Cumulative Frequency Distribution: Cumulative frequency is defined as a running total of frequencies. The frequency of an element in a set refers to how many times that particular element appears in the set. Cumulative frequency is also defined as the sum of all previous frequencies in a class interval before that specific class interval.

An ogive is ahand graph showing the curve of a cumulative frequency distribution. The term ‘ogive’ is derived from the word ‘ogee’, meaning a shape consisting of a concave flow into a convex arc, forming an \({\rm{S}}\)-shaped curve with vertical ends. The cumulative frequency curve (or ogive) is classified into less than ogive and more than ogive. In this article, we will discuss cumulative frequency.

Cumulative Frequency Distribution: Definition

The cumulative frequency will be calculated by adding each frequency from a frequency distribution table to its predecessors. The final value will always be equal to the total for all observations since all frequencies have already been added to the previous total.

1. If the frequency of the first-class interval is added to the frequency of the second class and this sum is added to the third class and so on, then frequencies so obtained are known as cumulative frequency \((c.f.)\).

2. A table that displays how cumulative frequencies are distributed over various classes is called a cumulative frequency distribution or cumulative frequency table.

3. There are two types of cumulative frequencies; they are less than type and more than type.

Cumulative Frequency Distribution: Examples

The cumulative frequency of a class interval is the sum of frequencies of all the classes up to this class interval.

Example:

Class interval \((C.I)\)	Frequency \((f)\)
\(20 – 25\)	\(3\)
\(25 – 30\)	\(6\)
\(30 – 35\)	\(10\)
\(35 – 40\)	\(8\)
\(40 – 45\)	\(4\)

Cumulative frequency table for above-given frequency distribution table:

Class interval \((C.I)\)	Frequency \((f)\)	Cumulative Frequency \((c.f.)\)
\(20 – 25\)	\(3\)	\(3\)
\(25 – 30\)	\(6\)	\(3 + 6 = 9\)
\(30 – 35\)	\(10\)	\(3+6+10=19\)
\(35 – 40\)	\(8\)	\(3+6+10+8=27\)
\(40 – 45\)	\(4\)	\(3+6+10+8+4=31\)

Representing a Cumulative Frequency Distribution: Graphically

In terms of graphics, we can represent a cumulative frequency distribution as either a cumulative frequency curve or an ogive of the less than type and the more than type.

If we plot the points taking the upper limits or lower limits of the class intervals as \(x\) coordinates and their corresponding cumulative frequencies as \(y\) coordinates and then join these points by ahand curve, the curve so obtained is called cumulative frequency curve.

(i) Construct a cumulative frequency table.
(ii) Mark the actual class limits along \({\rm{X}}\)-axis.
(iii) Mark the cumulative frequencies of respective classes along the \({\rm{Y}}\)-axis.
(iv) Plot the points corresponding to cumulative frequency at each upper limit point.
(v) Join the points plotted by ahand curve.

Less Than Type Ogive

We mark the upper-class limits along the \({\rm{X}}\)−axis and the corresponding cumulative frequencies along the \({\rm{Y}}\)−axis on graph paper.

On joining these points successively by smooth curves, we get a curve known as a cumulative frequency curve or an ogive.

More Than Type Ogive

We mark the lower-class limits along the \({\rm{X}}\)−axis and the corresponding cumulative frequencies along the \({\rm{Y}}\)−axis on graph paper.

On joining these points successively by smooth curves, we get a curve known as a cumulative frequency curve or an ogive.

Solved Examples – Cumulative Frequency Distribution

Q.1. Find the median of the following frequency distribution:

Weekly wages (in)	\(60 – 69\)	\(70 – 79\)	\(80 – 89\)	\(90 – 99\)	\(100 – 109\)	\(110 – 119\)
No. of days:	\(5\)	\(15\)	\(20\)	\(30\)	\(20\)	\(8\)

Ans: Here, the classes mentioned in the given frequency distribution table are discontinuous classes. In order to make them continuous classes, we shall subtract \(0.5\) from the lower class limits and add \(0.5\) to the upper-class limits for each class.

Transforming the above into exclusive form and preparing the cumulative frequency table, we get

Weekly wages	No. of workers	Cumulative frequency
\(59.5 – 69.5\)	\(5\)	\(5\)
\(69.5 – 79.5\)	\(15\)	\(20\)
\(79.5 – 89.5\)	\(20\)	\(40\)
\(89.5 – 99.5\)	\(30\)	\(70\)
\(99.5 – 109.5\)	\(20\)	\(90\)
\(1099.5 – 119.5\)	\(8\)	\(98\)
	\(N = \sum {{f_i}} = 98\)

We have, \(N = 98\). \(\therefore \frac{N}{2} = 49\). The cumulative frequency just greater than \(\frac{N}{2}\) is \(70\) and the corresponding class is \(89.5 – 99.5\). So, \(89.5 – 99.5\) is the median class.

