A brokerage survey reports that $25 \%$ of all individual investors have used a discount broker (one that does not charge the full commission). If a random sample of 65 individual investors is taken, approximate the probability that more than 14 have used a discount broker. Use the normal approximation to the binomial with a correction for continuity.
Round your answer to at least three decimal places. Do not round any intermediate steps.
(If necessary, consult a list of formulas.)

Step 1 ：This problem is a binomial problem where we are asked to find the probability of more than a certain number of successes, in this case, using a discount broker. However, since the sample size is large, we can use the normal approximation to the binomial distribution to simplify the problem. The normal approximation is appropriate when both np and n(1-p) are greater than 5, which is the case here.

Step 2 ：The mean (\(\mu\)) of the distribution is np and the standard deviation (\(\sigma\)) is \(\sqrt{np(1-p)}\). We are asked to find P(X > 14), but we need to apply a continuity correction because we are approximating a discrete distribution with a continuous one. So we actually need to find P(X > 14.5).

Step 3 ：To find this probability, we first convert 14.5 to a z-score, which is (X - \(\mu\)) / \(\sigma\). Then we find the area to the right of this z-score in the standard normal distribution, which is 1 - P(Z < z).

Step 4 ：Given that n = 65, p = 0.25, and X = 14.5, we can calculate \(\mu\) = np = 16.25 and \(\sigma\) = \(\sqrt{np(1-p)}\) = 3.491060010942235.

Step 5 ：Then, we calculate the z-score as z = (X - \(\mu\)) / \(\sigma\) = -0.5012804118276031.

Step 6 ：Finally, we find the probability as 1 - P(Z < z) = 0.6919131054913978.

Step 7 ：The approximate probability that more than 14 out of 65 individual investors have used a discount broker is \(\boxed{0.692}\).