(1) Use the normal approximation to the binomial to find the probability for \( n=10, p=0.5 \), and \( X \geq 7 \). Use \( \Theta \) The Standard Normal Distribution Table. Round \( z \) - value calculations to 2 decimal places and final answer to 4 decimal places. The probability is

We are given a binomial random variable \( X \) with parameters \( n = 10 \) and \( p = 0.5 \). We want to approximate \[ P(X \geq 7) \] using the normal approximation to the binomial. The steps are as follows. 1. The mean \( \mu \) and variance \( \sigma^2 \) of the binomial distribution are given by \[ \mu = np \quad \text{and} \quad \sigma^2 = np(1-p). \] For \( n = 10 \) and \( p = 0.5 \): \[ \mu = 10 \times 0.5 = 5, \quad \sigma^2 = 10 \times 0.5 \times 0.5 = 2.5. \] Thus, the standard deviation is \[ \sigma = \sqrt{2.5} \approx 1.58. \] 2. We use the continuity correction for the inequality \( X \geq 7 \). Since the normal approximation is continuous, we approximate \[ P(X \geq 7) \approx P\left( X \geq 6.5 \right). \] 3. Standardize using the transformation \[ z = \frac{x - \mu}{\sigma}. \] For \( x = 6.5 \): \[ z = \frac{6.5 - 5}{1.58} \approx \frac{1.5}{1.58} \approx 0.95. \] 4. Now, we find the probability using the standard normal distribution: \[ P(Z \geq 0.95) = 1 - \Phi(0.95). \] From the standard normal distribution table, we have \[ \Phi(0.95) \approx 0.8289. \] Thus, \[ P(Z \geq 0.95) \approx 1 - 0.8289 = 0.1711. \] 5. Rounding the final answer to four decimal places, we get: \[ \boxed{0.1711}. \]