## What is Median?

When elements in the data set are organised sequentially, that is, in either an ascending or descending order of magnitude, the **median **can be referred to as the middle value of the data set. Its value is located in a distribution in such a way that 50% of the items are below it and 50% are above it. It focuses on the center or middle of a distribution.

## Median in Individual Series

The steps required to determine the median of an individual series are as follows:

**Step 1: **Firstly, arrange the given data in ascending or descending order.

**Step 2: **Apply the following formula of the Median:

Median(M) =Â

Where,Â

N is the Number of Items

#### Median in case of Odd and Even Number of Items

- In the case of an
**odd**number of items, Median is the Middle term of the observation. - In the case of an
**even**number of items, the Median is the average of two middle terms and is determined by using the following formula:

Median(M)=Â

#### Example 1: (When there is an even number of items in the series)

Given below is the age of some students. Find out the median of their age: 40, 32, 38, 28, 20, 44, 22, 18.

#### Solution:

The number of items is 8. So,Â

Median =Â

Median =Â

=Â = 30

**Median = 30**

#### Example 2: (When there is an odd number of items in the series)

From the following data of the weekly wages of 7 employees (in â‚¹), compute the median wage.

450 | 600 | 800 | 250 | 430 | 200 | 400 |

#### Solution:

The number of items is 7. So,Â

Median =Â

Median =Â = Size of 4^{th} item = 430

**Median = 430**

## Median in Discrete Series

The steps required to determine the median of a discrete series are as follows:

**Step 1: **Arrange the given distribution in either ascending or descending order.

**Step 2: **Denote the variables as **X **and frequency as **f**.

**Step 3: **Determine the cumulative frequency; i.e., **cf.**

**Step 4: **Calculate the median item using the following formula:

Median(M)= **Size~of~[\frac{N+1}{2}]^{th}~item**

Where,Â

N = Total of Frequency

**Step 5: **Find out the value ofÂ . We can find it by first locating the cumulative frequency, which is equal toÂ or next higher to it, and then finding the value corresponding to this cumulative frequency. This value will be the Median value of the series.

#### Example:

Calculate the median of the following data.

#### Solution:

Median =Â

=Â = Size of 18^{th} item

Since the 18^{th} item falls under the cumulative frequency 28 the size of the distribution against this cumulative frequency value is 6.

**Median = 6**

## Median in Continuous Series

The value of the median cannot be easily located for continuous series. In this situation, the median is located between the lower and upper limits of a class interval.Â AÂ formula is usedÂ to interpolate (guess) the median to obtain the exact value. However, it should be kept in mind that when the median class of a series is first class, then the c.f. in the formula will be taken as zero. The steps required to determine the median of a continuous series are as follows:

**Step 1: **Arrange the given data in either descending or ascending order.

**Step 2: **Determine the cumulative frequency; i.e., **cf.**

**Step 3: **Calculate the median item using the following formula:

Median(M) =Â Â Â Â Â

Where,Â

N = Total of Frequency

**Step 4: **Now inspect the cumulative frequencies and find out the cf which is either equal to or just greater than the value determined in the previous step.

**Step 5: **Now, find the class corresponding to the cumulative frequency equal to or just greater than the value determined in the third step. This class is known as the **median class.Â **

**Step 6: **Now, apply the following formula for the median:

Median =Â Â Â Â Â

Where,

l_{1} = lower limit of the median class

c.f. = cumulative frequency of the class preceding the median class

f = simple frequency of the median class

i = class size of the median group or class

Note:While calculating the median of a given distribution, we have to assume that every class of the distribution is uniformly distributed in the class interval.

#### Example:Â

Calculate the median of the following data.

#### Solution:

Median(M) =Â =Â =Â

Hence, the median lies in the class 30-40.

l_{1 }= 30, f = 14, i = 10, c.f. = 18

Now apply the following formula:

Median =Â =Â

**Median = 35**

## Median in Special Cases

The calculation process of the median is different under some special circumstances. There are five types of special cases including:

- Cumulative Frequency Distribution
- Mid-Values are given
- Inclusive Class-Intervals
- Open-End Series
- Unequal Class Intervals

### 1. Cumulative Frequency Distribution (Less than or More than)

When the data is given in the form of less than or more than for all items in the series, then such data has to be converted into a simple frequency distribution, to find out the frequency of the median class. Once it is done, the rest of the procedure is the same as in any other continuous series.

