Financial Management and Business Data Analytics · Data Presentation: Visualisation and Graphical Presentation

Histogram, Frequency Polygon and Ogive Explained

Updated 10 October 2026 · Fact-checked

A histogram shows a continuous frequency distribution as touching bars. A frequency polygon joins the points above class marks. An ogive plots cumulative frequencies against class limits. To solve: make classes continuous, find class marks or cumulative totals, plot the points, join them, and read values such as the median at N/2.

Understand Histogram, Frequency Polygon and Ogive

A frequency distribution groups raw data into classes and counts how many items fall in each. A table is accurate but hard to read quickly. Graphs let you see the shape of the data at a glance: where it is concentrated, whether it is symmetrical, and how it is spread.

A histogram is used for a continuous variable such as marks, sales or salary. Each class is a vertical bar. The class limits go on the X-axis and the frequency on the Y-axis. The bars touch each other because the classes are continuous. The area of each bar is proportional to the class frequency. With equal class widths, the height is also proportional to the frequency.

A frequency polygon is drawn by plotting each frequency against the class mark (mid-point) of its class and joining the points with straight lines. The polygon is closed by adding one empty class at each end, with frequency zero, so the line touches the X-axis. You can draw it on top of a histogram by joining the mid-points of the tops of the bars. A frequency curve is the same polygon smoothed by hand into a curve.

An ogive (cumulative frequency curve) shows running totals. A less than ogive plots cumulative frequency against the upper limit of each class and rises from left to right. A more than ogive plots cumulative frequency against the lower limit of each class and falls from left to right. Both are drawn on the same axes, and the point where they cross gives the median.

A histogram is not a bar chart. A bar chart compares separate categories (such as sales by product), has gaps between the bars, and the width of the bars carries no meaning. A histogram shows a continuous scale, has no gaps, and bar area carries meaning.

Key rules to remember

Class mark (mid-point): Class mark = (Lower limit + Upper limit) ÷ 2; Used as the X-value for each point of a frequency polygon.
Adjustment for inclusive classes: Adjustment factor = (Lower limit of next class − Upper limit of this class) ÷ 2; Subtract it from every lower limit and add it to every upper limit to get true (exclusive) limits. For classes 10–19, 20–29 the factor is 0.5.
Class width: Class width = Upper limit − Lower limit (true limits); Check this before drawing a histogram. If widths are unequal, adjust the heights.
Frequency density (unequal widths): Frequency density = Class frequency ÷ Class width; Plot this as the bar height when class widths are unequal, so that area stays proportional to frequency.
Cumulative frequency: Less than: running total from the first class. More than: running total from the last class, upwards.; The last less than total and the first more than total both equal N, the total frequency.
Median from an ogive: Median = X-value at which cumulative frequency = N ÷ 2; Mark N ÷ 2 on the Y-axis, go across to the curve, then drop down to the X-axis. The same value comes from the intersection of the two ogives.
Median by interpolation (check): Median = L + [(N ÷ 2 − c) ÷ f] × h; L = lower limit of median class, c = cumulative frequency before it, f = its frequency, h = its width. Use it to verify your graph reading.

How to solve Histogram, Frequency Polygon and Ogive questions

Use this method for any question that asks you to draw or read a histogram, frequency polygon or ogive.

1Read what is asked: histogram, polygon, curve, less than ogive, more than ogive, or a value read from the graph (median, quartile, number of items).
2Check whether the classes are continuous (0–10, 10–20) or inclusive (10–19, 20–29). If inclusive, convert to true limits using the adjustment factor.
3Check whether class widths are equal. If not, compute frequency density for the histogram.
4Prepare a small table: true class limits, class mark, frequency, and less than and more than cumulative frequencies as required.
5Choose a scale with a clear title and labels. Mark the X-axis with class limits and the Y-axis with frequency or cumulative frequency. If the X-axis starts above zero, show a break.
6Plot: bars for a histogram, class marks for a polygon (add an empty class at each end), upper limits for a less than ogive, lower limits for a more than ogive. Join the points.
7To read a value, draw dotted lines from the Y-axis (for example N ÷ 2) to the curve and down to the X-axis. Write the value and the reading method.
8Verify any median or quartile with the interpolation formula, and state your final answer with units.

Quickest way: Table first, then plot with a pencil

When to use it: Use this in the exam when time is short and the question asks for a graph and a value read from it.

Write one compact table with true limits, frequency, class mark and both cumulative columns. Almost every point you need comes from this table.
Compute the median by formula first. This gives you the answer you should see on the graph and protects your marks if the plot is slightly off.
Plot the less than ogive using upper limits and the more than ogive using lower limits. Their crossing point should be at the median value and at a cumulative frequency of N ÷ 2.
Draw dotted guide lines to the axes, write the median value there, and label each curve.
For a polygon, plot the class marks, add the two empty end classes, and join with straight lines.

Common mistakes in Histogram, Frequency Polygon and Ogive

Leaving gaps between bars in a histogram or using inclusive limits as they are.
Students treat the histogram like a bar chart and forget that the classes must be continuous.
Fix: Convert inclusive classes to true limits first (9.5–19.5, 19.5–29.5). Draw the bars touching each other.
Plotting a less than ogive against class mark or lower limits.
Students mix up which limit goes with which ogive.
Fix: Less than uses the upper limit. More than uses the lower limit. Remember that a cumulative total is complete only at the end of the class.
Not closing the frequency polygon at both ends.
Students stop at the first and last class marks.
Fix: Add one imaginary class with frequency zero before the first class and after the last class. Join the polygon to the X-axis at their class marks.
Taking frequency as bar height when class widths are unequal.
The rule of equal widths is used without checking.
Fix: Compute frequency density (frequency ÷ class width) and plot that as the height. If asked, state that area, not height, represents frequency.
Reading the median at a cumulative frequency of N instead of N ÷ 2.
Students look for the top of the curve instead of the middle.
Fix: Mark N ÷ 2 on the Y-axis first. For N = 80 the median is read at 40.
Plotting the less than curve with the wrong cumulative totals.
Students add frequencies carelessly and the last total does not equal N.
Fix: Always check that the last less than total equals N and the first more than total equals N.

