Question bank — Cumulative frequency — ogive, median from graph

True — the last class accumulates every observation, so its running total is the whole population $N$. This is your standard sanity check on any CF table.

False — cumulative frequency is a running total of non-negative counts, so it can only stay flat or rise. A downward segment would mean a negative frequency, which is impossible.

True — it counts "how many are at or above this boundary." As the boundary moves right, fewer observations remain above it, so the count falls from $N$ down to $0$.

True — $C{F}_i$ counts everything below the upper boundary of class $i$, so the honest point is (upper boundary, $C{F}_i$). Using the midpoint would shift the whole curve left.

True — the more-than $CF$ of a class counts everything from its lower boundary onward, so the honest point is (lower boundary, more-than $CF$). Mirror image of the less-than rule, which uses upper boundaries.

False — that formula is for an ordered ungrouped list. A grouped ogive is treated as a smooth curve, so we cut it at height $N/2$ (halfway up), not $(N+1)/2$.

False — it's the class where the cumulative frequency first crosses $N/2$. The tallest bar can sit far from that crossing; see mode for where the tallest bar does matter.

True — at that point "number below" (less-than curve) equals "number above" (more-than curve), and both equal $N/2$ only at the median. Dropping a perpendicular gives the median's x-value (see figure s02).

False — the median simply sits at that boundary. The interpolation term just becomes zero, so $\text{Median}=L$ exactly (as in Example 1, where it landed on 30).

True — before the lower boundary of the first class, no observations have been counted yet, so the curve begins at $y=0$ at that lower boundary.

False — an ogive is a cumulative count curve, not a density. Nothing meaningful lives in the area under it; the useful information is the height at each x.

Wrong — the median class is the one inside which CF crosses $N/2$. CF rose from 13 (below 20) to 25 within 20–30, so the median class is 20–30, not the next one.

Wrong — these are discontinuous classes. Convert to continuous boundaries first (9.5–19.5, 19.5–29.5), so $L=19.5$. Skipping this shifts every boundary by 0.5.

Wrong for an ogive — CF counts everything below the upper boundary, so points must sit at upper boundaries. Midpoints belong to the frequency polygon, not the cumulative curve.

Wrong — $C{F}_b$ is the CF of the class before the median class (everything already counted at $L$). Using the median class's own CF double-counts and overshoots.

Wrong — $y=N$ is the very top of the curve (all data below it). The median cuts the data in half, so climb only to $y=N/2$ before sliding across (figure s01).

Wrong — they cross at height $N/2$, where "below" and "above" counts are equal. At $N$ the less-than curve has topped out but the more-than curve has hit zero.

Wrong — $f$ is the frequency of the median class only. It's the count spread across width $h$, giving the $f/h$ per-unit slope used in the interpolation.

Wrong — with $f=0$ the formula divides by zero and the ogive is flat there, so no crossing happens inside that class. A class with $f=0$ can never be the median class; the crossing must occur where the curve actually rises.

Wrong in general — an unbounded class has no genuine lower boundary. Assign a reasonable finite boundary using the common class width (e.g. treat '<20' as 10–20 if width is 10), or the interpolation base $L$ is arbitrary and the result unreliable.

Because $C{F}_i$ is the tally of everything strictly below the upper boundary of class $i$. The point (upper boundary, $C{F}_i$) is then a truthful "this many values lie below this x."

$C{F}_b$ observations were already counted when we reached $L$; we still need $N/2-C{F}_b$ more to reach the halfway mark. That leftover count is what we interpolate across the median class (figure s03).

The median class holds $f$ observations spread uniformly over its width $h$. So the fraction of the class we must enter is (count still needed)$/f$, giving the position within that class — a linear-interpolation step.

The less-than curve gives "how many below x" and the more-than gives "how many above x." These are equal only when both equal $N/2$, which by definition is the median.

We only know the total count per class, not where each value sits. The fairest, least-biased guess is even spacing, which makes the ogive a straight line across the class and lets us interpolate. This mirrors how the median is defined for grouped data.

The ogive models grouped data as a continuous quantity, so its output is a position on that smooth curve — a representative middle value, not necessarily an observed data point.

A quartile is just a different height: $N/4$ for $Q_1$, $3N/4$ for $Q_3$. You slide across at that height instead of $N/2$; the median is simply the $N/2$ (or $Q_2$) case.

Because the less-than $CF$ can only cross $N/2$ where it actually rises, and it rises only where the class frequency is positive. A flat ($f=0$) stretch is skipped over, never landed in.

The ogive becomes a straight line from $0$ to $N$, so its median is exactly the midpoint of the data range — the interpolation still runs, it just lands centrally.

Then $N/2=C{F}_b+f$ or $N/2=C{F}_b$, making the interpolation term hit $0$ or a full $h$, so the median lands on a boundary — no interpolation needed, as in Example 1.

The first class itself. Then $C{F}_b=0$ and $L$ is the first class's lower boundary; the formula still applies with those substitutions.

Yes — a class with frequency $0$ adds nothing to the running total, so the curve stays horizontal across that class's width before rising again. Such a class can never be the median class.

Give it a finite boundary using the shared class width — treat "<20" as one class-width wide (e.g. 10–20) and ">80" as 80–90 — before plotting or reading $L$ and $h$. Without a defined boundary the curve has no honest starting/ending x.

Usually yes — the median normally sits well inside the data, so an unbounded top class rarely touches $N/2$. You only need a finite boundary for the class that actually contains the crossing.

Only one can contain the crossing of $N/2$. Find where CF goes from below $N/2$ to at or above it; that single class is the median class, and the tie is resolved by the boundary rule above.

They generally differ. For a symmetric distribution all three coincide; for a skewed one the ogive median sits between the mean and mode, being resistant to extreme values.