The index of diversity

The index of diversity equation considers species richness and evenness to quantify biodiversity.

It is calculated as:

$\text{d }=\frac{\text{N}(\text{N}-1)}{\sum \text{n}(\text{n}-1)}$

Where:

• N = total individuals of all species.
• n = number of individuals of each species.

Higher values of d indicate greater diversity.

Worked example - Calculating the index of diversity

A biologist is assessing the biodiversity of two fields. Field A contains 150 individual plants, consisting of 85 daisies, 35 buttercups, and 30 clovers. Field B contains 150 individual plants, consisting of 55 daisies, 45 buttercups, and 50 clovers.

Calculate and compare the diversity indices for each habitat.

Step 1: Equation

$d=\frac{N(N-1)}{\sum n(n-1)}$

Step 2: Construct a table to calculate $\sum n(n-1)$ for field A

Step 3: Substitution and correct evaluation for field A

$\text{diversity of field A }=\frac{150(150-1)}{\mathrm{9,200}}$

$\text{diversity of field A }=\frac{150(149)}{\mathrm{9,200}}$

$\text{diversity of field A }=\frac{\mathrm{22,350}}{\mathrm{9,200}}$

$\text{diversity of field A }=2.43$

Step 4: Construct a table to calculate $\sum n(n-1)$ for field B

Step 5: Substitution and correct evaluation for Field B

$\text{diversity of field B }=\frac{150(150-1)}{\mathrm{7,400}}$

$\text{diversity of field B }=\frac{150(149)}{\mathrm{7,400}}$

$\text{diversity of field B }=\frac{\mathrm{22,350}}{\mathrm{7,400}}$

$\text{diversity of field B }=3.02$

Step 6: Interpretation of result

as the index of diversity for field A is 2.43 and for field B it is 3.02, this indicates that field B has a higher species diversity compared to field A