Let \(X\) be the number of days George goes swimming in 270 days. \(X\) follows a binomial distribution with parameters \(n = 270\) and \(p = \frac{1}{3}\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(np, npq)\), where \(q = 1 - p\).

Calculate \(np\) and \(npq\):

\(np = 270 \times \frac{1}{3} = 90\)

\(npq = 270 \times \frac{1}{3} \times \frac{2}{3} = 60\)

We use a continuity correction to approximate \(P(X > 100)\) as \(P(X > 99.5)\).

Standardize using the normal distribution:

\(P(X > 99.5) = P\left( z > \frac{99.5 - 90}{\sqrt{60}} \right) = P(z > 1.2264)\)

Using the standard normal distribution table, find \(P(z > 1.2264) = 1 - P(z < 1.2264) = 1 - 0.8899 = 0.110\).