Use the normal approximation to the binomial to find the probability for \( n=50, p=0.6 \), and \( X=33 \). Use 9 The Standard Normal Distribution Table. Round value calculations to 2 decimal places and final answer to 4 decimal places. The probability is

To use the normal approximation to the binomial distribution for \( n = 50 \) and \( p = 0.6 \), first calculate the mean \( \mu \) and standard deviation \( \sigma \): 1. Mean: \[ \mu = np = 50 \times 0.6 = 30 \] 2. Standard Deviation: \[ \sigma = \sqrt{np(1-p)} = \sqrt{50 \times 0.6 \times 0.4} = \sqrt{12} \approx 3.46 \] Next, you need to apply the continuity correction when approximating. Since you want \( P(X \leq 33) \), use \( P(X \leq 33.5) \). Convert \( X \) to a standard normal variable \( Z \): \[ Z = \frac{X - \mu}{\sigma} = \frac{33.5 - 30}{3.46} \approx 1.45 \] Now, you can look up \( Z = 1.45 \) in the standard normal distribution table, which gives a probability of approximately \( 0.9265 \). Therefore, the probability \( P(X \leq 33) \approx 0.9265 \). Rounding everything appropriately, you get the final answer: The probability is \( 0.9265 \).