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We know that the **median** divides the data into two halves. We also know that for a set of n **ordered** numbers the median is the (n + 1) ÷ 2 ^{th} value.

Similarly, the **lower quartile** divides the bottom half of the data into two halves, and the **upper quartile** also divides the upper half of the data into two halves.

Lower quartile is the (n + 1) ÷ 4 ^{th} value.

Upper quartile is the 3 (n + 1) ÷ 4 ^{th} value.

- Question
Find the median, lower quartile and upper quartile for the following data:

11, 4, 6, 8, 3, 10, 8, 10, 4, 12 and 31.

- Answer
Ordering the data, we get 3, 4, 4, 6, 8, 8,10, 10, 11, 12 and 31.

The median is the (11 + 1) ÷ 2 = 6

^{th}value.The lower quartile is the (11 + 1) ÷ 4 = 3

^{rd}value.The upper quartile is the 3 (11 + 1) ÷ 4 = 9

^{th}value.Therefore, the median is 8, the lower quartile is 4, and the upper quartile is 11.

3, 4,

**4**, 6, 8,**8**, 10, 10,**11**, 12, 31

The **interquartile range** is the difference between the upper quartile and lower quartile.

In this example, the interquartile range is 11 - 4 = 7.

- Question
A survey was carried out to find the number of pets owned by each child in a class.

The results are shown in the table:

Number of pets Frequency 0 3 1 5 2 2 3 7 4 10 5 3 6 1 >6 0 Find the interquartile range.

- Answer
3.

Remember that there is a total of 31 children in the class.

- The lower quartile is the 8
^{th}value, which is 1. - The upper quartile is the 24
^{th}value, which is 4. - Therefore, the interquartile range is 4 - 1 = 3.

- The lower quartile is the 8

Note that the interquartile range ignores extreme values. The range includes extreme values.

Look at this set of data:

- 1, 5, 7, 8, 9, 12, 13, 15, 17, 18, 35.
- The
**interquartile range**is 17 - 7 = 10. - The
**range**is 35 - 1 = 34.

In cases such as these, it is often preferable to use the interquartile range when comparing the data.

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