(ii) To approximate the probability, we use the normal approximation to the binomial distribution. Given that the sample size is 500, we calculate:

Mean: \(np = 340\)

Variance: \(npq = 108.8\)

We standardize the variable using continuity correction:

\(P(x > 337) = P\left( z > \frac{337.5 - 340}{\sqrt{108.8}} \right)\)

\(= P(z > -0.2396)\)

Using the standard normal distribution table, we find:

\(P(z > -0.2396) = 0.595\)

(iii) The normal approximation is justified because both \(np = 340\) and \(nq = 160\) are greater than 5.