# Cumulative Frequency Graphs

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Cumulative frequency is adding up frequencies to provide the running total. It helps us find the median, lower quartile range and upper quartile range in a set of data.
Example:
Look at the table. Compute the cumulative frequencies by adding them up to have a running balance, then get the upper-class boundaries. Upper-class boundaries are found by adding 0.5 to the upper-class limit.

The table shows the income for 40 days on threads of The Threads and Needles Shop.

After completing the table, we can now graph it. The graph below shows the x-axis is the upper-class boundaries and the cumulative frequencies are on the y-axis. Cumulative frequency graphs is an o-give or usually have an S-shape.

The Median Value
The median is the value of the middle term or the number in the middle. It is easy to find the median if the number is arranged. It can be increasing or decreasing order.

For large quantities of data, the median is computed using the cumulative frequency column. The middle number is determined by n/2 where n is the number of cases or respondents.

To find the median:

The median is the sum of the lower limit (L) of median class and the fractional part of the class interval size. As shown in the formula, L is the lower limit of median class. N is the number of cases. F is less than or equal to cumulative frequency before the class interval which contain the median or below the middle class. f is the frequency of the median class and i is the size class interval.

Lower limit (L) is gained by subtracting the lowest value in the range by 0.5.
Example:
172 – 0.5 = 171.5
163 – 0.5 = 162.5
154 – 0.5 = 153.5
136 – 0.5 = 135.5
127 – 0.5 = 126.5
118 – 0.5 = 117.5

Solution
To get the middle point or median, the total number of frequency is divided by 2.
n / 2 = 40 / 2 = 20
The answer is 20. The cumulative frequency is close to 20 but not exceeding 20. It is between 154 – 162 and 145 – 153. 144.5 is the lower limit.

Let’s try to use the formula:
Median =L + [(n/2- F)/f]i

L = 144.5 ; F = 34 ; f = 12 ; i = 9

Solution
Median = 144.5 + [(40 / 2- 34) / 12]9
Median = 144.5 + [(-14) / 12]9
Median = 144.5 + (-10.5)
Median = 134

The median is 134. This means 134 is one-half of the income in selling threads.

Quartiles
Quartiles divide the distribution into four equal parts. Q_1is the 25%, Q_2is the 50% and Q_3is the 75%. For group data, it uses same procedure to get the median. If we follow the formula and solve, we will get the following values of quartile, which are Q_1 = 167.9, Q_2 = 134 and Q_3 = 144.5.