The upper boundary $30$ . The CF of that class counts everything below $30$ , so the honest point is $(30, C F)$ . Plotting at the midpoint would slide the whole curve left and give a wrong median. See the parent note's "midpoint mistake."

Exercise 1.3. An ogive passes through the point $(40, 32)$ . In plain words, what does that dot mean?

Recall Solution 1.3

"== $32$ observations are less than $40$ ==." The y-value of a less-than ogive at $x = 40$ is literally the count of data below $40$ .

Level 2 — Application

(Plug into the median formula correctly.)

Exercise 2.1. Marks of $40$ students:

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

Find the median.

Recall Solution 2.1

Step 1 — CF table. $4, 10, 20, 32, 40$ . So $N = 40$ and $N /2 = 20$ . Step 2 — median class. We need where CF first reaches $20$ . CF is $10$ below $20$ , then climbs to $20$ inside the $20 - 30$ class → median class is $20 - 30$ . Step 3 — read the letters. $L = 20, C F_{b} = 10, f = 10, h = 10$ . $Median = 20 + (\frac{20 - 10}{10}) \times 10 = 20 + 1 \times 10 = 30.$ This is exactly the slide-across drawn in Figure s01: $y = 20$ , right to the curve, down to $x = 30$ . ✓

Exercise 2.2. Frequencies for the same class limits become $5, 8, 12, 10, 5$ (total $40$ ). Find the median.

Recall Solution 2.2

CF: $5, 13, 25, 35, 40$ . $N = 40, N /2 = 20$ . First CF $\geq 20$ is $25$ (at "less than $30$ "), and CF jumps from $13$ (below $20$ ) to $25$ inside $20 - 30$ → median class $20 - 30$ . $L = 20, C F_{b} = 13, f = 12, h = 10$ : $Median = 20 + (\frac{20 - 13}{12}) \times 10 = 20 + \frac{7}{12} \times 10 = 20 + 5.8 \overline{3} = 25.83.$ Here the median lands mid-class, so Linear interpolation does real work. Look at Figure s02, which zooms into just the $20 - 30$ class: the teal segment climbs straight from $C F_{b} = 13$ (at $x = 20$ ) to $25$ (at $x = 30$ ) — straight because of the uniform-spread assumption. We needed $7$ more counts past $13$ to reach $20$ , and the $12$ counts fill the width evenly, so the plum marker sits $7/12$ of the way across, at $x \approx 25.83$ . ✓

Level 3 — Analysis

(Spot the median class carefully; handle boundaries and unequal widths.)

Exercise 3.1. Given CF values $8, 20, 20, 35, 50$ for classes $0 - 10, \dots, 40 - 50$ (width $10$ , $N = 50$ ), find the median. Watch the repeated $20$ .

Recall Solution 3.1

$N /2 = 25$ . Scan the CFs: $8, 20, 20, 35, 50$ . The first CF that is $\geq 25$ is $35$ , in the $30 - 40$ class. The class before it has $C F_{b} = 20$ . Recover $f$ for the median class from CF: $35 - 20 = 15$ . $L = 30, C F_{b} = 20, f = 15, h = 10$ : $Median = 30 + (\frac{25 - 20}{15}) \times 10 = 30 + \frac{5}{15} \times 10 = 30 + 3.3 \overline{3} = 33.33.$ The flat stretch (CF $20 \to 20$ ) is a class with zero frequency — the ogive is horizontal there, no observations. It never reaches $25$ , so it can't be the median class. ✓

Exercise 3.2. Discontinuous classes $1 - 10, 11 - 20, 21 - 30$ with frequencies $6, 14, 10$ ( $N = 30$ ). Find the median after converting to continuous boundaries.

Recall Solution 3.2

Convert boundaries first: subtract $0.5$ from each lower limit, add $0.5$ to each upper limit → $0.5 - 10.5, 10.5 - 20.5, 20.5 - 30.5$ . Width $h = 10$ . CF: $6, 20, 30$ . $N /2 = 15$ . First CF $\geq 15$ is $20$ (class $10.5 - 20.5$ ); $C F_{b} = 6$ . $L = 10.5, C F_{b} = 6, f = 14, h = 10$ : $Median = 10.5 + (\frac{15 - 6}{14}) \times 10 = 10.5 + \frac{9}{14} \times 10 = 10.5 + 6.4286 = 16.93.$ Skipping the $0.5$ shift would give $\approx 16.43$ — wrong by exactly $0.5$ . ✓

Exercise 3.3 (unequal class widths). Ages of $50$ people grouped into deliberately uneven classes:

Age	$f$	width $h$
0–10	8	10
10–30	22	20
30–35	12	5
35–50	8	15

Find the median, taking care to use the median class's own width.

