## Outliers for ungrouped data

An outlier is any observation that is $1.5(IQR)$ away from the first quartile ($Q_1$) or third quartile ($Q_3$).

## Formula

$x$ is an outlier if $x$ is below $Q_1 -1.5\times IQR$ or above $Q_3+1.5\times IQR$,

where,

- $Q_1$ is the first quartile,
- $Q_3$ is the third quartile,
- $IQR = Q_3-Q_1$ is an inter-quartile range.

$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$

where $n$ is the total number of observations.

## Example 1

Following is the data about hourly wages (in dollars) of sample of 15 workers working in a company

20,21,24,23,25,12,22,34,24,22,20,22,19,22.

Check whether any outlier exists in the data.

### Solution

The formula for $i^{th}$ quartile is

$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$

where $n$ is the total number of observations.

**Arrange the data in ascending order**

12, 19, 20, 20, 21, 22, 22, 22, 22, 23, 23, 24, 24, 25, 34

**First Quartile $Q_1$**

The first quartle $Q_1$ can be computed as follows:

```
$$
\begin{aligned}
Q_1 &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ observation}\\
&=\text{Value of }\bigg(\dfrac{1(15+1)}{4}\bigg)^{th} \text{ observation}\\
&= \text{ Value of }\big(4\big)^{th} \text{ observation}\\
&=20
\end{aligned}
$$
```

**Third Quartile $Q_3$**

The third quartile $Q_3$ can be computed as follows:

```
$$
\begin{aligned}
Q_3 &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ observation}\\
&=\text{Value of }\bigg(\dfrac{3(15+1)}{4}\bigg)^{th} \text{ observation}\\
&= \text{Value of }\big(12\big)^{th} \text{ observation}\\
&=24
\end{aligned}
$$
```

**Inter-quartile range**

```
$$
\begin{aligned}
IQR & = Q_3 - Q_1\\
&= 24 - 20\\
& = 4.
\end{aligned}
$$
```

$Q_1-1.5*IQR = 14$ and $Q3+1.5*IQR = 30$.

The observation $12$ is less than $14$ and the observation $34$ is greater than $30$.

Thus the outliers are $12, 34$.

## Example 2

Blood sugar level (in mg/dl) of a sample of 20 patients admitted to the hospitals are as follows:

75,89,72,78,87, 85, 73, 75, 97, 87, 84, 76,73,79,99,86,83,76,78,73.

Check whether any outlier exists in the data.

### Solution

The formula for $i^{th}$ quartile is

$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$

where $n$ is the total number of observations.

**Arrange the data in ascending order**

72, 73, 73, 73, 75, 75, 76, 76, 78, 78, 79, 80, 82, 83, 84, 85, 86, 87, 97, 99

**First Quartile $Q_1$**

The first quartle $Q_1$ can be computed as follows:

```
$$
\begin{aligned}
Q_1 &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ observation}\\
&=\text{Value of }\bigg(\dfrac{1(20+1)}{4}\bigg)^{th} \text{ observation}\\
&= \text{ Value of }\big(5.25\big)^{th} \text{ observation}\\
&= \text{Value of }\big(5\big)^{th} \text{ obs.}+0.25 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\
&=75+0.25\big(75 -75\big)\\
&=75
\end{aligned}
$$
```

**Third Quartile $Q_3$**

The third quartile $Q_3$ can be computed as follows:

```
$$
\begin{aligned}
Q_3 &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ observation}\\
&=\text{Value of }\bigg(\dfrac{3(20+1)}{4}\bigg)^{th} \text{ observation}\\
&= \text{Value of }\big(15.75\big)^{th} \text{ observation}\\
&= \text{Value of }\big(15\big)^{th} \text{ obs.}+0.75 \big(\text{Value of } \big(16\big)^{th}\text{ obs.}-\text{Value of }\big(15\big)^{th} \text{ obs.}\big)\\
&=84+0.75\big(85 -84\big)\\
&=84.75
\end{aligned}
$$
```

**Inter-quartile range**

```
$$
\begin{aligned}
IQR & = Q_3 - Q_1\\
&= 84.75 - 75\\
& = 9.75.
\end{aligned}
$$
```

$Q_1-1.5*IQR = 60.375$ and $Q3+1.5*IQR = 99.375$.

All the observations are within $Q_1-1.5*IQR$ and $Q3+1.5*IQR$.

So no outliers found in the data.