Homework \( 6-4 \) Ouestion 2 of 7 (3 points) I Question Attempt 1 of 5 Use the normal approximation to the binomial to find the probability for \( n=10, p=0.5 \), and \( X \geq 7 \). Use 9 The Standard Normal Distribution Table. Round \( z \) - value calculations to 2 decimal places and final answer to 4 decimal places. The probability is

1. The binomial distribution has parameters \( n=10 \) and \( p=0.5 \). Its mean and variance are: \[ \mu = np = 10 \times 0.5 = 5, \] \[ \sigma^2 = np(1-p) = 10 \times 0.5 \times 0.5 = 2.5. \] The standard deviation is: \[ \sigma = \sqrt{2.5} \approx 1.58. \] 2. We are asked to approximate the probability \( P(X \geq 7) \) using the normal distribution. Applying the continuity correction, we have: \[ P(X \geq 7) \approx P(Y \geq 6.5), \] where \( Y \) is normally distributed with mean \(\mu = 5\) and standard deviation \(\sigma \approx 1.58\). 3. To find \( P(Y \geq 6.5) \), we convert to the standard normal variable \( Z \) as follows: \[ Z = \frac{Y - \mu}{\sigma} = \frac{6.5 - 5}{1.58} \approx \frac{1.5}{1.58} \approx 0.95. \] 4. Using the standard normal distribution table, the cumulative probability for \( Z = 0.95 \) is approximately: \[ \Phi(0.95) \approx 0.8289. \] Therefore, the probability we are looking for is: \[ P(Z \geq 0.95) = 1 - \Phi(0.95) \approx 1 - 0.8289 = 0.1711. \] 5. The final answer, rounded to four decimal places, is: \[ 0.1711. \]