#### Example:Â

Calculate the median from the following data.

#### Solution:

Median(M) =Â =Â =Â

Hence, the median lies in the class 20-25.

l_{1 }= 20, f = 50, i = 5, c.f. = 60

Now apply the following formula:

Median =Â =Â

**Median = 24**

### 2. Mid-Values are given

When mid values of class intervals are given, then the class intervals are found; i.e., to calculate the median, we need to first convert it into a continuous series.

**Step 1:** First of all, calculate the difference between the two mid-values.

**Step 2:** Then, half of the difference is subtracted and added to each mid-value to find the lower and upper limits respectively of the class intervals.

#### Example:

Compute the median from the following data:

#### Solution:

**Step 1:** The difference between the two mid-values is 10.

**Step 2:** Half of the difference is 5. Now, 5 is reduced and added to each mid value to determine the lower limit and upper limit.

Median(M) =Â =Â =Â

Hence, the median lies in the class 150-160.

l_{1 }= 150, f = 58, i = 10, c.f. = 75

Now apply the following formula:

Median =Â **Â **

**=Â **

**Median = 153.7 or 154**

**3. **Inclusive Class-Intervals

**Step 1:** Find the difference between the upper limit of a class interval and the lower limit of the next class interval.

**Step 2: **Add half of this difference to the upper limit of a class interval and subtract the remaining half from the lower limit of each class interval. This procedure fills up the gap between the two classes and thereby we get the exclusive classes.

#### Example:

Calculate the median from the following data:

#### Solution:

Median(M) =Â =Â =Â

Hence, the median lies in the class 79.5-89.5.

l_{1 }= 79.5, f = 120, i = 10, c.f. = 210

Now apply the following formula:

Median =Â =Â

**Median = 84.**

### 4. Open-End Series

In the case of open-end classes, the lower limit of the first class and the upper limit of the last class are not given. Median is known to be the best average in open-end class-interval series. In this case, there is no need to complete the class intervals and the formula remains the same.

#### Example:

Calculate the median from the following data:

#### Solution:

Median(M) =Â **Â **

Hence, the median lies in the class 20-30.

l_{1 }= 20, f = 36, i = 10, c.f. = 32

Now apply the following formula:

Median =Â =Â

**Median = 25**

### 5. Unequal Class Intervals

When the class intervals are unequal, there is no need to make the class intervals equal. The frequencies need not be adjusted and the same formula will be applied.

#### Example:

Calculate the median from the following data:

#### Solution:

Median(M) =Â =Â =Â

Hence, the median lies in the class 30-60.

l_{1 }= 30, f = 60, i = 30, c.f. = 40

Now apply the following formula:

Median =Â =Â

**Median = 40**

**Missing Frequency**

Steps to determine the missing frequency if the value of median is already given, are as follows:

**Step 1: **Represent missing frequencies by f_{1 }or f_{2} as the case may be.

**Step 2:** Apply the formula for the calculation of median. In this process, we get an equation that gives us the missing frequencies.

**Example:**

The following table gives the distribution of the monthly salary of 1,800 employees. However, the frequencies of the classes 40-50 and 60-70 are missing. If the median of the distribution is â‚¹59.25, find the missing frequencies.

#### Solution:

Let f_{1 }and f_{2} be the frequencies of the classes 40-50 and 60-70, respectively.

As the median of the above series is given as 59.25, the median lies in the class 50-60Â

l_{1} = 50, c.f. = 240+f_{1}, f = 400, i = 10, and N = 1,800 (given)

Median =Â Â =Â

59.25 =Â

9.25 Ã— 40 = 660 – f_{1}

f_{1} = 660 – 370

**f _{1} = 290**

Also, 1,010 + f_{1 }+ f_{2} = 1,800 (given)

1,010 + 290 + f_{2} = 1,800

f_{2} = 1,800 – 1,300

**f _{2} = = 500**