# Interquartile Range

The smallest of all the measures of dispersion in statistics is called the Interquartile Range. Interquartile Range is the difference between the two extreme observations or terms of the distribution.

In Interquartile Range quartiles are the partitional values that divide the whole number series into 4 equal parts. These quartiles are represented by Q1( the lower quartile), Q2 (the median), and Q3 (the upper quartile).

Therefore, the difference between the upper and lower quartile is known as the interquartile range.

## What is Interquartile Range?

The Interquartile Range can be given as:

**Interquartile range = Upper Quartile – Lower Quartile = Q _{3} – Q_{1}**

where,

Q_{1} is the first quartile and Q_{3} is the third quartile of the series

Q_{2} is the second quartile or median of the series.

Half of the interquartile range formula is given by:

**Semi Interquartile Range = (Q _{3}– Q_{1}) / 2 **

**Few steps to calculate the interquartile range:**

- Arrange the number series into increasing or decreasing order.
- If the number series is odd, then the center value is median otherwise for even number series the median is the mean value for two center values.
- The median of the given number series equally cuts the given values into two equal parts, we refer to this as Q
_{2}. - Q
_{1 }can be calculated using the formula:

Q_{1} = \( \dfrac{(n+1)}{4}^{\text{th}}\) term

- Q
_{3}_{ }can be calculated using the formula:

Q_{3} = \( \dfrac{3(n+1)}{4}^{\text{th}}\) term

where,

n = Number of terms

- Finally, we have to subtract the median values of Q
_{1}and Q_{3}. The resulting value is the interquartile range of series.

## Solved Examples Using Interquartile Range

**Example 1:**

### Determine the interquartile range value for the first ten odd numbers.

**Solution:**

To find: The interquartile range value for the first ten odd numbers

Given:

The first ten odd numbers:1, 3, 5, 7, 9, 11, 13, 15, 17, 18

Number of values = 10

8 is an even number. Therefore, the median is the mean of 5th and 6th terms.

That is Q_{2} = (9+11)/2.

Q_{2} = 10.

Now

Q_{1} part is {1, 3, 5, 7, 9}

Here the number of values = 5

5 is an odd number. Therefore, the center value Q_{1} is 5

Q_{3} part : {11, 13, 15, 17, 19}

Here the number of values = 5

5 is an odd number. Therefore, the center value Q_{3} is 15

Using Interquartile Range,

**Interquartile Range = Upper Quartile – Lower Quartile = Q _{3} – Q_{1}**

Q_{3} - Q_{1} is 15 – 5 = 10

**Answer:** 15 is the interquartile range for the given set of first 10 odd numbers.

**Example 2:**

### Using the interquartile range formula, calculate the range of the following set of data:

### {4, 17, 7, 14, 18, 12, 3, 16, 10, 4, 4, 11}

**Solution:**

To find: Interquartile range

Given:

Number of terms = 12

Set = {4, 17, 7, 14, 18, 12, 3, 16, 10, 4, 4, 11}

Ordered set = {3, 4, 4, 4, 7, 10, 11, 12, 14, 16, 17, 18}

Dividing the set into quartiles, each quarter will have 3 terms as: {3, 4, 4}, {4, 7, 10}, {11, 12, 14}, {16, 17, 18}

First Quartile,

Q1 = (4 + 4)/ 2 = 4

Third Quartile,

Q3 = (14 + 16)/2 = 15

Using Interquartile Range Formula,

**Interquartile Range = Upper Quartile – Lower Quartile = Q _{3} – Q_{1}**

= 15 - 4

= 11

**Answer: **Interquartile range of the given set = 11

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