## Mean, median and mode for ungrouped data

Let $x_i, i=1,2, \cdots , n$ be $n$ observations.

## Formula

The mean of $X$ is denoted by $\overline{x}$ and is given by

`$\overline{x} =\dfrac{1}{n}\sum_{i=1}^{n}x_i$`

Arrange the data in ascending order of magnitude.

Median of $X$ is given by

### Median

`$$ \begin{equation*} M= \left\{ \begin{array}{ll} \text{value of }\big(\frac{n+1}{2}\big)^{th}\text{ observation}, & \hbox{if $n$ is odd;} \\ \text{average of }\big(\frac{n}{2}\big)^{th}\text{ and }\big(\frac{n}{2}+1\big)^{th} \text{ observation}, & \hbox{if $n$ is even.} \end{array} \right. \end{equation*} $$`

### Mode is the value of $X$ that occurs maximum number of times.

## Example 1

A random sample of 15 patients yielded the following data on the length of stay (in days) in the hospital.

`5, 6, 9, 10, 15, 10, 14, 12, 10, 13, 13, 9, 8, 10, 12`

.

Find the mean, median and mode.

### Solution

**Mean**

The sum of observations is $\sum x_i =156$ days.

The mean of the length of stay in the hospital is

`$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{156}{15}\\ &=10.4 \text{ days}. \end{aligned} $$`

**Median**

The data in ascending order of magnitude is $5, 6, 8, 9, 9, 10, 10, 10, 10, 12, 12, 13, 13, 14, 15$.

Here $n = 15$ which is odd.

Sample median = value of $(\frac{n+1}{2})^{th}$ observations.

Thus the median of the length of stay in the hospital is

`$$ \begin{aligned} M &= \bigg(\frac{15+1}{2}\bigg)^{th}\text{Obs.} \\ &= \big(8\big)^{th}\text{Obs.} \\ &=10 \text{ days}. \end{aligned} $$`

**Mode**

The observation $10$ occurs with a highest frequency of 4.

The mode of the length of stay in the hospital is $10$ days.

## Example 2

The age (in years) of 6 randomly selected students from a class are

`22, 25, 24, 23, 24, 20`

.

Find the mean, median and mode.

### Solution

**Mean**

The sum of observations is $\sum x_i =138$ days.

The mean age of students is

`$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{138}{6}\\ &=23 \text{ years}. \end{aligned} $$`

**Median**

The data in ascending order of magnitude is $20, 22, 23, 24, 24, 25$.

Here $n = 6$ which is even.

Sample median = average of $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations.

Thus the median age of students is

`$$ \begin{aligned} M &= \frac{\big(\frac{6}{2}\big)^{th}\text{Obs.} +\big(\frac{6}{2}+1\big)^{th}\text{Obs.}}{2}\\ &= \frac{\big(3\big)^{th}\text{Obs.} +\big(4\big)^{th}\text{Obs.}}{2}\\ &=\frac{23 +24}{2} \\ &= 23.5 \text{ years}. \end{aligned} $$`

**Mode**

The observation $24$ occurs with a highest frequency of 2.

The mode of age of students is $24$ years.