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### Course: AP®︎/College Biology>Unit 8

Lesson 4: Community ecology

# Simpson's index of diversity

Learn how to calculate Simpson's Diversity Index, a valuable tool for quantifying species diversity in communities. By comparing two communities with equal populations but different species distributions, the index reveals that a more evenly distributed community exhibits higher diversity. This mathematical method offers one way to evaluate biodiversity.

## Want to join the conversation?

• Just a question about the formula; in my text book I have a formula D = 1 - sum of (n/N)^2.

Why are these formulas different?
• They are the same formula, just written differently.
• I tried the last question "Community #4" that has 1 organism from 5 different species and I got -1. Is it possible to get an answer less than 0?
• Hey Sophie!! Not entirely sure how you got -1, but let's work through it here. So the total number of organisms N in your example would be 5, and n(i) would be one; then by the formula, the diversity index is 1 - (1(0) + 1(0) + 1(0) + 1(0) + 1(0))/(5*4) , which simplifies to 1 - 5/20 = 0.75.

In fact to answer your original question, it's not possible to get an index less than 0(check out the Community 3 in the example in the video) because no matter the population size, having just one species will yield a diversity of 0. Similarly, it's not possible to have an index greater than 1, since if you have an infinitely large population, the value of the second term minimizes at 0 and the index becomes 1.
Hope this helps!!
• Does the equation have to be 1 - (Σn(n-1))/(N(N-1))
or can it just be (Σn(n-1))/(N(N-1))? And by JUST having (Σn(n-1))/(N(N-1)), will it just mean the values are interpreted the other way? (i.e. the smaller the value the higher the biodiversity)
• I guess it's just easier to understand with the 1- so that the higher the value the higher the diversity.
(1 vote)
• Is this related to standard deviation or is it different since it has to do with multiple data points?
(1 vote)
• / species EH
(1 vote)