Let the random variable \(X\) represent the number of plants producing pink flowers in a box of 100. \(X\) follows a binomial distribution with parameters \(n = 100\) and \(p = 0.3\).

To approximate this binomial distribution, we use a normal distribution with mean \(\mu\) and variance \(\sigma^2\).

Calculate the mean: \(\mu = np = 100 \times 0.3 = 30\).

Calculate the variance: \(\sigma^2 = np(1-p) = 100 \times 0.3 \times 0.7 = 21\).

Thus, \(X \sim N(30, 21)\).

We need to find \(P(X < 35)\). Using a continuity correction, this becomes \(P(X < 34.5)\).

Standardize the variable: \(Z = \frac{34.5 - 30}{\sqrt{21}} = \frac{4.5}{\sqrt{21}} \approx 0.9820\).

Using the standard normal distribution table, \(\Phi(0.9820) \approx 0.837\).

Therefore, the probability that a box contains fewer than 35 plants which produce pink flowers is \(0.837\).

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.

Let the random variable \(X\) represent the number of residents who rate the service as 'good'. \(X\) follows a binomial distribution with parameters \(n = 125\) and \(p = 0.52\).

To approximate this binomial distribution, we use a normal distribution with mean \(\mu = np = 0.52 \times 125 = 65\) and variance \(\sigma^2 = np(1-p) = 0.52 \times 0.48 \times 125 = 31.2\).

Thus, \(X \sim N(65, 31.2)\).

We need to find \(P(X > 72)\). Using a continuity correction, this is approximated by \(P(X > 72.5)\).

Standardize the variable: \(Z = \frac{X - \mu}{\sigma} = \frac{72.5 - 65}{\sqrt{31.2}} \approx 1.343\).

Thus, \(P(X > 72.5) = P(Z > 1.343) = 1 - P(Z \leq 1.343)\).

Using standard normal distribution tables, \(P(Z \leq 1.343) \approx 0.9104\).

Therefore, \(P(Z > 1.343) = 1 - 0.9104 = 0.0896\).

Hence, the probability that more than 72 residents rate the service as 'good' is approximately between 0.0896 and 0.0897.