Statistics · Measures of Location and Spread · Year 1

Measures of Central Tendency

Mean, median and mode all answer the same question — "where's the centre?" — but they answer it in different ways, and knowing which to trust is as important as knowing how to calculate them.

9 min read 4 worked examples

Mode, median and mean

A measure of location is a single value that describes a position within a dataset. When that value describes the centre of the data, it's called a measure of central tendency. There are three in regular use:

Mode: the value (or class) that occurs most often. The only one of the three that makes sense for qualitative data.
Median: the middle value once the data is put in order. For $n$ ordered values, it sits at position $\dfrac{n+1}{2}$.
Mean: $\bar{x} = \dfrac{\sum x}{n}$ — every value summed, then divided by how many there are.
Choosing the right measure

Mode suits qualitative data, or quantitative data with a clear single (or double, "bimodal") peak — it's uninformative if every value occurs only once. Median suits quantitative data, especially when extreme values are present, since it isn't affected by them. Mean also suits quantitative data and uses every value, giving a genuinely representative average — but that same property means it's pulled around by extreme values.

The mean from a frequency table

When discrete data is presented in a frequency table, the mean is found using

$\bar{x} = \dfrac{\sum xf}{\sum f},$

where $\sum xf$ is the sum of each data value multiplied by its frequency, and $\sum f$ is the total frequency (equivalently, $n$).

Estimating the mean from grouped data

When continuous data has been grouped into classes, the individual values are no longer known — so every value in a class is treated as if it sat exactly at that class's midpoint, giving an estimate of the mean:

$\bar{x} \approx \dfrac{\sum xf}{\sum f},$

where $x$ now stands for each class midpoint. This is always described as an estimate, since the true values within each class are unknown.

Why "estimate" is the right word

Grouping data means you know how many observations fall inside a class, but not exactly where. Using the midpoint assumes an even spread across the class — a sensible approximation, but not a fact about the data. Any statistic calculated from grouped data inherits this same uncertainty.

Worked example 1

Mean, median and mode from raw data
Find the mean, median and mode of the dataset: $6, 9, 6, 11, 7, 6, 10$.
MethodSum and count for the mean; order and locate for the median
1
Mean: $\bar{x} = \dfrac{6+9+6+11+7+6+10}{7} = \dfrac{55}{7} \approx 7.86$.
Sum divided by count.
2
Ordered: $6,6,6,7,9,10,11$. With $n=7$, the median sits at position $\dfrac{7+1}{2}=4$, which is $7$.
Middle value of 7 ordered numbers.
3
The mode is $6$, appearing three times — more than any other value.
Most frequent value.
▪ solved
Examiner's eye

Always order the data before finding the median — applying the position formula $\frac{n+1}{2}$ to unordered data is one of the most common, and entirely avoidable, errors.

Worked example 2

Mean from a discrete frequency table
The table shows the number of pets owned by 40 households.
Number of pets0123
Frequency916114

Find the mean number of pets per household.

MethodBuild Σxf and Σf, then divide
1
$\sum xf = 0(9)+1(16)+2(11)+3(4) = 0+16+22+12 = 50$.
Each value multiplied by its frequency, summed.
2
$\sum f = 9+16+11+4 = 40$.
Total number of households.
3
Mean $= \dfrac{50}{40} = 1.25$ pets.
Σxf divided by Σf.
▪ solved

Worked example 3

Estimating the mean from grouped data
The table shows the delivery times, in minutes, of 50 parcels. Estimate the mean delivery time.
Delivery time (min)10–1920–2930–3940–49
Frequency621176
MethodFind each midpoint, then apply Σxf / Σf
1
Midpoints: $14.5,\ 24.5,\ 34.5,\ 44.5$.
Halfway between each class's boundaries.
2
$\sum xf = 6(14.5)+21(24.5)+17(34.5)+6(44.5) = 87+514.5+586.5+267 = 1455$.
Each midpoint times its frequency, summed.
3
$\sum f = 6+21+17+6 = 50$.
Total number of parcels.
4
Estimated mean $= \dfrac{1455}{50} = 29.1$ minutes.
Σxf divided by Σf.
▪ solved
The key insight

Every parcel in the "10–19" class is treated as if it took exactly $14.5$ minutes — none actually did. That approximation is unavoidable once data is grouped, which is why the answer must be reported as an estimate.

Worked example 4

Choosing the right measure · Challenge
A local estate agent reports the "average" house price on a street of 20 houses. 19 of the houses sold for between £180,000 and £260,000; one much larger house sold for £1.2 million. Explain why the median is a more useful measure of location here than the mean.
MethodConsider how the extreme value affects each measure
1
The mean uses every value in its sum, so the £1.2 million sale pulls it well above what a typical house on the street sold for.
The mean is sensitive to extreme values.
2
The median depends only on the middle-ranked value, so one extreme price — however large — barely affects it.
Robust to extreme values.
3
The median therefore gives a far more representative picture of what a typical house on the street actually costs.
Median matches intuition; the mean would mislead.
▪ explained
Look deeper

Whenever a dataset is skewed by one or two extreme values — house prices, incomes, company salaries — the median gives a far more representative "typical value" than the mean, which is dragged toward the extreme end.

Common mistake

For grouped data, the mode is usually reported as the modal class — the class with the highest frequency — since no individual value inside a class is known exactly. Don't report a single number as "the mode" of grouped data unless the question specifically asks for the midpoint of the modal class.

Check yourself

Find the mean and median of the dataset: $14, 18, 11, 25, 18, 21, 13$.
(Hint: order the data first before finding the median's position.)

Method
1
Mean $= \dfrac{14+18+11+25+18+21+13}{7} = \dfrac{120}{7} \approx 17.14$.
2
Ordered: $11,13,14,18,18,21,25$. Median (position 4) $= 18$.
n=7, position (n+1)/2 = 4.
▪ solved