The probability that Ana is on time on any given day is:

\(p = 1 - 0.05 - 0.75 = 0.2\)

We are looking for the probability that she is on time on fewer than 20 out of 96 days. This is a binomial distribution problem with parameters \(n = 96\) and \(p = 0.2\).

We approximate the binomial distribution with a normal distribution:

\(\mu = np = 96 \times 0.2 = 19.2\)

\(\sigma^2 = np(1-p) = 96 \times 0.2 \times 0.8 = 15.36\)

Using continuity correction, we find:

\(P(X < 20) = P\left( Z < \frac{19.5 - 19.2}{\sqrt{15.36}} \right) = P(Z < 0.07654)\)

Using standard normal distribution tables or a calculator, we find:

\(P(Z < 0.07654) = 0.531\)