Let A and B be the ASDR of locality A and Locality B P – the standard population Then, SDR(A) =\(\frac{∑PA}{∑P}\)
SDR (B) =\(\frac{∑PB}{∑P}\)

SDR (A) = \(\frac{1945000}{105000}\) = 18.5238
SDR (B) =\(\frac{1900000}{1050000}\) = 18.095
Here, SDR(B) < SDR (A)
Locality B is healthier.