Ogive and graphical median

An ogive (pronounced "oh-jive") is a graph of cumulative frequencies. It is a smooth curve drawn through plotted points of cumulative frequency against class boundaries. The shape is roughly S-shaped, rising from $(0, 0)$ to $(\text{last upper limit}, n)$.

Two kinds of ogives exist, depending on which cumulative frequencies you plot:

• Less-than ogive: Plot $cf$ against the upper limit of each class. The curve rises from left to right.
• More-than ogive: Plot the more-than cumulative frequency against the lower limit of each class. The curve falls from left to right.

If both curves are drawn on the same axes, they intersect , and the $x$-coordinate of their intersection is the median.

Drawing a less-than ogive

Suppose we have

Plot the points $(10, 4), (20, 10), (30, 20), (40, 25), (50, 28)$ on graph paper. Optionally include $(0, 0)$ at the left. Join the points with a smooth curve (or, for the board exam, with straight line segments , both are acceptable).

The horizontal axis is "Marks" (or whatever variable is being measured); the vertical axis is "Cumulative frequency".

Drawing a more-than ogive

For the same data, the more-than cumulative frequencies are computed from the bottom up:

Plot $(0, 28), (10, 24), (20, 18), (30, 8), (40, 3)$ and (optionally) $(50, 0)$. Join with a smooth curve.

Reading the median

The median is the $x$-coordinate of the point where the two ogives meet. Equivalently, draw a horizontal line from $y = n/2$ on the vertical axis; where it meets the less-than ogive, drop a perpendicular to the $x$-axis. The $x$-coordinate is the median.

For the example: $n/2 = 14$. Drawing $y = 14$ on the less-than ogive (which rises from $(20, 10)$ to $(30, 20)$), we interpolate: $14$ falls between $10$ and $20$, at fraction $(14-10)/(20-10) = 0.4$. So the $x$-coordinate is $20 + 0.4(30 - 20) = 24$.

This matches the formula-derived median of $24$.

Why ogives are useful

Beyond finding the median, ogives let you visualise:

• The spread of the data (steepness of the curve).
• The proportion of data below or above any chosen value.
• The quartiles (Q1 at $n/4$, Q3 at $3n/4$).
• The percentile rank of any observation.

These uses are not directly tested in Class X, but they justify why ogives matter.

Worked examples

Example 1. Draw the less-than ogive for the data: classes $0-10, 10-20, 20-30, 30-40, 40-50$ with frequencies $5, 10, 15, 12, 8$. Find the median.

$cf$: $5, 15, 30, 42, 50$. $n/2 = 25$.

Plot points $(10, 5), (20, 15), (30, 30), (40, 42), (50, 50)$.

For median: $y = 25$ falls between $(20, 15)$ and $(30, 30)$. Interpolate: $(25 - 15)/(30 - 15) = 10/15 = 2/3$ of the way. $x = 20 + (2/3)(10) = 20 + 6.67 = 26.67$.

Formula check: $L = 20, F = 15, f = 15, h = 10$. $\text{Median} = 20 + (25 - 15)/15 \times 10 = 26.67$. Matches.

Example 2. Draw both ogives and find the median for the data: classes $0-20, 20-40, 40-60, 60-80, 80-100$ with frequencies $4, 12, 24, 20, 10$.

$n = 70, n/2 = 35$.

Less-than $cf$: $4, 16, 40, 60, 70$. Points $(20, 4), (40, 16), (60, 40), (80, 60), (100, 70)$.

More-than $cf$: $70, 66, 54, 30, 10$. Points $(0, 70), (20, 66), (40, 54), (60, 30), (80, 10)$.

Median: $y = 35$ on the less-than ogive falls between $(40, 16)$ and $(60, 40)$. Interpolate: $(35 - 16)/(40 - 16) = 19/24$. $x = 40 + (19/24)(20) = 40 + 15.83 = 55.83$.

Formula: $L = 40, F = 16, f = 24, h = 20$. Median $= 40 + (35 - 16)/24 \times 20 = 40 + 380/24 = 40 + 15.83 = 55.83$. Matches.

Example 3. A less-than ogive passes through $(10, 8), (20, 20), (30, 36), (40, 50)$. Find the median.

$n = 50, n/2 = 25$. $y = 25$ falls between $(20, 20)$ and $(30, 36)$. Interpolate: $(25 - 20)/(36 - 20) = 5/16$. $x = 20 + (5/16)(10) = 20 + 3.125 = 23.125$.

Formula: median class is $20-30$, $L = 20, F = 20, f = 36 - 20 = 16, h = 10$. Median $= 20 + (25 - 20)/16 \times 10 = 20 + 50/16 = 23.125$. Matches.

Example 4. Find the modal class from the less-than ogive: cumulative frequencies $10, 15, 30, 38, 45, 50$ for classes $0-10, 10-20, 20-30, 30-40, 40-50, 50-60$.

Class frequencies: $10, 5, 15, 8, 7, 5$. Maximum frequency $15$, in class $20-30$. So modal class is $20-30$.

Try it yourself

1. Draw the less-than ogive for frequencies $5, 10, 15, 8, 7$ in classes of size $10$ starting from $0$.
2. Read the median from your ogive in (1).
3. Draw the more-than ogive for the same data.
4. Find where the two ogives meet.
5. Construct the less-than ogive for: classes $20-30, 30-40, 40-50, 50-60$ with frequencies $5, 12, 8, 5$.
6. Find the median graphically and confirm with the formula.
7. A less-than ogive passes through $(5, 4), (10, 9), (15, 14), (20, 20)$. Find the median.
8. Given less-than $cf$: $<10: 6, <20: 18, <30: 30, <40: 38, <50: 42$. Find the median graphically and by formula.
9. A more-than ogive passes through $(0, 40), (10, 32), (20, 22), (30, 10), (40, 0)$. Find the median.
10. Sketch what a less-than ogive looks like for a symmetric (bell-shaped) distribution.
11. A class of $30$ students. Less-than ogive's coordinates: $(40, 5), (50, 12), (60, 20), (70, 27), (80, 30)$. Find the median.
12. Two ogives (less-than and more-than) intersect at $x = 45$. What does this tell you?

Pitfalls / Insight

(1) For the less-than ogive, plot against the upper limit. For the more-than ogive, plot against the lower limit. Don't mix them up.

(2) On graph paper, use a consistent scale on each axis. Label the axes clearly.

(3) The two ogives' intersection gives the median exactly (assuming linear interpolation), the same as the formula.

(4) Always start the less-than ogive at $(0, 0)$ , or at the first lower limit with cumulative frequency $0$ , and end at $(\text{last upper}, n)$. This makes the curve complete.

(5) For board exam, show the median on the graph with dashed lines to both axes.