Recall Solution 3.3

CF: $8, 30, 42, 50$ . $N = 50, N /2 = 25$ . First CF $\geq 25$ is $30$ , so the median class is $10 - 30$ ; $C F_{b} = 8$ . Here is the whole point of the exercise: the median class's width is $h = 30 - 10 = 20$ , not $10$ from the first row. Grab the width of the median class itself. $L = 10, C F_{b} = 8, f = 22, h = 20$ : $Median = 10 + (\frac{25 - 8}{22}) \times 20 = 10 + \frac{17}{22} \times 20 = 10 + 15.4545 = 25.45.$ The uniform-spread assumption still applies, but the counts-per-unit slope is now $f / h = 22/20$ , gentler than a width-10 class would give — exactly why the correct $h$ matters. ✓

Level 4 — Synthesis

(Combine the ogive graph, twin-ogive idea, and other statistics.)

Before Exercise 4.1 we need the second curve, which the exercise relies on:

Exercise 4.1. Twin-ogive reasoning. For $N = 60$ , the less-than ogive and more-than ogive intersect at $x = 27$ . (a) What is the y-value at the intersection? (b) What is the median?

Recall Solution 4.1

(a) At the crossing, "number below" $=$ "number above" $= N /2 = 60/2 = 30$ . So the y-value is $30$ — read this straight off Figure s03, where the two curves meet at the halfway height. (b) Drop a perpendicular from the crossing to the x-axis → the median is $x = 27$ . Why it works: the less-than curve gives "count below," the more-than gives "count above"; those are equal only at $N /2$ , which is precisely the median position. See Quartiles and percentiles from ogive for the same slide-across trick applied to $N /4$ and $3 N /4$ .

Exercise 4.2. A distribution has classes of width $5$ : $0 - 5, 5 - 10, 10 - 15, 15 - 20$ with frequencies $7, 13, 20, 10$ ( $N = 50$ ). Find (a) the median, and (b) the lower quartile $Q_{1}$ (found by sliding across at height $N /4$ instead of $N /2$ ).

Recall Solution 4.2

CF: $7, 20, 40, 50$ . (a) $N /2 = 25$ . First CF $\geq 25$ is $40$ (class $10 - 15$ ), $C F_{b} = 20$ , $f = 20$ , $L = 10$ , $h = 5$ : $Median = 10 + (\frac{25 - 20}{20}) \times 5 = 10 + \frac{5}{20} \times 5 = 10 + 1.25 = 11.25.$ (b) $N /4 = 12.5$ . First CF $\geq 12.5$ is $20$ (class $5 - 10$ ), $C F_{b} = 7$ , $f = 13$ , $L = 5$ , $h = 5$ : $Q_{1} = 5 + (\frac{12.5 - 7}{13}) \times 5 = 5 + \frac{5.5}{13} \times 5 = 5 + 2.1154 = 7.12.$ Same interpolation machine, just a different target height. ✓

Level 5 — Mastery

(Reverse the machinery: unknowns, missing frequencies.)

Exercise 5.1. A grouped table (width $10$ ) has frequencies $5, x, 20, 15, y$ over classes $0 - 10, \dots, 40 - 50$ . The total is $N = 70$ and the median is known to be $28.5$ . The median class is $20 - 30$ . Find $x$ and $y$ .

Recall Solution 5.1

Equation 1 (total): $5 + x + 20 + 15 + y = 70 \Rightarrow x + y = 30$ . Equation 2 (median): median class $20 - 30$ gives $L = 20, f = 20, h = 10$ , and $C F_{b} = 5 + x$ (CF before the median class). $N /2 = 35$ . $28.5 = 20 + (\frac{35 - ( 5 + x )}{20}) \times 10.$ Solve: $8.5 = \frac{30 - x}{20} \times 10 = \frac{30 - x}{2}$ , so $30 - x = 17 \Rightarrow x = 13$ . Then $y = 30 - 13 = 17$ . Check the median class assumption: CF is $5, 18, 38, \dots$ — CF crosses $35$ inside $20 - 30$ (goes $18 \to 38$ ). Consistent. ✓

Exercise 5.2. For the completed table in 5.1 ( $f = 5, 13, 20, 15, 17$ ), also state the modal class and confirm it is not the same as the median class — illustrating that "biggest bar $\neq =$ median class."

Recall Solution 5.2

The modal class is the class with the largest frequency: $20 - 30$ has $f = 20$ , the maximum. Here the modal and median classes coincidentally both land in $20 - 30$ — so this table is a counter-counterexample: they can agree. The general lesson still holds: the median class is fixed by where CF crosses $N /2$ , not by bar height. For the deeper contrast see Mean and mode of grouped data. (If we shifted $f$ to $5, 25, 20, 15, 5$ , the modal class would be $10 - 20$ while CF $= 5, 30, 50, \dots$ still puts the median class at $20 - 30$ — different classes.)

Active Recall

Recall One-line self-tests

In Exercise 3.1, why was the flat CF stretch ignored? ::: It's a zero-frequency class; the ogive is horizontal and never reaches $N /2$ .
What $h$ do you use in Exercise 3.3? ::: The median class's own width, $20$ (not $10$ ).
What target height finds $Q_{1}$ ? ::: $N /4$ .
In 5.1, what were $x$ and $y$ ? ::: $x = 13, y = 17$ .
Why is straight-line interpolation across a class even valid? ::: The uniform-spread assumption makes the ogive a straight segment there.