Therefore, \(l = 89.5,\,h = 10,\,f = 30\)

Now, \({\rm{Median}} = l + \frac{{\frac{N}{2} – F}}{f} \times h\)

\( \Rightarrow {\rm{Median}} = 89.5 + \frac{{49 – 40}}{{30}} \times 10 = 92.5\)

Hence, the median of the given frequency distribution table is \(92.5\)

Q.2. Draw an ogive for the following frequency distribution by less than method.

Marks:	\(0 – 10\)	\(10 – 20\)	\(20 – 30\)	\(30 – 40\)	\(40 – 50\)	\(50 – 60\)
Number of students	\(7\)	\(10\)	\(23\)	\(51\)	\(6\)	\(3\)

Ans: We first prepare the cumulative frequency distribution table by less than the method as given below:

Marks	Marks less than	No. of Students	Cumulative Frequency
\(0 – 10\)	\(10\)	\(7\)	\(7\)
\(10 – 20\)	\(20\)	\(10\)	\(17\)
\(20 – 30\)	\(30\)	\(23\)	\(40\)
\(30 – 40\)	\(40\)	\(51\)	\(91\)
\(40 – 50\)	\(50\)	\(6\)	\(97\)
\(50 – 60\)	\(60\)	\(3\)	\(100\)

Here, since the first class interval is \(0 – 10\) and there are \(7\) students in that class interval, the ogive should ideally be drawn from the origin \((0,\,0)\). Now, we mark the upper-class limits (including the imagined class) along \(X\)-axis on a suitable scale and the cumulative frequencies along \(Y\)-axis on an appropriate scale.

Thus, we plot the points \((0,\,0),\,(10,\,7),\,(20,\,17),\,(30,\,40),\,(40,\,91),\,(50,\,97)\) and \(\left( {60,\,100} \right).\)

Now, we join the plotted points by ahand curve to obtain the required ogive as shown below:

Q.3. The frequency distribution of scores obtained by \(230\) candidates in a medical entrance test is as follows.

Draw cumulative frequency curves by less than and more than the method on the same axes.
Ans: Less than method: Let us prepare the cumulative frequency table by less than the method as given below:

Scores	Number of Candidates
\(400-450\)	\(20\)
\(450-500\)	\(35\)
\(500-550\)	\(40\)
\(550-600\)	\(32\)
\(600-650\)	\(24\)
\(650-700\)	\(27\)
\(700-750\)	\(18\)
\(750-800\)	\(24\)

Scores	Number of Candidates	Scores less than	Cumulative Frequency
\(400 – 450\)	\(20\)	\(450\)	\(20\)
\(450 – 500\)	\(35\)	\(500\)	\(55\)
\(500 – 550\)	\(40\)	\(550\)	\(95\)
\(550 – 600\)	\(32\)	\(600\)	\(127\)
\(600 – 650\)	\(24\)	\(650\)	\(151\)
\(650 – 700\)	\(27\)	\(700\)	\(178\)
\(700 – 750\)	\(18\)	\(750\)	\(196\)
\(750 – 800\)	\(34\)	\(450\)	\(230\)

Other than the given class intervals, we assume a class interval \(350 – 400\) prior to the first-class interval \(400 – 450\) with zero frequency. Now, we mark the upper-class limits on \(X\)-axis and the cumulative frequencies along \(Y\)-axis on suitable scales.

Thus, we plot the points \((400,\,0),\,(450,\,20),\,(500,\,55),\,(550,\,95),\,(600,\,127),\,(650,\,151),\,(700,\,178),\,(750,\,196)\) and \((800,\,230)\).

By joining these points by a hand smooth curve, we obtain an ogive by less than method, as shown in the figure below.

More than method: Let us prepare the cumulative frequency table by more than method as given below:

Scores	Number of Candidates	Scores more than	Cumulative Frequency
\(400 – 450\)	\(20\)	\(400\)	\(230\)
\(450 – 500\)	\(35\)	\(450\)	\(210\)
\(500 – 550\)	\(40\)	\(500\)	\(175\)
\(550 – 600\)	\(32\)	\(550\)	\(135\)
\(600 – 650\)	\(24\)	\(600\)	\(103\)
\(650 – 700\)	\(27\)	\(650\)	\(79\)
\(700 – 750\)	\(18\)	\(700\)	\(52\)
\(750 – 800\)	\(34\)	\(750\)	\(34\)

Other than the given class intervals, we assume class interval \(800 – 850\) after the last class interval \(750 – 800\) with zero frequency.

Now, we mark the lower-class limits on \(X\) axis and the cumulative frequencies along \(Y\) axis on suitable scales.

Thus, we plot the points \({\rm{(400,}}\,{\rm{230),}}\,{\rm{(450,}}\,{\rm{210),}}\,{\rm{(500,}}\,{\rm{175),}}\,{\rm{(550,}}\,{\rm{135), (600,}}\,{\rm{103),}}\,{\rm{(650,}}\,{\rm{79),}}\,{\rm{(700,}}\,{\rm{52),}}\,{\rm{(750,}}\,{\rm{34)}}\) and \({\rm{(800,}}\,{\rm{0)}}\)

By joining these points on the same graph by a hand smooth curve, we obtain an ogive by more than method. Therefore, ogive for both less than type and more than the type of given frequency distribution table is shown below.

