One of the essential concepts in mathematics and statistics is the measure of central tendency, which gives a single value that attempts to describe a set of data by identifying the central position within that set of data. It is also known as the measure of central location. The median, which represents the middle value for any data set, is one of the measures of central tendency, apart from the mean and the mode.

The **median**, in statistics, is the value at which half the data is greater than it, and half is less than it. It is the most straightforward statistical measure to compute and is useful in representing a large number of values with a single value. For calculation of median, the entries in a data set have to be arranged in either ascending or descending order, and then the middlemost value represents the median of the given data set.

For example, the median of 6, 7, 8 is 7.

Moreover, the calculation of the median is affected by the number of observations in a data set. If we have an odd number of observations in a data set, the median is the middlemost value. On the other hand, if the number of observations is even, the median is the average of the two middle values. In the following sections, we will learn more about the median, how to calculate the median for an even or odd number of data, and how to use the median formula.

**What is median?**

Median is one of the measures of central tendency. It is defined as the value of the middlemost observation obtained after arranging the data in ascending or descending order. In some instances, it is difficult to consider all of the data for representation, so this is where the median is useful. Median is also known as the **place average**, as the data placed in the middle of a sequence is taken as the median.

Consider the following example in determining the median for a given set of data.

Suppose a data set contains 2, 5, 3, 3, 7, 2, and 1.

*Step 1:* We must arrange the values in the data set in ascending order: 1, 2, 2, 3, 3, 5, 7.

*Step 2:* Count the number of values in the data set. Hence, there are seven values.

*Step 3:* Look for the middle value. Since we have seven values, the middle value is located in the 4^{th} place. Thus, **the median is 3**.

**The Median Formula**

We can determine the middle value of the arranged set of numbers using the median formula. In finding this measure of central tendency, it is necessary to write down all the values in the data set in increasing order. The median formula differs depending on the number of observations there are in the data set and whether the number of observations they odd or even. The following formulas will assist you in determining the median of a given set of data.

**Median Formula for Ungrouped Data**

The steps below can help you apply the median formula to ungrouped data.

Step 1: Arrange the values in the data set in ascending or descending order.

Step 2: Count the total number of observations and represent it as n.

Step 3: Determine whether the number of observations is even or odd.

To compute for the median of a given set of data having an odd number of observations, we use the formula:

Median= $(\frac{n+1}{2})$^{th} term

As an illustration, let us consider finding the median of the data set {2, 3, 5, 8, 9}. The values are already arranged in increasing order. And since there are 5 values in the data set, and we know that 5 is an odd number, then the median can be computed as follows.

Median= $(\frac{5+1}{2})$^{th} term=3^{rd} term

The result tells us that the median is the third value in our data set, or the median can be found in the third place.

On the other hand, we can compute for the median of a given set of data having an even number of observations by using the formula:

Median=$\frac{\frac{n}{2}^{th} term+(\frac{n}{2}+1)^{th} term }{2}$

For example, let us consider finding the median of the data set {2, 3, 5, 9}. The values are already arranged in increasing order. And since there are 4 values in the data set, and we know that 4 is an even number, then the median can be computed as follows.

Median=$\frac{\frac{4}{2}^{th} term+(\frac{4}{2}+1)^{th} term}{2}$=4

Therefore, the data set {2, 3, 5, 9} has a median of 4.

**Median Formula for Grouped Data**

If you are working on continuous data that is in the form of a frequency distribution, the median can be computed by following the steps listed below.

Step 1: Count the total number of observations and let it be n.

Step 2: Establish the class size and let it be h; then, divide the data into different classes.

Step 3: Determine the cumulative frequency of each class.

Step 4: Look for the class in which the median falls. The median class is the class where $\frac{n}{2}$ lies.

Step 5: Identify the lower limit of the median class (l) and the cumulative frequency of the class preceding the median class (c).

Now, calculate the median value using the following formula

Median=I+ [$\frac{\frac{n}{2}-c}{f}$] ×h

To illustrate how the formula is used, let us consider the following set of data in the frequency distribution table and compute for the median.

Test scores | 0 – 10 | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 |

Number of students | 6 | 20 | 36 | 10 | 6 |

We need to calculate the cumulative frequencies to find the median.

