Let \(X\) be the number of days Wenjie goes out with her friends. \(X\) follows a binomial distribution with \(n = 252\) and \(p = \frac{1}{7}\).

For normal approximation, calculate the mean \(\mu\) and variance \(\sigma^2\):

\(\mu = np = 252 \times \frac{1}{7} = 36\)

\(\sigma^2 = npq = 252 \times \frac{1}{7} \times \frac{6}{7} = 30.857\)

Standard deviation \(\sigma = \sqrt{30.857}\).

Using continuity correction, find:

\(P(X < 30) = P\left( z < \frac{29.5 - 36}{\sqrt{30.857}} \right)\)

\(P(X > 44) = P\left( z > \frac{44.5 - 36}{\sqrt{30.857}} \right)\)

Calculate the z-scores:

\(z_1 = \frac{29.5 - 36}{\sqrt{30.857}} = -1.170\)

\(z_2 = \frac{44.5 - 36}{\sqrt{30.857}} = 1.530\)

Using standard normal distribution tables:

\(P(z < -1.170) = 0.1209\)

\(P(z > 1.530) = 0.0630\)

Total probability: \(P(X < 30) + P(X > 44) = 0.1209 + 0.0630 = 0.184\)

For part (ii), the use of a normal approximation is justified because both \(np\) and \(nq\) are greater than 5.