Upper Quartile and the Method of Finding it for Raw Data

If the data are arranged in ascending or descending order then the variate lying at the middle between the largest and the median is called the upper quartile (or the third quartile), and it denoted by Q3.


In order to calculate the upper quartile of raw data, follow these steps.

Step I: Arrange the data in ascending order.

Step II: Finding the number of variates in the data. Let it be n. Then find the upper quartile as follows. If n is not divisible by 4 then the mth variate is the upper quartile, where m is the integer just greater than \(\frac{3n}{4}\).

If n is divisible by 4 then the upper quartile is the mean of the \(\frac{3n}{4}\)th variate and the variate just greater then it.

Solved Problems on Upper Quartile and the Method of Finding it for Raw Data:

1. Find the upper quartile of the first thirteen natural numbers.

Solution:

The variates in ascending order are

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13.

Here n = 13.

So, \(\frac{3n}{4}\) = \(\frac{3 × 13}{4}\) = \(\frac{39}{4}\) = 9\(\frac{3}{4}\)

So, m = 10.

Therefore, the tenth variates is the upper quartile.

Hence, the upper quartile Q3 = 10.

 

2. If the variate 13 is removed from the above example, what will be the upper quartile?

Solution:

The variates in ascending order are

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Here, n = 12.

So, \(\frac{3n}{4}\) = \(\frac{3 × 12}{4}\) = \(\frac{36}{4}\) = 9, i.e., \(\frac{3n}{4}\) is an integer.

Therefore, the mean of the 9th and 10th variates is Q3 (the upper quartile).

Therefore, Q3 = \(\frac{9 + 10}{2}\) = \(\frac{19}{2}\) = 9.5.

Upper Quartile and the Method of Finding it for Raw Data

3. The following data represent the number of books issued by a library on 12 different days.

96, 180, 98, 75, 270, 80, 102, 100, 94, 75, 200, 610.

Find the upper quartile

Solution:

Write the data in ascending order, we have

75, 75, 80, 94, 96, 98, 100, 102, 180, 200, 270, 610.

Here, n = 12.

So, \(\frac{3n}{4}\) = \(\frac{3 × 12}{4}\) = \(\frac{36}{4}\) = 9, i.e., \(\frac{3n}{4}\) is an integer.

Therefore, the mean of the 9th and 10th variates is Q3 (the upper quartile).

Therefore, Q3 = \(\frac{180 + 200}{2}\) = \(\frac{380}{2}\) = 190.







9th Grade Math

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