The problem involves a binomial distribution with parameters:

- Number of trials, \(n = 250\)

- Probability of success, \(p = 0.65\)

We approximate the binomial distribution with a normal distribution using the following parameters:

- Mean, \(\mu = np = 250 \times 0.65 = 162.5\)

- Variance, \(\sigma^2 = np(1-p) = 250 \times 0.65 \times 0.35 = 56.875\)

- Standard deviation, \(\sigma = \sqrt{56.875}\)

We need to find \(P(X < 179)\). Using the continuity correction, we approximate this as \(P(X < 178.5)\).

Standardize the variable:

\(Z = \frac{178.5 - 162.5}{\sqrt{56.875}} = \frac{16}{\sqrt{56.875}} \approx 2.122\)

Using the standard normal distribution table, find \(P(Z < 2.122)\).

The probability is approximately \(0.983\).