Oucstion 5 of 7 (3 points) I Question Attempt 1 of 5 \( P(X>21)= \) Single Americans In a recent year, about \( 22 \% \) of Americans 14 years and older are single. What is the probability that in a random sample of 150 Americans 14 or older, more than 21 are single? Round the final answer to at least 4 decimal places and intermediate \( z \)-value calculations to 2 decimal places. P1)

Let \( X \) be the number of single Americans in a sample of 150. We have \[ X \sim \text{Binomial}(n=150,\; p=0.22) \] The mean and variance of \( X \) are \[ \mu = np = 150 \times 0.22 = 33 \quad \text{and} \quad \sigma^2 = np(1-p) = 150 \times 0.22 \times 0.78 \approx 25.74. \] Thus, the standard deviation is \[ \sigma \approx \sqrt{25.74} \approx 5.07. \] We wish to compute \[ P(X>21). \] Using the normal approximation with a continuity correction, we approximate \( P(X>21) \) as \[ P(X>21) \approx P\left(Z > \frac{21.5-\mu}{\sigma}\right) \] where \[ \frac{21.5-33}{5.07} \approx \frac{-11.5}{5.07} \approx -2.27. \] Thus, \[ P(X>21) \approx P(Z > -2.27). \] Using the symmetry of the normal distribution, \[ P(Z > -2.27) = 1 - P(Z \leq -2.27) = \Phi(2.27). \] From standard normal tables, we find \[ \Phi(2.27) \approx 0.9884. \] Therefore, the probability that more than 21 Americans in the sample are single is approximately \[ \boxed{0.9884}. \]