The number of times the gymnast performs the routine correctly follows a binomial distribution with parameters \(n = 50\) and \(p = 0.65\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = n \times p = 50 \times 0.65 = 32.5\)

\(\sigma^2 = n \times p \times (1-p) = 50 \times 0.65 \times 0.35 = 11.375\)

We use a continuity correction to find the probability of fewer than 29 occasions:

\(P(X < 29) \approx P\left(Z < \frac{28.5 - 32.5}{\sqrt{11.375}}\right)\)

\(= P\left(Z < \frac{-4}{\sqrt{11.375}}\right)\)

\(= P(Z < -1.186)\)

Using the standard normal distribution table, \(P(Z < -1.186) = 1 - P(Z < 1.186) = 1 - 0.8822 = 0.118\).