Test scores | Number of students | Cumulative frequency | |

0 – 10 | 6 | 0 + 6 | 6 |

11 – 20 | 20 | 6 + 20 | 26 |

21 – 30 | 35 | 26 +35 | 61 |

31 – 40 | 11 | 61 +11 | 72 |

41 – 50 | 6 | 72 + 8 | 80 |

N = sum of cf = 80 ; $\frac{N}{2}= \frac{80}{2}$= 40

The cumulative frequency greater than 40 is 61, and the class is 21 – 30. Thus, the median class is 21 – 30.

l = 21, f = 35, c = 26, h = 10

Using Median formula:

Median=$I+ [\frac{\frac{n}{2}-c}{f}] ×h$

Median=$21+ [\frac{\frac{80}{2}-26}{35}] ×10 =21+ [\frac{14}{35}] ×10=25$

Therefore, the **median is 25.**

**How to apply the median formula in solving real-life problems?**

Let us use the median in solving real-life problems to understand and grasp its significance.

**Problem #1**

The salaries given to the employees who belong to the 5 top management employees in an organization are $3,000, $8,000, $5,000, $4,500, and $7,500. Determine the median salary using the median formula.

**Solution**

We will use the median formula for ungrouped data.

*Step 1:* Arrange the given data from lowest to highest.

$3,000, $4,500, $5,000, $7,500, and $8,000.

*Step 2:* Determine the total number of observations.

n = 5 ; 5 is odd

*Step 3:* Use the median formula for odd observation.

Median= $\frac{n+1}{2}$^{th} term

Median= $\frac{5+1}{2}$^{th} term

Median= $\frac{6}{2}$^{th} term

Median= 3rd term

The third term is $5,000

Therefore, the **median salary is $5,000**.

**Problem #2**

The coach of a school basketball team recorded the height of the members (in centimeters) of the members. Find the median of the above set if the coach’s record is shown below.

{144, 138, 135, 148, 155,165, 168,152}

**Solution**

Step 1: Arrange the values in the data set in ascending order.

Original set: {144, 138, 135, 148, 155,165, 168,152}

Ordered Set: {135, 138, 144, 148, 152, 155, 165, 168}

Step 2: Determine the total number of observations.

n = 8 ; 5 is even

Step 3: Use the median formula for even observation.

Median = $\frac{\frac{n^{th}}{2} term + (\frac{n}{2} + 1)^{th} term}{2}$

Median = $\frac{\frac{8^{th}}{2} term + (\frac{8}{2} + 1)^{th} term}{2}$

Median = $\frac{4^{th} term + (4 + 1)^{th} term}{2}$

Median = $\frac{4^{th} term + 5^{th} term}{2}$

The 4^{th} term is 148, and the 5^{th} term is 152. So,

Median = $\frac{148 + 152}{2}$

Median = $\frac{300}{2}$ = 150

Therefore, the **median is 150 centimeters**.

**Problem #3**

The following table gives the weekly expenditure of 200 families. Find the median of the weekly expenditure.

Weekly Expenditure ($) | 0 – 1000 | 1001 – 2000 | 2001 – 3000 | 3001 – 4000 | 4001 – 5000 |

Number of Families | 30 | 16 | 45 | 55 | 54 |

**Solution**

*Step 1:* Count the total number of observations.

n = 200

*Step 2:* Establish the class size and divide the data into different classes.

h = 1000

*Step 3:* Determine the cumulative frequency of each class.

Weekly expenditure ($) | Number of Families | Cumulative frequency | |

0 – 1000 | 30 | 0 + 30 | 30 |

1001 – 2000 | 16 | 30 + 16 | 46 |

2001 – 3000 | 45 | 46 + 45 | 91 |

3001 – 4000 | 50 | 91 + 50 | 141 |

4001 – 5000 | 59 | 141 + 59 | 200 |

*Step 4:* Identify the median class. It is the class where $\frac{n}{2}$ lies.

The cumulative frequency of where $\frac{n}{2}$=100 lies are in 141.

Thus, the median class is 3001 – 4000

*Step 5:* Identify the lower limit of the median class (l)

*I * = 3001

Identify the cumulative frequency of the class preceding the median class

*c* = 91

Now, calculate the median using the formula for grouped data.

