The median is a fundamental measure of central tendency representing the exact middle value of an ordered data set. When numerical observations are sorted in ascending order from smallest to largest, the median splits the distribution into two equal halves: exactly 50% of the data points lie below or equal to the median, and 50% lie above or equal to it.

In statistical theory, the median is also designated as the second quartile ( $Q_2$ ) and the $50\text{th}$ percentile ( $P_{50}$ ).

Why the Median Matters

Robustness Against Outliers: Unlike the arithmetic mean, which is pulled disproportionately by extreme high or low values, the median is an outlier-resistant (non-parametric) statistic. For example, national economic statistics report median household income and median house prices rather than mean averages, because a handful of multi-millionaires would distort the mean upward.
Academic Relevance: The median forms the centre of box plots (box-and-whisker diagrams), defines the boundaries of the Interquartile Range ( $\text{IQR} = Q_3 - Q_1$ ), and serves as the baseline for evaluating distribution skewness.
Exam Importance: The median is assessed across all UK exam boards (Edexcel, AQA, OCR). At GCSE, students are tested on finding the median from raw lists, frequency tables, stem-and-leaf diagrams, and cumulative frequency curves (typically 2–4 marks). At A-Level Mathematics (Statistics), questions demand linear interpolation for grouped continuous data and comparing the median with the mean to evaluate skewness (regularly worth 4–6 marks).

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Ungrouped Discrete Data

To find the median of a list of raw numerical observations:

Sort: Arrange the data in ascending order (smallest to largest).
Count: Determine the total number of observations, denoted by n.
Locate: Use the position formula:

\text{Position of Median} = \dfrac{n + 1}{2}

Depending on whether the sample size n is odd or even, two distinct cases arise:

Case A: Odd Number of Elements

When n is odd, $\dfrac{n + 1}{2}$ yields a whole integer. The median is the single observed value occupying that exact position.

\text{Data: } [x_1, x_2, \dots, \mathbf{x_{\frac{n+1}{2}}}, \dots, x_n] \implies \text{Median} = x_{\frac{n+1}{2}}

Case B: Even Number of Elements

When n is even, $\dfrac{n + 1}{2}$ yields a half-integer (e.g., 4.5, 8.5). There is no single middle number. The median is the arithmetic mean of the two central values:

\text{Median} = \dfrac{x_{\left(\frac{n}{2}\right)} + x_{\left(\frac{n}{2} + 1\right)}}{2}

Skewness: Comparing the Median and the Mean

Comparing the median to the mean provides immediate insight into the shape and skewness of a statistical distribution:

Symmetric Distribution: $\text{Mean} \approx \text{Median} \approx \text{Mode}$

Positively Skewed (Right-Skewed): $\text{Mean} > \text{Median}$ . A tail of unusually large positive values pulls the mean to the right, while the median remains anchored.

Negatively Skewed (Left-Skewed): $\text{Mean} < \text{Median}$ . A tail of unusually small negative values pulls the mean to the left.

Worked Example

Ungrouped Discrete Data

Calculate the median of the following raw test scores:

7, \; 9, \; 10, \; 4, \; 3, \; 2, \; 1, \; 8, \; 12

Step 1: Arrange the data in ascending order

1, \; 2, \; 3, \; 4, \; 7, \; 8, \; 9, \; 10, \; 12

Step 2: Determine n and find the median position

There are n = 9 items.

\text{Position} = \dfrac{n + 1}{2} = \dfrac{9 + 1}{2} = 5

The median is the 5th value in the ordered list.

Step 3: Identify the value

x_1 = 1, \; x_2 = 2, \; x_3 = 3, \; x_4 = 4, \; \mathbf{x_5 = 7}, \; x_6 = 8, \; x_7 = 9, \; x_8 = 10, \; x_9 = 12

\text{Median} = 7

Practice Questions & Solutions

Score:

A list of six positive integers is arranged in ascending order:

$\text{[math]}$

Given that the median of this data set is 9:

(a) Find the value of $\text{[math]}$ .

(b) Determine the arithmetic mean of the completed list.

Solution

(a)

Since the data set contains n = 6 items (an even sample size), the median lies halfway between the $\text{[math]}$ and $\text{[math]}$ ordered values:

$\text{[math]}$

The $\text{[math]}$ term is $\text{[math]}$ and the $\text{[math]}$ term is $\text{[math]}$ .

