Mode, Median and Arithmetic Mean

Three ways to describe a typical value

An average summarises the centre or typical value of a dataset. Mode, median and arithmetic mean do this in different ways, so they need not give the same answer.

The mode is the value or category occurring most often. The median is the middle value when the data is put in order. The arithmetic mean shares the total equally between all observations: add every value and divide by how many values there are.

In exams, mean means arithmetic mean unless another type is explicitly named.

The mode can describe non-numerical categories, such as favourite ice-cream flavours. A mean cannot be calculated from flavour names. For numerical data, the mean uses every value, so an unusually large or small value can pull it towards that extreme. The median depends on the middle positions and is less affected by extreme values.

Finding the mode and median from individual values

For the mode, count how often each value appears. If two values share the highest frequency, both are modes: the dataset is bimodal. If every value occurs equally often, there is no mode.

For the median, first arrange the values in ascending order. Write $n$ for the number of observations.

• If $n$ is odd, take the value in position $\frac{n+1}{2}$.
• If $n$ is even, take the arithmetic mean of the values in positions and .

For example, the ordered times, in seconds, taken by 15 students to solve a maths problem were:

$9,\ 10,\ 11,\ 12,\ 12,\ 13,\ 14,\ 14,\ 15,\ 16,\ 17,\ 19,\ 20,\ 21,\ 23$.

There are two modes, 12 seconds and 14 seconds, because each occurs twice and every other time occurs once. The middle position is $(15+1)/2=8$, so the median is 14 seconds, the eighth value.

With an even number of observations, the median may lie between two observed values; it does not have to be a value that actually occurs.

Calculating the arithmetic mean

The arithmetic mean is

$$\bar{x}=\frac{\sum x}{n}.$$

Here $\bar{x}$ means the mean, and $\sum x$ means the sum of all the data values. The symbol $\sum$ tells you to add.

The 15 problem-solving times total 226 seconds, so their mean is $226/15=15.0666\ldots$ seconds, or 15.1 seconds to three significant figures. This is the time each student would have taken if the total time were shared equally. It is not necessarily anyone's actual time.

The same relationship can be rearranged as $\text{total}=\text{mean}\times n$. This is useful when a mean is known but a total or a missing value is needed.

Using a discrete frequency table

Discrete data takes separate values, such as a number of pets. A frequency table records each value $x$ and how many times it occurs, its frequency $f$. Unlike grouping into intervals, this preserves the individual values and their counts.

To find the mode, choose the value with the highest frequency. To find the median, add the frequencies to obtain $n$, then use running totals, called cumulative frequencies, to locate the middle observation or observations.

For the mean, multiply each value by its frequency. This accounts for its repeated contribution to the total. For example, the time 12 seconds occurs twice in the problem-solving data, contributing $12\times2=24$ seconds. Add all these products and divide by the total frequency:

$$\bar{x}=\frac{\sum fx}{\sum f}.$$

Here $\sum fx$ is the total of the products, and $\sum f$ is the number of observations. These methods give exact averages because the values have not been lost through grouping.

Grouped data: modal class and estimated mean

Grouped data records how many observations fall within each interval, but not their individual values. Consider these weights of 20 three-week-old Labrador puppies:

The exact mode is unknown, but the modal class is $4\leq w<4.5$ kg: it contains the most puppies.

To estimate the mean, represent every observation in a class by its midpoint:

$$\text{midpoint}=\frac{\text{lower boundary}+\text{upper boundary}}{2}.$$

Multiply each midpoint by its frequency to estimate that class's total, then add the products and divide by the total frequency:

$$\text{estimated mean}=\frac{\sum fm}{\sum f}.$$

For the puppies, this gives $87/20=4.35$ kg. It is an estimate because a puppy within an interval need not weigh exactly the midpoint amount.

Estimating a grouped median by interpolation

A grouped table cannot reveal the exact middle value either. Linear interpolation estimates it by assuming observations are evenly spread across the class containing the median.

The following table records weekly music-listening time for 50 students:

For grouped interpolation, use the halfway count $n/2=25$. The cumulative frequency reaches 22 before the $20<x\leq30$ class and 34 at its end, so this is the median class.

The halfway count is $25-22=3$ observations into a class containing 12 observations. Assuming an even spread, the estimated median lies $3/12=1/4$ of the way across that class. Its width is $30-20=10$ hours, so move hours above its lower boundary. The estimated median is therefore hours.

The general calculation is

$$\text{estimated median}=L+\frac{n/2-C}{f}\times w,$$

where $L$ is the median class's lower boundary, $C$ is the cumulative frequency before that class, $f$ is its frequency and $w$ is its width. The fraction identifies how far through the class to move; adding the lower boundary converts that distance into an estimated data value.