Two woodlands can both contain six species and be nothing like each other. One has them in balance; the other is almost entirely one species with five hangers-on. Counting species alone cannot tell those two apart. A diversity index can, and this page shows you how to calculate one.
📚 What you need to know
Species richness is the number of species in a community or defined area.
Richness alone can be misleading, because it ignores how many individuals of each species there are.
Species evenness is how evenly individuals are shared between the species present.
Species diversity combines richness and evenness, and is what conservationists usually prefer.
Simpson’s diversity index puts a number on it: D = N(N − 1) ÷ Σn(n − 1).
A higher D means greater diversity. The lowest possible value is 1.
Only compare D between similar habitats, or the same habitat over time.
Richness is not enough
Species richness is the simplest measure there is: count how many different species you find. It is useful for a quick comparison, but it treats a species with 200 individuals and a species with 1 individual as exactly the same thing.
Species evenness fixes that. It describes the relative abundance of the species present — whether the individuals are spread out or piled into one dominant species.
Species richness is 4 in both areas. If richness were the whole story, these two would be identical — and they clearly are not.
Simpson’s diversity index
A diversity index is a mathematical tool that turns richness and evenness into one number, so communities can be compared with each other or tracked over time. The one ESS uses is Simpson’s index.
Simpson’s diversity index
D = N(N − 1) ÷ Σn(n − 1)
D = Simpson’s diversity index
N = total number of individuals of all species sampled
n = number of individuals of each species
Σ means “add up”, so you work out n(n − 1) for every species and total the column
You are given this formula in the exam, so do not waste revision time memorising it. What you must be able to do is use it quickly and without arithmetic slips, because that is the only part being tested.
🧩 The method, every time
Draw a table with three columns: species, n, and n(n − 1).
Fill in n for each species from the data.
Work out n(n − 1) for each row. Any species with n = 1 gives 0, and n = 0 also gives 0.
Total the n column to get N, and total the last column to get Σn(n − 1).
Put both totals into the formula and divide.
Quote D to two decimal places, with no units. The index is a ratio.
A full worked calculation
Students sampled ground invertebrates in two grassland sites using pitfall traps: an unmanaged meadow and a regularly mown lawn. Both sites gave 6 species and 52 individuals, so the species richness is identical.
Site A — unmanaged meadow
Species
Number (n)
n(n − 1)
Ground beetle
12
12 × 11 = 132
Rove beetle
10
10 × 9 = 90
Wolf spider
9
9 × 8 = 72
Woodlouse
8
8 × 7 = 56
Springtail
7
7 × 6 = 42
Centipede
6
6 × 5 = 30
Total
N = 52
Σn(n − 1) = 422
Site A
D = (52 × 51) ÷ 422 = 2652 ÷ 422 = 6.28
Site B — regularly mown lawn
Species
Number (n)
n(n − 1)
Ground beetle
40
40 × 39 = 1560
Rove beetle
4
4 × 3 = 12
Wolf spider
3
3 × 2 = 6
Woodlouse
2
2 × 1 = 2
Springtail
2
2 × 1 = 2
Centipede
1
1 × 0 = 0
Total
N = 52
Σn(n − 1) = 1582
Site B
D = (52 × 51) ÷ 1582 = 2652 ÷ 1582 = 1.68
What the two numbers tell you
Species richness is the same in both sites: 6 species.
N is the same in both sites: 52 individuals.
But D is 6.28 in the meadow and 1.68 in the lawn, so the meadow has far higher species diversity.
The whole difference is evenness. In the lawn one species makes up 40 of the 52 individuals.
Higher diversity suggests the meadow is likely to be more resilient to disturbance than the lawn.
There is no fixed upper limit. The maximum depends on how many species could be present, which is why cross-habitat comparisons are meaningless.
Getting the data in the first place
The index is only as good as the sampling behind it. The method has to suit the organism.
