Let the random variable \(X\) represent the number of callers connected immediately. \(X\) follows a binomial distribution with parameters \(n = 80\) and \(p = 0.9\).

To approximate, use a normal distribution: \(X \sim N(\mu, \sigma^2)\).

Calculate the mean: \(\mu = np = 80 \times 0.9 = 72\).

Calculate the variance: \(\sigma^2 = npq = 80 \times 0.9 \times 0.1 = 7.2\).

Use continuity correction for \(P(X > 69)\):

\(P(X > 69) = P(Z > \frac{69.5 - 72}{\sqrt{7.2}})\).

Calculate the standard score:

\(Z = \frac{69.5 - 72}{\sqrt{7.2}} = -0.9317\).

Find the probability using the standard normal distribution:

\(P(Z > -0.9317) = \Phi(0.9317) \approx 0.824\).

For part (c), justify the approximation:

Both \(np = 72\) and \(nq = 8\) are greater than 5, so the normal approximation is valid.