Is Q_1 the same as the median?

No. Q_1 is the 25th percentile (one-quarter of the data lies below it), while the median is the 50th percentile (half the data lies below it). The median is also called Q_2. They are three different positions.

Can the IQR ever be negative?

No. By construction, Q_3 always sits further up the sorted list than Q_1, so Q_3 \geq Q_1. Therefore, \text{IQR} = Q_3 - Q_1 is always zero or positive.

How do I use IQR to find outliers?

Use the 1.5 × IQR rule: any value below Q_1 - 1.5 \cdot \text{IQR} or above Q_3 + 1.5 \cdot \text{IQR} is flagged as a potential outlier. This rule identifies points far outside the central cluster.

Should I include the median in both halves when I split the data?

No. If your data has an odd count, exclude the median from both halves—it is the dividing line, not part of either half. For even-sized data, there is no single median, so you split cleanly at the midpoint without double-counting.

Interquartile Range

The interquartile range measures the spread of the middle 50% of your data, telling you how compressed or spread out typical values are.

\text{IQR} = Q_3 - Q_1

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What each symbol means

What Interquartile Range takes

Symbol	Meaning
$Q_1$	The first quartile, or 25th percentile—the value below which one quarter of the dataset falls, measured in the same units as your data.
$Q_3$	The third quartile, or 75th percentile—the value below which three quarters of the dataset falls, measured in the same units as your data.

When to use it

Use IQR when comparing variability between datasets while ignoring extreme outliers, or when describing how concentrated the central half of your data is.

Level

Usually taught in: Statistics

Worked examples

1. Find the IQR of a small dataset with odd count

Problem

Find the interquartile range of

3, 5, 7, 9, 11, 13, 15

$3, 5, 7, 9, 11, 13, 15$
The dataset is already sorted in ascending order, which is the first step for any quartile calculation.
$n = 7$
We have 7 values. Since $n$ is odd, the median sits exactly in the middle.
$Q_2 = 9$
The median (4th value) is 9, which divides the data into two halves.
$\text{Lower half: } 3, 5, 7$
Take all values strictly below the median.
$Q_1 = 5$
The median of the lower half is 5.
$\text{Upper half: } 11, 13, 15$
Take all values strictly above the median.
$Q_3 = 13$
The median of the upper half is 13.
$\text{IQR} = 13 - 5 = 8$
Subtract $Q_1$ from $Q_3$ to get the interquartile range.

Answer: $\text{IQR} = 8$

This dataset has a compact IQR of 8, meaning the middle 50% of values span only 8 units. The calculation split the data at the median, then found the median of each half, which is the most intuitive method for odd-sized datasets.

2. Find the IQR with negatives and even count

Problem

Find the interquartile range of

-8, -3, 0, 2, 5, 7, 12, 14

$-8, -3, 0, 2, 5, 7, 12, 14$
The data is already sorted, which is essential before calculating quartiles.
$n = 8$
We have 8 values (even count), so we split the data cleanly in half without a middle median.
$\text{Lower half: } -8, -3, 0, 2$
The first 4 values form the lower half.
$\text{Upper half: } 5, 7, 12, 14$
The last 4 values form the upper half.
$Q_1 = \frac{-3 + 0}{2} = -1.5$
For the lower half, $Q_1$ is the average of the two middle values ( $-3$ and $0$ ).
$Q_3 = \frac{7 + 12}{2} = 9.5$
For the upper half, $Q_3$ is the average of the two middle values ( $7$ and $12$ ).
$\text{IQR} = 9.5 - (-1.5) = 11$
When subtracting a negative number, you add its absolute value instead.

Answer: $\text{IQR} = 11$

Even-sized datasets require averaging the two middle values of each half to find quartiles. This example shows how negative values are handled naturally— $Q_1$ is negative, and subtracting it adds to the range, making the IQR larger. The IQR of 11 shows much wider variability than in Example 1.