Median = $I + [\frac{\frac{n}{2} – c}{f}] × h$

Median = $3001 + [ \frac{\frac{200}{2} – 91}{50}] × 1000 = 3181$

Therefore, the median of the weekly expenditure is **$3181**.

**Some Important Points on Median**

Below are some important points about the concept of the median that you must always keep in mind.

- The median, which is also known as the positional average, is the central value of a data set
- The values in data must be arranged first in either ascending or descending order to determine the middle value or median.
- Not all values in the data set are considered while calculating the median.
- The median does not get affected by extreme points or outliers.

**Median VS. Mean: Pros and Cons**

The mean and the median are both statistical measures that are used to describe the location of a set of data. But then, what are their advantages and disadvantages?

As we all know, the mean is the more commonly used measure of central tendency than the median. It is also the foundation for many advanced statistical methods. The mean, for example, is required to calculate the standard deviation, which is the most prominent measure of variability in a set of data. It’s also required for many statistical testing procedures, such as the t-test.

If so, what are the advantages of the median?

To illustrate this, let us consider the following five measurements of systolic blood pressure (mmHg)

**142, 124, 121, 151, 132**

Suppose 151 is the correct value, but a device failure shows 171 instead, which is a false measurement. Let’s see what happens to mean and median?

The mean of the resulting five values is now 138 rather than 134 as calculated from the original data, demonstrating the significant effect of the incorrect measurement.

We have learned that in determining the median, we must sort the data first in increasing order:

**121, 124, 132, 142, 171**

As before, the value 132 is in the center of the data row, so the false measurement does not affect the median.

That is why the median is referred to as “robust against outliers,” whereas the mean is “sensitive to outliers.”

Another advantage of the median that is associated with this type of robustness can be seen in “skewed” distributions.

In the context of an observational study, one example of such a distribution is the time since the onset of a specific disease. The date of diagnosis is frequently close to the time of reporting, i.e., at or just a few days before the baseline visit. However, patients who have had the disease for a long time are frequently included in the study group.

When the mean of the individual time spans since disease onset is calculated, such large values have a huge impact, making the mean larger than the actual data distribution would suggest.

The good news is that outliers have little effect on the median. As a result, the median provides a more realistic picture of the data in this case.

So, which one should we go with?

Calculating both measures is the best strategy.

If they are not too dissimilar, use the mean for data discussion because almost everyone is familiar with it.

If both measures are significantly different, this indicates that the data are skewed (that is, they are not normally distributed), and the median generally provides a more accurate picture of the data distribution.

**What is the application of median in our lives?**

Median is used to find the middlemost data. It is used to determine the point at which 50% of the data is more and 50% of the data is less. It is used in situations where extreme cases can be ignored. Below are some situations in which the concept of the median is applied.

**Choosing a movie to watch**

Suppose you and your family went to a cinema to watch a movie. You have to choose a movie from three different movies genre( for children, teenagers, and adults) that will entertain all members of the family. Your ages are 6, 13, 15, 17, and 60. The mean or average of the ages is 22, which is adult age, so you go see a movie intended for adults. But in the theatre, only one of you enjoys it while the others are bored. In this case, calculating the median is preferable to calculate the mean. As in this case, the median is 15, which is the middle of the data. More people enjoy a cinema day out when they watch a thriller. You can use median to decide which movie to watch next time.

**Home budget**

We all plan a monthly budget, whether we stick to it or not. Suppose you spend $700 on personal care, $100 on water, $800 on snacks, $500 on electricity, and $6000 on rent. The average expenditure is $1,620 using the mean concept and $700 using the median concept. Because the data is skewed and asymmetrical, the median concept is more efficient.

**Median salary**

Workplace rules and regulations vary. Some of them use the average salary as a benchmark, while others use the median salary. Median salary informs employees of their career’s middle point. As the name implies, half of the employees earn more than the median salary, while the other half earn less. This promotes healthy competition and growth.

**Explicating the poverty line**

The concept of the median is widely used in the calculation of the poverty line. Assume we have data on the monthly income of different groups of people in the range of $15, $30, $45, $450, and $110. The mean is $130. It is highly inappropriate to use the concept of mean to set the poverty line. So we move on to the median. The above data shows that the median is $45, indicating that those below this mark are poverty-ridden.