Set up the median equation:

$\text{[math]}$

(b)

Substitute x = 8 into the list to identify all six integers:

$\text{[math]}$

Calculate the arithmetic mean:

$\text{[math]}$

A survey recorded the number of books read in a month by 50 secondary school students. The results are summarized below:

0 books: 14 students
1 book: 17 students
2 books: 11 students
3 books: 5 students
4 books: 3 students

(a) Find the median number of books read.

(b) Determine the lower quartile ( $\text{[math]}$ ) and the upper quartile ( $\text{[math]}$ ).

Solution

First, determine the cumulative frequencies ( $\text{[math]}$ ) for the discrete distribution:

Up to $\text{[math]}$ books: $\text{[math]}$
Up to $\text{[math]}$ book: $\text{[math]}$
Up to $\text{[math]}$ books: $\text{[math]}$
Up to $\text{[math]}$ books: $\text{[math]}$
Up to $\text{[math]}$ books: $\text{[math]}$

(a)

Total frequency $\text{[math]}$ .

$\text{[math]}$

The $\text{[math]}$ and $\text{[math]}$ observations both fall within the group of students reading $\text{[math]}$ book (covering positions $\text{[math]}$ to $\text{[math]}$ ).

$\text{[math]}$

(b)

Lower Quartile ( $\text{[math]}$ ):

$\text{[math]}$

The $\text{[math]}$ position falls into the $\text{[math]}$ books category (since $\text{[math]}$ ).

$\text{[math]}$
Upper Quartile ( $\text{[math]}$ ):

$\text{[math]}$

The $\text{[math]}$ position falls into the $\text{[math]}$ books category (since the cumulative frequency covers positions $\text{[math]}$ to $\text{[math]}$ ).

$\text{[math]}$
(c)

$\text{[math]}$

The times, $\text{[math]}$ minutes, taken by $\text{[math]}$ athletes to complete a $\text{[math]}$ race are summarized as follows:

$\text{[math]}$ minutes: $\text{[math]}$ athletes
$\text{[math]}$ minutes: $\text{[math]}$ athletes
$\text{[math]}$ minutes: $\text{[math]}$ athletes
$\text{[math]}$ minutes: $\text{[math]}$ athletes

Use linear interpolation to calculate an estimate for the median race time.

Solution

Compute the cumulative frequencies for the race times:

$\text{[math]}$ : $\text{[math]}$
$\text{[math]}$ : $\text{[math]}$
$\text{[math]}$ : $\text{[math]}$
$\text{[math]}$ : $\text{[math]}$

Step 1: Determine the median position

For continuous grouped data:

$\text{[math]}$

The $\text{[math]}$ observation falls at the end of the interval $\text{[math]}$ .

Step 2: Identify interpolation variables for the class $\text{[math]}$

Lower class boundary ( $\text{[math]}$ ) $\text{[math]}$
Cumulative frequency before this class ( $\text{[math]}$ ) $\text{[math]}$
Frequency of the class ( $\text{[math]}$ ) $\text{[math]}$
Class width ( $\text{[math]}$ ) $\text{[math]}$
Step 3: Apply the linear interpolation formula

$\text{[math]}$

The daily maximum temperatures (in $\text{[math]}$ ) recorded over $\text{[math]}$ consecutive days in Town A are:

$\text{[math]}$

In Town B, the temperatures over the same $\text{[math]}$ -day period have a recorded median of $\text{[math]}$ and a calculated mean of $\text{[math]}$ .

(a) Calculate the median and the mean temperature for Town A.

(b) State, with reasons, the type of skewness shown by each town's temperature distribution.

Solution

(a)

Town A has $\text{[math]}$ recorded values.

The data is already sorted in ascending order:

$\text{[math]}$

Median of Town A:

$\text{[math]}$

The $\text{[math]}$ value is $\text{[math]}$ .

$\text{[math]}$

Mean of Town A:

$\text{[math]}$

$\text{[math]}$
(b)

Town A: The mean ( $\text{[math]}$ ) is greater than the median ( $\text{[math]}$ ). This indicates that the distribution has a positive skew (skewed to the right), caused by higher temperature values such as $\text{[math]}$ pulling the mean upward.
Town B: The mean ( $\text{[math]}$ ) is greater than the median ( $\text{[math]}$ ). Since $\text{[math]}$ , Town B also exhibits a positive skew.