Statistics • Topic 3 of 3

Histograms, Frequency Polygons, and Ogives

What are histograms, frequency polygons, and ogives? These are graphical representations of frequency distributions that let you see the shape of the data and, crucially for Class 10, read the median straight off a graph.

1. Histogram

A histogram is a bar graph for continuous grouped data. Its features: bars are drawn without gaps (they touch, because the classes are continuous); the width of each bar equals the class width; and the height equals the frequency. The x-axis shows class boundaries and the y-axis shows frequency. A histogram differs from an ordinary bar graph precisely in this no-gap property — bar graphs are for categorical data and leave spaces between bars. If class widths are unequal, you must plot frequency density (frequency ÷ width) as the height so that area, not height, represents frequency.

2. Frequency polygon

A frequency polygon is a line graph obtained by plotting each frequency against its class mark (midpoint) and joining the points with straight segments. To "close" the polygon, add a point with zero frequency at the midpoint of an imagined class just before the first and just after the last. You can draw it on top of a histogram (join the midpoints of the bar tops) or directly from the table. Its handy property: the area under a frequency polygon equals the area of the corresponding histogram, so it conveys the same total frequency while making it easy to compare two distributions on one set of axes.

3. Ogive (cumulative frequency curve)

An ogive is a smooth curve drawn from cumulative frequencies. Two types:

  • Less-than ogive: plot the less-than cumulative frequency against the upper class limit, then join with a free-hand smooth curve. It rises from left to right.
  • More-than ogive: plot the more-than cumulative frequency against the lower class limit, then join smoothly. It falls from left to right.

Reading the median from an ogive

There are two standard graphical routes, both examinable:

  • Two-ogive method: draw both ogives on the same axes. They intersect at one point; drop a perpendicular from that point to the x-axis. The foot of the perpendicular is the median.
  • Single-ogive method: on a less-than ogive, locate $\dfrac{N}{2}$ on the y-axis, move horizontally to the curve, then drop a perpendicular to the x-axis. Where it lands is the median.

Both methods give the same value because the point where cumulative frequency equals $\dfrac{N}{2}$ is, by definition, the median. The graphical median should agree with the formula $\text{Median}=l+\left(\dfrac{\frac{N}{2}-cf}{f}\right)h$ to within the accuracy of your drawing — checking one against the other is a good exam habit.

Common mistakes to avoid. (1) Leaving gaps between histogram bars (only bar graphs have gaps). (2) Plotting a less-than ogive against lower limits — it must be the upper limits. (3) Forgetting the two zero-frequency end points that close a frequency polygon. (4) Reading the median off the y-axis instead of the x-axis — the median is always an x-value.

Histogram with touching bars and a frequency polygon joining class-mark pointsHistogram & frequency polygonFrequency f01020304050bars touch (continuous data); the red polygon joins the class marksOgive with a dashed line from N over 2 across to the curve and down to read the medianReading the median from the ogivecumulative f5152535N/2 = 17.501020304050Median ≈ 23.75
Solved Examples
1
Worked Example
Draw a histogram for the data: Class $0$–$10\ (f=4)$, $10$–$20\ (f=6)$, $20$–$30\ (f=10)$, $30$–$40\ (f=5)$.
Solution
  1. Step 1: Mark class boundaries $0,10,20,30,40$ on the x-axis and frequency on the y-axis.
  2. Step 2: Each bar has width $10$ (the class width).
  3. Step 3: Heights equal the frequencies: $4,6,10,5$.
  4. Step 4: Draw the bars touching each other (no gaps), since the data is continuous.

Answer: A histogram with touching bars of heights $4,6,10,5$ over the four classes.

2
Worked Example
From the following less-than ogive points, estimate the median: $(10,4),(20,10),(30,22),(40,30),(50,35)$.
Solution
  1. Step 1: Total frequency $N=35$ (the last cf).
  2. Step 2: $\dfrac{N}{2}=17.5$.
  3. Step 3: On the curve, $17.5$ lies between $(20,10)$ and $(30,22)$.
  4. Step 4: Interpolate: $\text{Median}=20+\left(\dfrac{17.5-10}{22-10}\right)10$.
  5. Step 5: $=20+\dfrac{7.5}{12}\times10=20+6.25=26.25$.

