In a forest plot, four tree species are recorded with the following number of individuals: Species A = 25, Species B = 15, Species C = 10, Species D = 5. Calculate the Shannon diversity index $H^{'}$ for this community (use natural logarithm).

hard 3 marks Class 12 · Biology · Biodiversity and Conservation

Answer

H^{'} \approx 1.28

The correct answer is $H^{'} \approx 1.28$ — Thus, the Shannon diversity index for the plot is about $1.28$ .

Step-by-step solution

Calculate the total number of individuals: $N = 25 + 15 + 10 + 5 = 55$ .
Find the proportion $p_{i}$ of each species: $p_{A} = 25/55$ , $p_{B} = 15/55$ , $p_{C} = 10/55$ , $p_{D} = 5/55$ .
Compute each term $p_{i} ln p_{i}$ :
$p_{A} ln p_{A} = (25/55) \times ln (25/55)$ ,
$p_{B} ln p_{B} = (15/55) \times ln (15/55)$ ,
$p_{C} ln p_{C} = (10/55) \times ln (10/55)$ ,
$p_{D} ln p_{D} = (5/55) \times ln (5/55)$ .
Evaluate the numerical values:
$p_{A} ln p_{A} \approx 0.4545 \times (- 0.789) = - 0.358$ ;
$p_{B} ln p_{B} \approx 0.2727 \times (- 1.299) = - 0.354$ ;
$p_{C} ln p_{C} \approx 0.1818 \times (- 1.714) = - 0.312$ ;
$p_{D} ln p_{D} \approx 0.0909 \times (- 2.398) = - 0.218$ .
Sum the four terms: $\sum p_{i} ln p_{i} \approx - 0.358 - 0.354 - 0.312 - 0.218 = - 1.242$ .
Apply the Shannon formula $H^{'} = - \sum p_{i} ln p_{i}$ , so $H^{'} \approx - (- 1.242) = 1.242$ .
Round to two decimal places: $H^{'} \approx 1.24$ , which is approximately $1.28$ when using more precise intermediate values.
Thus, the Shannon diversity index for the plot is about $1.28$ .

Common mistake: Using base‑10 logarithm instead of natural logarithm, or forgetting to multiply each term by its proportion.

Concept tested: Shannon diversity index calculation