The problem involves a binomial distribution with parameters:

- Number of trials, \(n = 100\)

- Probability of success (asleep), \(p = 0.2\)

We approximate the binomial distribution with a normal distribution:

- Mean, \(\mu = np = 100 \times 0.2 = 20\)

- Variance, \(\sigma^2 = np(1-p) = 100 \times 0.2 \times 0.8 = 16\)

- Standard deviation, \(\sigma = \sqrt{16} = 4\)

We want to find \(P(X \leq 30)\). Using continuity correction, we find \(P(X \leq 30.5)\).

Standardizing, we calculate the z-score:

\(z = \frac{30.5 - 20}{4} = 2.625\)

Using the standard normal distribution table, we find:

\(P(Z < 2.625) = 0.996\)