Answer: Median $\approx 26.25$.

3
Worked Example
For the data in Example 1, list the class marks needed to draw the frequency polygon, including the two end points that close it.
Solution
  1. Step 1: Class marks of the four classes: $5,15,25,35$.
  2. Step 2: Add a class before the first ($-10$ to $0$) with class mark $-5$ and frequency $0$.
  3. Step 3: Add a class after the last ($40$ to $50$) with class mark $45$ and frequency $0$.
  4. Step 4: Plot $(-5,0),(5,4),(15,6),(25,10),(35,5),(45,0)$ and join with straight lines.

Answer: Points: $(-5,0),(5,4),(15,6),(25,10),(35,5),(45,0)$.

4
Worked Example
Draw a less-than ogive for the data: $0$–$10\ (f=5)$, $10$–$20\ (f=8)$, $20$–$30\ (f=12)$, $30$–$40\ (f=7)$, $40$–$50\ (f=3)$. List the points.
Solution
  1. Step 1: Less-than cumulative frequencies: $5,13,25,32,35$.
  2. Step 2: Pair each with the upper class limit: $10,20,30,40,50$.
  3. Step 3: Plot $(10,5),(20,13),(30,25),(40,32),(50,35)$.
  4. Step 4: Join with a smooth rising curve — the less-than ogive.

Answer: Points: $(10,5),(20,13),(30,25),(40,32),(50,35)$.

5
Worked Example
For the same data as Example 4, list the points of the more-than ogive.
Solution
  1. Step 1: More-than cumulative frequencies: $35,30,22,10,3$.
  2. Step 2: Pair each with the lower class limit: $0,10,20,30,40$.
  3. Step 3: Plot $(0,35),(10,30),(20,22),(30,10),(40,3)$.
  4. Step 4: Join with a smooth falling curve — the more-than ogive.

Answer: Points: $(0,35),(10,30),(20,22),(30,10),(40,3)$.

6
Worked Example
Using the two ogives from Examples 4 and 5, explain how to read the median and state $\dfrac{N}{2}$.
Solution
  1. Step 1: $N=35$, so $\dfrac{N}{2}=17.5$.
  2. Step 2: Draw both ogives on the same axes.
  3. Step 3: They intersect at a single point (here at cumulative frequency $17.5$).
  4. Step 4: Drop a perpendicular from the intersection to the x-axis; its foot is the median (about $25.4$ for this data).

Answer: $\dfrac{N}{2}=17.5$; the x-coordinate of the ogives' intersection (≈$25.4$) is the median.

7
Worked Example
Verify the graphical median from Example 6 against the formula.
Solution
  1. Step 1: $N=35$, $\dfrac{N}{2}=17.5$; less-than cf $5,13,25,32,35$.
  2. Step 2: First $cf\ge17.5$ is $25$, so median class is $20$–$30$.
  3. Step 3: $l=20$, $cf=13$, $f=12$, $h=10$.
  4. Step 4: $\text{Median}=20+\left(\dfrac{17.5-13}{12}\right)10=20+\dfrac{4.5}{12}\times10=20+3.75=23.75$.
  5. Step 5: This is close to the graphical reading, confirming both methods agree within drawing accuracy.

Answer: Median $=23.75$ by formula, consistent with the ogive reading.

8
Worked Example
Using the single-ogive method on the less-than ogive of Example 4, describe how to locate the median on the graph.
Solution
  1. Step 1: Compute $\dfrac{N}{2}=\dfrac{35}{2}=17.5$ and mark it on the y-axis.
  2. Step 2: Draw a horizontal line from $17.5$ until it meets the ogive.
  3. Step 3: From that meeting point, drop a vertical perpendicular to the x-axis.
  4. Step 4: The x-value at the foot of the perpendicular is the median (≈$23.75$).

