Run a cumulative total down the frequency column, each row adding on to the one before. The last cumulative frequency must equal the total number of data points — always check this before plotting.

Always at the upper class boundary — a cumulative frequency of 18 means "18 values are at most 2 kg," so it belongs at $x=2$, never at the midpoint. Discrete data grouped as "$10$–$19$", "$20$–$29$" has a genuine gap, so the true upper boundary sits halfway across it, at $19.5$.

Plot one extra "anchor" point at (lowest lower bound, $0$) — before any data exists, cumulative frequency is zero — then join every point with a smooth curve (the ogive), not straight lines.

Don't use the discrete-list formula $\frac{n+1}{2}$ here — a cumulative frequency graph models a continuous spread, so quartile positions on the y-axis are $\frac{N}{4}$, $\frac{N}{2}$, $\frac{3N}{4}$ for $Q_1$, median, $Q_3$. For each: find that value on the y-axis, draw across to the curve, then down to the x-axis — and show these construction lines, since they often carry their own marks.

$N=120$: median position $=60$th, $Q_1$ position $=30$th, $Q_3$ position $=90$th.

"Fewer than" a value is a direct read: go up from that x-value to the curve, across to the y-axis.

"More than" a value needs one extra step — the curve only ever shows totals up to a point, so subtract from $N$: if 77 of 80 people waited up to 70 minutes, then $80-77=3$ waited more.

"Between" two values is a subtraction of two direct reads: $CF(\text{high}) - CF(\text{low})$. E.g. between 50 and 70 minutes $=77-49=28$ people.

A "bottom 30%" position is simply $0.30\times N$ on the y-axis, read across then down as usual. A "top X%" question is the trap: since the graph only counts "up to" a value, the top 10% corresponds to the 90th percentile, not the 10th.

"Top 10% of insects are 'large' — find the minimum mass", $N=80$: position $=0.90\times80=72$ on the y-axis; read across to the curve, down to the x-axis (e.g. 47 g). Any insect $\ge47$ g is "large."

A cumulative frequency graph shows the time (minutes) for 120 students to finish a puzzle. Estimate the median and the interquartile range.

1. $N=120$; median position $=60$, $Q_1$ position $=30$, $Q_3$ position $=90$.
2. Reading across and down the curve: median $\approx42$ min, $Q_1\approx35$ min, $Q_3\approx51$ min.
3. $\text{IQR} = Q_3-Q_1 = 51-35 = 16$ minutes.

The IQR describes the spread of the middle 50% of students and, unlike the full range, ignores any extreme stragglers — always state what it represents if asked to interpret it, not just calculate it.