Q.4. Calculate the cumulative frequency for the following frequency distribution:

Marks obtained	\(50 – 60\)	\(60 – 70\)	\(70 – 80\)	\(80 – 90\)	\(90 – 100\)
No. of students	\(4\)	\(8\)	\(12\)	\(6\)	\(10\)

Ans: The cumulative frequency table for the given (continuous) frequency distribution is:

Marks obtained (class interval)	No. of students (frequency)	Cumulative frequency
\(50 – 60\)	\(4\)	\(4\)
\(60 – 70\)	\(8\)	\(12(4 + 8)\)
\(70 – 80\)	\(12\)	\(24(12 + 12)\)
\(80 – 90\)	\(6\)	\(30(24 + 6)\)
\(90 – 100\)	\(10\)	\(40(30 + 10)\)

Q.4. Compute the median for the following cumulative frequency distribution:

Less than \(20\)	Less than \(30\)	Less than \(40\)	Less than \(50\)	Less than \(60\)	Less than \(70\)	Less than \(80\)	Less than \(90\)	Less than \(100\)
\(0\)	\(4\)	\(16\)	\(30\)	\(46\)	\(66\)	\(82\)	\(92\)	\(100\)

Ans: We are given the cumulative frequency distribution. So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute the median.

	Class intervals	Cumulative frequency \((c.f.)\)	Frequency \((f)\)
Less than \(20\)	\(10 – 20\)	\(0\)	\(0\)
Less than \(30\)	\(20 – 30\)	\(4\)	\(4\)
Less than \(40\)	\(30 – 40\)	\(16\)	\(12\)
Less than \(50\)	\(40 – 50\)	\(30\)	\(14\)
Less than \(60\)	\(50 – 60\)	\(46\)	\(16\)
Less than \(70\)	\(60 – 70\)	\(62\)	\(20\)
Less than \(80\)	\(70 – 80\)	\(82\)	\(16\)
Less than \(90\)	\(80 – 90\)	\(92\)	\(10\)
Less than \(100\)	\(90 – 100\)	\(100\)	\(8\)
			\(N = \sum {{f_i}} = 100\)

Here, \(N = \sum {{f_i}} = 100\)

\(\therefore \,\frac{N}{2} = 50\)

We observe that the cumulative frequency just greater than \(\frac{N}{2} = 50\) is \(66\) and the corresponding class is \(60 – 70\).

So, \(60 – 70\) is the median class.

\(\therefore \,l = 60,\,f = 20,\,cf = 46\) and \(h = 10\)

Now, \({\rm{Median}} = l + \frac{{\frac{N}{2} – cf}}{f} \times h\)

\( \Rightarrow {\rm{Median}} = 60 + \frac{{50 – 46}}{{20}} \times 10 = 62\)

Hence, the median of the given frequency distribution table is \(62\).

Summary

In this article, we learnt about the definition of cumulative frequency distribution, cumulative frequency distribution example, representation of a cumulative frequency distribution graphically, solved examples on cumulative frequency distribution, frequently asked questions on cumulative frequency distribution.

This article’s learning outcome is that cumulative frequency graphs are used to find the median of the given set of knowledge. If both less than and greater than cumulative frequency curves drawn on the same graph, we can easily find the median value.

Frequently Asked Questions (FAQ) – Cumulative Frequency Distribution

Q.1. Explain cumulative frequency distribution with example?
Ans: The cumulative frequency of a class interval is the sum of frequencies of all the classes up to the specified class interval.

Example:

Class interval \((C.I)\)	Frequency \((f)\)
\(5 – 15\)	\(7\)
\(15 – 25\)	\(8\)
\(25 – 35\)	\(13\)
\(35 – 45\)	\(12\)
\(45 – 55\)	\(10\)

Cumulative frequency table for above-given frequency distribution table:

Class interval \((C.I)\)	Cumulative frequency \((c.f.)\)
\(5 – 15\)	\(7\)
\(15 – 25\)	\(8 + 7 = 15\)
\(25 – 35\)	\(15 + 13 = 28\)
\(35 – 45\)	\(28 + 12 = 40\)
\(45 – 55\)	\(40 + 10 = 50\)

Q.2. What is a less than ogive graph?
Ans: On a graph paper, we mark the upper-class limits along the \(X\)-axis and the corresponding cumulative frequencies along the \(Y\)-axis.
On joining these points successively by smooth curves, we get a curve known as a cumulative frequency curve or an ogive.

Q.3. What is a cumulative frequency distribution?
Ans: The cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors. The last value will always be equal to the total for all observations since all frequencies will already have been added to the previous total.

Q.4. How do you get the cumulative frequency?
Ans: If the frequency of the first-class interval is added to the frequency of the second class and this sum is added to the third class and so on, then frequencies so obtained are known as cumulative frequency \((c.f.)\).

Q.5. What is a more than ogive graph?
Answer: On a graph paper, we mark the lower-class limits along the \(X\)-axis and the corresponding cumulative frequencies along the \(Y\)-axis.
On joining these points successively by smooth curves, we get a curve known as a cumulative frequency curve or an ogive.