Method
Used for
Watch out for
Quadrats
Plants and other non-moving organisms
Quadrats must be placed randomly to avoid bias
Transects
Recording how communities change along a gradient, such as up a shore
Deliberately not random, so it shows change rather than average abundance
Pitfall traps
Ground-living invertebrates
Catches active species more often than sedentary ones
Kick sampling
Invertebrates in a river bed
Effort must be standardised: same time, same area, same person if possible
Capture–mark–release–recapture
Mobile animals such as small mammals
Assumes marks do not harm the animal or change its behaviour
Sample size matters. One quadrat tells you almost nothing. Repeating the sampling and using the mean reduces the effect of chance, and lets you say something about the whole site rather than one lucky patch of it. If a question asks how to improve a study, “take more samples and calculate a mean” is nearly always a valid mark.
What the index cannot do
It is only meaningful when comparing similar habitats, or the same habitat at different times. Comparing a pond with a hedgerow tells you nothing useful.
It says nothing about which species are present. A community of six common species and a community of six rare, threatened species can score the same D.
It ignores habitat and genetic diversity entirely — it is a species-level measure only.
The result depends completely on how well the sampling was done.
🧠
Easy way to remember the formula
The top is about the whole sample: big N, big N minus one. The bottom is about each species in turn: little n, little n minus one, all added up. Big over little.
Worked examples
WE 1
Calculate Simpson’s diversity index
A quadrat contains four plant species with 10, 8, 6 and 4 individuals. Calculate Simpson’s diversity index for this quadrat, showing your working. (3 marks)
Step 1: find N
N = 10 + 8 + 6 + 4 = 28Step 2: work out n(n − 1) for each species
10 × 9 = 90, 8 × 7 = 56, 6 × 5 = 30, 4 × 3 = 12
Step 3: total the column
Σn(n − 1) = 90 + 56 + 30 + 12 = 188Step 4: substitute
D = (28 × 27) ÷ 188 = 756 ÷ 188
D = 4.02show every step — method marks are awarded even if the final arithmetic slips
WE 2
Interpret two index values
Site X has D = 5.9 and Site Y has D = 1.4. Both sites contain seven species. Explain what this shows. (3 marks)
Point 1: richness
Species richness is the same in both sites, so the difference cannot be explained by the number of species.
Point 2: evenness
Site X has a much higher D, so its individuals are spread far more evenly between the species. Site Y is dominated by one or two species.
Point 3: the implication
Site X has higher species diversity and is therefore likely to be more resilient to disturbance.
Same richness, very different evennessalways state that richness is equal first — it proves you understand what the index adds
WE 3
Evaluate a sampling method
A student places five quadrats along the edge of a field, next to the footpath, and calculates D. Suggest two ways to improve the study. (3 marks)
Problem 1: bias
All quadrats were placed in one part of the field, next to a footpath, so the sample is not representative of the whole site. Trampling also affects that edge.
Improvement 1
Place the quadrats randomly across the whole field, for example using random coordinates.
Improvement 2
Take more quadrats and calculate a mean, which reduces the effect of chance variation.
Random placement plus a larger sample sizename the source of bias, not just the fix — the mark is usually for spotting why it matters
💡 Exam tips
Set out the three-column table even in an exam. It stops arithmetic errors.
Remember N is the total of all individuals, not the number of species.
Give D to two decimal places with no units.
State that richness is equal before you talk about evenness when comparing sites.
Say the index only compares similar habitats or the same habitat over time.
For sampling questions, “random placement” and “repeat and take a mean” are reliable marks.
⚠ Common mistakes
Using the number of species as N. N is the total number of individuals.
Forgetting the minus one. Both N(N − 1) and n(n − 1) need it.
Skipping species with n = 1. Include the row; it simply contributes 0.
Giving D units. It is a ratio and has none.
Comparing very different habitats. The comparison is meaningless.
Thinking a high D means healthy. It says nothing about whether the species present are rare, native or invasive.
Up next: Managing Biodiversity in Practice — once you can measure it and explain how it arose, the last question is what to do to keep it.
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