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THE QUARTILES AND MEDIAN OF GROUPED DATA











In this post,  we will learn how to determine the quartiles when some quantitative data are presented in a frequency distribution table. For example, we have the following data, showing Flesch Readability Score of 80 monthly bulletin articles published by Britt and Co. Ltd.
Find the quartiles of these readability scores.

To answer this, first augment the table with a new column to the right of the frequency column, namely Data Numbers column. There are 5 data in the first class, so the class contains data no. 1 to  no. 5. There are 7 data in the second class, so the class contains data no. 6 to no. 12. There are 13 data in the third class, so the class contains data no. 13 to no. 25. Continuing this way, we get the following:


In this case, finding the first quartile means finding the 20th data, after the data have been ordered from the smallest to the highest (20 = ¼ x 80). Note that the 20th data is in the third class (20 is in the range of 13 - 25, as seen in the rightmost column). We call this class the “first quartile class”. The lower class boundary of the class is L1 = 50 - 0.5 = 49.5. The frequency of the class is f1 = 13. The class width is c = 59.5 - 49.5 = 10. Note that there are 12 data in the preceding classes (12 = 5 + 7), written as (Σf)1 = 12. Finally, substitute the values into this formula: $Q_{1} = L_{1} + \frac{n/4-(\Sigma f)_{1}}{f_{1}} \cdot c$.
Then we have: $ Q_{1} = 49.5 + \frac{20 - 12}{13} \cdot 10 \approx  55.65$. The first quartile of these readability data is 55.65.

Similarly, to find the third quartile, we use $Q_{3} = L_{3} + \frac{3n/4-(\Sigma f)_{3}}{f_{3}} \cdot c$.
Firstly, we have to find the “third quartile class”, that is the class containing the third quartile. To do this, calculate 3n/4. In this case, $\frac{3n}{4} = \frac{3 \cdot 80}{4} = 60$. Looking at the Data Numbers column, we know that the “third quartile class” is the fifth class  (60 is in the range of 49 - 70).

The lower class boundary of the class is L3 = 70 - 0.5 = 69.5. The frequency of the class is f3 = 22. The class width is c = 79.5 - 69.5 = 10. Note that there are 48 data in the preceding classes (48 = 5 + 7 + 13 + 23), written as (Σf)3 = 48. Finally, substitute the values into the formula for Q3.
$Q_{3} = 69.5 + \frac{60 - 48}{22} \cdot 10 \approx  74.95$
The third quartile of these readability data is 74.95.

(to be continued)



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