The number of days Julie's train is late follows a binomial distribution with parameters:

- Number of trials, \(n = 90\)

- Probability of success (train being late), \(p = 0.3\)

The mean \(\mu\) and variance \(\sigma^2\) of the distribution are given by:

\(\mu = np = 90 \times 0.3 = 27\)

\(\sigma^2 = np(1-p) = 90 \times 0.3 \times 0.7 = 18.9\)

We approximate the binomial distribution with a normal distribution \(N(27, 18.9)\).

To find \(P(X > 35)\), apply continuity correction:

\(P(X > 35) = 1 - \Phi \left( \frac{35.5 - 27}{\sqrt{18.9}} \right)\)

\(= 1 - \Phi(1.955) = 0.0253\)

To find \(P(X < 27)\), apply continuity correction:

\(P(X < 27) = \Phi \left( \frac{26.5 - 27}{\sqrt{18.9}} \right)\)

\(= \Phi(-0.115) = 0.4542\)

The total probability is:

\(P(X > 35) + P(X < 27) = 0.0253 + 0.4542 = 0.480\)