3. Calculate IQR for work hours to measure consistency

Problem

A construction supervisor records daily work hours for Crew A over a week:

6, 7, 8, 8, 9, 10, 11

hours. Find the IQR to understand how consistent their schedule is.

$6, 7, 8, 8, 9, 10, 11$
The data is sorted in ascending order.
$n = 7$
Crew A has 7 recorded workdays.
$Q_2 = 8 \text{ (position 4)}$
With 7 values, the median is the 4th value, which is 8.
$\text{Lower half: } 6, 7, 8$
All values less than or equal to the first occurrence below the median.
$Q_1 = 7$
The median of the lower half is 7.
$\text{Upper half: } 9, 10, 11$
All values greater than the median.
$Q_3 = 10$
The median of the upper half is 10.
$\text{IQR} = 10 - 7 = 3$
A small IQR shows that hours are fairly consistent day to day.

Answer: $\text{IQR} = 3 \text{ hours}$

The IQR of 3 hours tells you that the middle 50% of Crew A's workdays fall within a predictable 3-hour window (7 to 10 hours). This is valuable for construction scheduling because it lets you forecast resource needs and crew availability without being thrown off by the occasional short day or overtime day.

Common mistakes

Where Interquartile Range usually goes wrong

The mistake	Why it is wrong	The fix
Calculating quartiles before sorting the data.	Unsorted data puts values in the wrong positions, giving you incorrect $Q_1$ and $Q_3$ values and a useless IQR.	Always sort your dataset in ascending order before you begin—this is the single most important first step.
Including the median in both the lower half and upper half when finding $Q_1$ and $Q_3$ .	The median is the dividing line; if you count it twice, you distort both quartiles and make your IQR unreliable.	Exclude the median from both halves when $n$ is odd. When $n$ is even, split cleanly at the midpoint with no overlap.
Using the range ( $\text{max} - \text{min}$ ) instead of IQR when asked for spread.	The range captures only extremes and is destroyed by a single outlier, whereas IQR ignores the outer 25% and shows you typical spread.	Use IQR when outliers might be present or when you want to describe the central cluster; use range only if you specifically need the full extent and have no extreme values.

Tips and when to use something else

Always sort your data first—more errors come from skipping this step than from any arithmetic mistake.
IQR is robust to outliers because it ignores the extreme 25% on each end; use it instead of range whenever you see unusually large or small values.
If you need to account for all values in your spread measure (not just the middle 50%), switch to standard deviation or variance instead.
Check whether your data has an odd or even count before you divide it; even-sized datasets require you to average middle pairs to find quartiles.

Related formulas

MedianThe median is the middle value of ordered data, most useful when your data contains outliers that would skew the mean value.Range of a Data SetRange measures the spread of a data set by finding the difference between the maximum and minimum values, telling you how dispersed your data is.Standard DeviationStandard deviation measures how spread out data points are from the average—use it to understand variability in a dataset.Outlier RuleThe Outlier Rule identifies data values farther than 1.5 times the interquartile range from the quartiles—use it to flag unusual observations.

Frequently asked questions

Is $Q_1$ the same as the median?: No. $Q_1$ is the 25th percentile (one-quarter of the data lies below it), while the median is the 50th percentile (half the data lies below it). The median is also called $Q_2$ . They are three different positions.
Can the IQR ever be negative?: No. By construction, $Q_3$ always sits further up the sorted list than $Q_1$ , so $Q_3 \geq Q_1$ . Therefore, $\text{IQR} = Q_3 - Q_1$ is always zero or positive.
How do I use IQR to find outliers?: Use the 1.5 × IQR rule: any value below $Q_1 - 1.5 \cdot \text{IQR}$ or above $Q_3 + 1.5 \cdot \text{IQR}$ is flagged as a potential outlier. This rule identifies points far outside the central cluster.
Should I include the median in both halves when I split the data?: No. If your data has an odd count, exclude the median from both halves—it is the dividing line, not part of either half. For even-sized data, there is no single median, so you split cleanly at the midpoint without double-counting.

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Reviewed 2026-09-18