Ogive (graphical representation)

An ogive is built from a cumulative frequency table, and there are exactly two kinds, distinguished by which class boundary is plotted and which direction the curve moves.

• Less-than ogive. Plot the upper class limit of each interval on the $x$-axis against the cumulative frequency of observations less than that limit on the $y$-axis. Successive cumulative frequencies only grow, so the curve rises monotonically from left to right, starting at $(\text{lower limit of first class},\,0)$.
• More-than ogive. Plot the lower class limit of each interval against the cumulative frequency of observations greater than or equal to that limit. This curve falls monotonically from left to right, starting near the total frequency $n$ at the left edge.

Both curves are built from the same raw frequency table; they are simply two different bookkeeping directions through it. Drawn on the same axes, they necessarily cross once, because one is rising from near zero and the other is falling from near $n$.

The value of $n/2$ marked on the $y$-axis of the less-than ogive, carried across to the curve and then down to the $x$-axis, lands at the same point as the intersection of the two curves — both are graphical readings of the median. Algebraically that reading is computed by the cumulative-frequency formula:

where: $l$ is the lower boundary of the median class (the class in whose cumulative frequency $n/2$ first falls), $n$ is the total number of observations, $cf$ is the cumulative frequency before the median class, $f$ is the frequency of the median class, and $h$ is the class width. Notice that only the less-than table is needed for the formula — the more-than ogive is not required for calculation, but it is useful as a visual cross-check: since both ogives are built from the same data, their curves must cross inside the same median class, at (approximately) the same $x$-value the formula produces. If they crossed anywhere else, it would be a sign of a bookkeeping error in one of the two cumulative frequency tables.

The median class is, by definition, the class interval that contains the $\frac{n}{2}$-th observation when all observations are arranged in order. Inside that class we don't know exactly where the individual values sit, so we make the standard assumption used throughout grouped-data statistics: the observations are spread uniformly across the class width $h$.

Step 1. Before the median class, $cf$ observations have already been counted. We still need $\frac{n}{2}-cf$ more observations to reach the halfway mark, and these must come from inside the median class, which itself contains $f$ observations spread evenly across width $h$.

Step 2. Under the uniform-spread assumption, each unit of frequency inside the median class corresponds to a slice of width $h/f$ along the number line. Advancing $\frac{n}{2}-cf$ units of frequency from the class's lower boundary $l$ therefore means advancing a distance of $\left(\frac{n/2-cf}{f}\right)\times h$ along the $x$-axis, which is exactly the correction term added to $l$ in the formula.

Step 3. This is precisely what the ogive intersection (or the single-ogive-plus-horizontal-line construction) reads off visually: the curve's rise from $cf$ to $n/2$ within the median class is a straight-line approximation of that same uniform spread, and the $x$-value where the curve reaches height $n/2$ is the graphical median — numerically identical, in principle, to the formula answer.

Consider the grouped frequency distribution used in the "Median of grouped data" article, so the two methods can be compared directly.

Step 1. Total observations: $n = 5+8+15+16+6 = 50$, so $\frac{n}{2} = 25$.

Step 2. Build the less-than cumulative frequency table by running totals: $10\to5$, $20\to13$, $30\to28$, $40\to44$, $50\to50$. Since $25$ first appears (is exceeded) inside the class $20$–$30$ (cf jumps from $13$ to $28$), this is the median class: $l=20$, $cf=13$, $f=15$, $h=10$.

Step 3. Apply the formula: $\text{Median} = 20 + \left(\dfrac{25-13}{15}\right)\times 10 = 20 + \dfrac{12}{15}\times 10 = 20+8 = \mathbf{28}$.

Step 4 — the graphical check. Plotting the less-than ogive points $(10,5),(20,13),(30,28),(40,44),(50,50)$ and drawing a horizontal line from cumulative frequency $25$ on the $y$-axis to the curve, then dropping a vertical to the $x$-axis, lands close to $x=28$ — the same class $20$–$30$, the same neighbourhood as the formula answer, confirming the graphical and numerical methods agree.

Competitive stretch

If instead you were given only the more-than cumulative frequencies — $50,45,37,22,6$ at boundaries $0,10,20,30,40$ — you would still identify the median class as $20$–$30$ (where the more-than count drops through $25$, going from $37$ to $22$), and the same formula machinery applies once you convert back to the ordinary (less-than-style) $l$, $cf$, $f$, $h$ for that class. Concretely, the more-than value at boundary $20$ is $37$, and the more-than value at boundary $30$ is $22$; the difference $37-22=15$ recovers the class frequency $f=15$ for the interval $20$–$30$, exactly matching the original table — a useful consistency check when only one of the two ogive tables has been supplied.

• Assuming ogive construction is examinable in the current CBSE Class 10 syllabus — it is not; spend exam-prep time on the formula method instead.
• Swapping less-than and more-than plotting rules: less-than uses upper limits and rises, more-than uses lower limits and falls.
• Forgetting the less-than ogive must start at $(\text{lower limit of first class},\,0)$, not at the first plotted cumulative-frequency point.
• Treating the graphical median reading as more exact than the formula answer — both rest on the same uniform-spread assumption inside the median class, so neither is more "true" than the other.

Ask the tutor: Want to see how this same uniform-spread idea also underlies the mode formula for grouped data?