Use the normal approximation to the binomial to find the probability for \( n=50, p=0.6 \), and \( X=33 \). Use The Standard Normal Distribution Table. Round \( z \) - value calculations to 2 decimal places and final answer to 4 decimal places. The probability is

Let \( X \) be the binomial random variable with parameters \( n = 50 \) and \( p = 0.6 \). The mean and standard deviation are \[ \mu = np = 50 \times 0.6 = 30, \] \[ \sigma = \sqrt{npq} = \sqrt{50 \times 0.6 \times 0.4} = \sqrt{12} \approx 3.4641. \] Using the normal approximation for \( P(X = 33) \), we apply the continuity correction. This gives \[ P(X = 33) \approx P(32.5 < Y < 33.5). \] Standardize the bounds using \[ z = \frac{y - \mu}{\sigma}. \] For the lower bound: \[ z_1 = \frac{32.5 - 30}{3.4641} \approx \frac{2.5}{3.4641} \approx 0.72. \] For the upper bound: \[ z_2 = \frac{33.5 - 30}{3.4641} \approx \frac{3.5}{3.4641} \approx 1.01. \] Thus, the probability is \[ P(32.5 < Y < 33.5) \approx P(0.72 < Z < 1.01) = \Phi(1.01) - \Phi(0.72), \] where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution. Using the Standard Normal Distribution Table: \[ \Phi(1.01) \approx 0.8438, \] \[ \Phi(0.72) \approx 0.7642. \] Subtracting, \[ P(0.72 < Z < 1.01) \approx 0.8438 - 0.7642 = 0.0796. \] Thus, the probability is \[ \boxed{0.0796}. \]