Using mean=np, variance=np(1-p) where n=30 (sample size), p=0.66, we get mean=19.8, variance=6.732, standard deviation=√variance=2.5946 approx.
If X=15, then the Z-score is (15-19.8)/2.5946=-1.85 approx. corresponding to 0.0322, which means that there is a 3.22% probability that there will be less than 15 females in the group of 30. Therefore, there’s a 96.78% probability that there will be at least 15 females. (This compares favourably with 97.72% obtained through binomial distribution tables.)