Simpson's Index of Diversity

Simpson's Index of Diversity is the cornerstone quantitative measure used by A-Level Biology students to compare the biodiversity of two habitats. Unlike species richness (which is just a count), Simpson's Index combines richness and evenness into a single number between 0 and 1, where higher values mean greater diversity. OCR A-Level Biology A specification 4.2.1 (e) explicitly requires you to be able to calculate Simpson's Index and interpret the value.

Key Definitions:

• Simpson's Index of Diversity (D) — a measure of biodiversity combining species richness and evenness.
• n — the number of individuals of a particular species.
• N — the total number of individuals of all species.
• Richness — the number of different species present.
• Evenness — how equally abundant each species is.

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The Formula

$D = 1 - \sum\left(\frac{n}{N}\right)^2$D=1−∑(Nn​)2

The formula works by:

1. Dividing the number of individuals of each species (n) by the total number of all individuals (N), giving the proportion each species contributes.
2. Squaring that proportion.
3. Summing the squared proportions across all species.
4. Subtracting from 1.

The result is a number between 0 (no diversity — all individuals belong to one species) and nearly 1 (high diversity — many species, each with similar abundance).

Exam Tip: Be careful: some textbooks present Simpson's Index as $D = \sum(n/N)^2$D=∑(n/N)2, which runs the other way (lower = more diverse). OCR uses the Index of Diversity form ($1 - \sum(n/N)^2$1−∑(n/N)2), where higher values mean higher diversity. Always state clearly which version you are using.

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Worked Example 1: Two Meadows

A student samples two meadows of equal area and counts the flowering plants in ten quadrats in each.

Meadow A:

Species	n
Daisy	40
Buttercup	35
Clover	30
Dandelion	25
Plantain	20
Total (N)	150

Meadow B:

Species	n
Daisy	130
Buttercup	10
Clover	5
Dandelion	3
Plantain	2
Total (N)	150

Both meadows have five species (same richness), but Meadow A has high evenness and Meadow B is dominated by daisies.

Meadow A calculation:

$\sum(n/N)^2 = (40/150)^2 + (35/150)^2 + (30/150)^2 + (25/150)^2 + (20/150)^2$∑(n/N)2=(40/150)2+(35/150)2+(30/150)2+(25/150)2+(20/150)2

$= 0.0711 + 0.0544 + 0.0400 + 0.0278 + 0.0178 = 0.2111$=0.0711+0.0544+0.0400+0.0278+0.0178=0.2111

$D_A = 1 - 0.2111 = 0.789$DA​=1−0.2111=0.789

Meadow B calculation:

$\sum(n/N)^2 = (130/150)^2 + (10/150)^2 + (5/150)^2 + (3/150)^2 + (2/150)^2$∑(n/N)2=(130/150)2+(10/150)2+(5/150)2+(3/150)2+(2/150)2

$= 0.7511 + 0.0044 + 0.0011 + 0.0004 + 0.0002 = 0.7572$=0.7511+0.0044+0.0011+0.0004+0.0002=0.7572

$D_B = 1 - 0.7572 = 0.243$DB​=1−0.7572=0.243

Interpretation: Meadow A (D = 0.789) is far more diverse than Meadow B (D = 0.243), despite both having the same species richness. This is exactly what our eyes told us — Meadow B is dominated by daisies — but now we have a single number to quantify it.

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Worked Example 2: Comparing Rocky Shores

Two rocky shores are surveyed with ten 0.5 m² quadrats each; the total counts of mobile invertebrates are:

Species	Shore 1 (n)	Shore 2 (n)
Common limpet	42	88
Dog whelk	18	5
Common periwinkle	27	7
Shore crab	9	0
Beadlet anemone	14	0
Total (N)	110	100

Shore 1: $\sum(n/N)^2 = (42/110)^2 + (18/110)^2 + (27/110)^2 + (9/110)^2 + (14/110)^2$∑(n/N)2=(42/110)2+(18/110)2+(27/110)2+(9/110)2+(14/110)2 $= 0.1458 + 0.0268 + 0.0602 + 0.0067 + 0.0162 = 0.2557$=0.1458+0.0268+0.0602+0.0067+0.0162=0.2557 $D_1 = 1 - 0.2557 = 0.744$D1​=1−0.2557=0.744