Answer: Median ≈ $23.75$ read from $\dfrac{N}{2}=17.5$ on the single ogive.

9
Worked Example
State two differences between a histogram and a bar graph.
Solution
  1. Step 1: A histogram represents continuous grouped data; a bar graph represents discrete/categorical data.
  2. Step 2: Histogram bars touch (no gaps) because the classes are continuous; bar-graph bars have gaps between them.
  3. Step 3: In a histogram, area (width × height) can represent frequency; in a bar graph only the height matters.

Answer: Histogram: continuous data, bars touch, area meaningful. Bar graph: categorical data, gaps between bars, only height meaningful.

10
Worked Example
The frequency polygon for a distribution has vertices at midpoints $(5,2),(15,5),(25,8),(35,4),(45,1)$. Find the total frequency and the modal class.
Solution
  1. Step 1: Total frequency $=$ sum of all the frequencies $=2+5+8+4+1$.
  2. Step 2: $=20$.
  3. Step 3: The highest frequency is $8$, at midpoint $25$.
  4. Step 4: Midpoint $25$ corresponds to the class $20$–$30$, so that is the modal class.

Answer: Total frequency $N=20$; modal class $20$–$30$.

11
Worked Example
A less-than ogive passes through $(20,6),(40,18),(60,40),(80,52),(100,60)$. Use $\dfrac{N}{2}$ to estimate the median.
Solution
  1. Step 1: $N=60$ (the last cf), so $\dfrac{N}{2}=30$.
  2. Step 2: $30$ lies between $(40,18)$ and $(60,40)$, so the median class is $40$–$60$.
  3. Step 3: $l=40$, $cf=18$, $f=40-18=22$, $h=20$.
  4. Step 4: $\text{Median}=40+\left(\dfrac{30-18}{22}\right)20=40+\dfrac{12}{22}\times20$.
  5. Step 5: $=40+10.91=50.91$.

Answer: Median $\approx 50.91$.

12
Worked Example
Why do the two ogives of a distribution always intersect, and what does the intersection's height tell you?
Solution
  1. Step 1: The less-than ogive rises from $0$ to $N$; the more-than ogive falls from $N$ to $0$.
  2. Step 2: A continuously rising curve and a continuously falling curve over the same range must cross exactly once.
  3. Step 3: At the crossing, the less-than cf equals the more-than cf.
  4. Step 4: Since (less-than cf) + (more-than cf) $=N$, each equals $\dfrac{N}{2}$ at the crossing — its height is $\dfrac{N}{2}$ and its x-coordinate is the median.

Answer: They cross once; at the crossing both cumulative frequencies equal $\dfrac{N}{2}$, and the x-coordinate there is the median.

Key Points

  • Histogram: Touching bars for continuous data; area of bar represents frequency
  • Frequency polygon: Points at midpoints joined by lines; starts and ends at zero frequency
  • Less than ogive: Upper limits vs less than cumulative frequency
  • Greater than ogive: Lower limits vs greater than cumulative frequency
  • Median can be found at intersection of the two ogives
  • Ogives are useful for finding percentiles and quartiles
  • Histograms show distribution shape (symmetric, skewed left, skewed right)
  • Less-than ogive uses UPPER limits and rises; more-than ogive uses LOWER limits and falls
  • Two-ogive method: the x-coordinate of the intersection is the median; single-ogive method: read across from $\dfrac{N}{2}$ then drop to the x-axis
  • Frequency polygon: plot frequency vs class mark and add two zero-frequency end points to close it
  • Graphical median should match $\text{Median}=l+\left(\dfrac{\frac{N}{2}-cf}{f}\right)h$ within drawing accuracy
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.A histogram represents grouped data using:
Explanation: Adjacent bars.
Q2.An ogive is a:
Explanation: Cumulative frequency curve.
Q3.A frequency polygon is drawn by joining the:
Explanation: Midpoints of class tops.
Q4.In a histogram, the bars are drawn:
Explanation: No gaps (continuous data).
Practice more MCQs → 📝 Assignment · 35 marks