
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: \[ H \approx 1.19 \]

