This is a binomial distribution because there are only two states. Let n=950, p=0.19 and q=1-p=0.81. mean, μ, =np=180.5, standard deviation, σ =√(npq)=12.09 approx. 25% of 950=237.5.
The binomial distribution is approximated by the normal distribution.
In a normal distribution Z-score, Z=(μ-X)/σ, where X=237.5. However, if we want 25% exactly using the normal distribution we have to calculate Z-scores for 237 and 238.
Z1=(237-180.5)/12.09=4.673, Z2=(238-180.5)/12.09=4.755 corresponding to 0.999998516 and 0.9999990078, so the difference is 0.00000049=0.00005% approx. This is very close to a zero probability.