Worked examples

Example 1

The marks of 50 students are: 10–19: 4; 20–29: 8; 30–39: 15; 40–49: 12; 50–59: 7; 60–69: 4. Prepare the data needed to draw (a) a histogram and (b) a frequency polygon.

Show the solution

The classes are inclusive. Adjustment factor = (20 − 19) ÷ 2 = 0.5.
Subtract 0.5 from each lower limit and add 0.5 to each upper limit. True classes: 9.5–19.5, 19.5–29.5, 29.5–39.5, 39.5–49.5, 49.5–59.5, 59.5–69.5.
Check the total: 4 + 8 + 15 + 12 + 7 + 4 = 50.
Class width is 10 for every class, so bar height = frequency.
Histogram: draw touching bars over the true classes with heights 4, 8, 15, 12, 7 and 4.
Class marks: (9.5 + 19.5) ÷ 2 = 14.5, then 24.5, 34.5, 44.5, 54.5, 64.5.
Polygon: plot (14.5, 4), (24.5, 8), (34.5, 15), (44.5, 12), (54.5, 7), (64.5, 4).
Close the polygon by adding the empty classes 0 to 9.5 and 69.5 to 79.5, with class marks 4.5 and 74.5 and frequency 0. Plot (4.5, 0) and (74.5, 0) and join.

Answer: Histogram bars: true classes 9.5–69.5 with heights 4, 8, 15, 12, 7, 4. Polygon points: (4.5, 0), (14.5, 4), (24.5, 8), (34.5, 15), (44.5, 12), (54.5, 7), (64.5, 4), (74.5, 0).

Example 2

The daily sales of a shop over 80 days (in ₹ thousand) are: 0–10: 6; 10–20: 14; 20–30: 25; 30–40: 20; 40–50: 10; 50–60: 5. Draw the less than and more than ogives and find the median graphically. Verify by calculation and estimate the number of days with sales below ₹35 thousand.

Show the solution

N = 6 + 14 + 25 + 20 + 10 + 5 = 80, so N ÷ 2 = 40.
Less than cumulative frequencies: below 10 = 6; below 20 = 20; below 30 = 45; below 40 = 65; below 50 = 75; below 60 = 80. The last value equals N.
More than cumulative frequencies: 0 and above = 80; 10 and above = 74; 20 and above = 60; 30 and above = 35; 40 and above = 15; 50 and above = 5; 60 and above = 0.
Plot the less than points (10, 6), (20, 20), (30, 45), (40, 65), (50, 75), (60, 80). Start the curve at (0, 0).
Plot the more than points (0, 80), (10, 74), (20, 60), (30, 35), (40, 15), (50, 5), (60, 0). Join each set of points smoothly.
Draw a line from 40 on the Y-axis to the curves. The two ogives cross at cumulative frequency 40. Drop a perpendicular to the X-axis to read the median.
Verify: the median class is 20–30 because the cumulative frequency reaches 45 there. Median = 20 + [(40 − 20) ÷ 25] × 10 = 20 + 8 = 28.
For sales below 35: on the less than ogive, between 30 (45) and 40 (65) the curve rises 20 over 10 units. At 35 the value is 45 + 10 = 55 days (by straight-line interpolation).

Answer: Median daily sales = ₹28 thousand, found where the two ogives cross at cumulative frequency 40. About 55 days had sales below ₹35 thousand.

Exam tips

In MCQs, watch for the keywords: continuous data and no gaps means histogram; separate categories means bar chart; running totals mean ogive. There is no negative marking, so always attempt every MCQ.
For a written answer, show the adjusted class limits, class marks and cumulative columns in a table. These earn step marks even if your drawing is rough.
Always label the axes, give the graph a title and state the scale. State the answer read from the graph in a final line, with units.
Do the interpolation check for the median. If the graph reading and the formula differ slightly, write the formula value and mention that the graph is approximate.
Remember the pairing: less than goes with upper limits, more than goes with lower limits, and the intersection of the two gives the median.

Practice questions from Data Presentation: Visualisation and Graphical Presentation

Histogram, Frequency Polygon and Ogive in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Histogram, Frequency Polygon and Ogive: frequently asked questions

What is the difference between a histogram and a bar chart?

A histogram shows a continuous variable using touching bars, and the bar area represents the frequency. A bar chart compares separate categories with gaps between the bars, and only the height matters. Use a histogram for grouped data such as marks or income classes.

How do you draw a less than and more than ogive?

Prepare cumulative frequency columns. For a less than ogive, plot cumulative frequency against the upper class limits. For a more than ogive, plot it against the lower class limits. Join the points in a smooth curve. The two curves cross at the median.

How do you find the median from an ogive graph?

Calculate N ÷ 2 and mark it on the Y-axis. Draw a horizontal line to the less than ogive and then a vertical line down to the X-axis. The value on the X-axis is the median. Quartiles are read the same way, at N ÷ 4 and 3N ÷ 4.

Why do we add an empty class at each end of a frequency polygon?

It brings the polygon down to the X-axis at both ends, so the figure is closed. The empty classes have the same width as the other classes. The area under the closed polygon then equals the area of the histogram. Without these empty classes the polygon is left open.