Question

The Shannon-Weaver index 𝐻 is a measure of diversity and is calculated as H= -∑𝑝𝑖 𝑙𝑛(𝑝𝑖) where S is the total number of species and 𝑝𝑖 is the proportional abundance of a species i.The table below gives the abundance of different species in a community. The Shannon-Weaver index of reptile diversity in this community is ___________.(Rounded off to two decimal places)Species Abundance Indian gliding lizard 270 Malabar flying frog 325 Travancore tortoise 180 Malabar hornbill 160 Forest cane turtle 120 Malabar pit viper 30

Hint:

When compositions are constrained by categories, count choices within each category using combinations and then multiply — the multiplication rule for independent selections.

Answer 1

Correct Answer: 1.18

Step 1: Identify the focal group (reptiles) and total counts.

\[ N_{\text{reptiles}} = 270 + 180 + 120 + 30 = 600 \]

Step 2: Convert abundances to proportions.

\[ p_{\text{lizard}} = \tfrac{270}{600}=0.45,\quad p_{\text{tortoise}} = \tfrac{180}{600}=0.30,\quad p_{\text{turtle}} = \tfrac{120}{600}=0.20,\quad p_{\text{viper}} = \tfrac{30}{600}=0.05 \] 

Step 3: Compute Shannon terms \(p_i \ln p_i\).

\[ \begin{aligned} 0.45 \ln 0.45 &\approx -0.3593, \\ 0.30 \ln 0.30 &\approx -0.3612, \\ 0.20 \ln 0.20 &\approx -0.3219, \\ 0.05 \ln 0.05 &\approx -0.1498 \end{aligned} \] 

Step 4: Sum and apply the negative sign.

\[ \sum p_i \ln p_i \approx -1.1922 \] \[ H = -\sum p_i \ln p_i \approx 1.1922 \] 

Step 5: Round and sanity check.

\[ \boxed{H \approx 1.19} \] Maximum possible for \(S=4\) is \(\ln 4 \approx 1.386\). Our value is slightly lower, consistent with uneven abundances.

Final Answer: \[ H \approx 1.19 \]