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)

To solve for \( P(X > 21) \) where \( X \) is the number of single Americans in a random sample of 150, we first identify relevant values. Given that \( p = 0.22 \) (the proportion of single Americans), we can use the normal approximation to the binomial distribution since our sample size is large. The mean \( \mu \) and standard deviation \( \sigma \) can be calculated as: \[ \mu = np = 150 \times 0.22 = 33 \] \[ \sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.22 \times 0.78} \approx 6.51 \] Next, we convert \( X = 21 \) to a z-score: \[ z = \frac{X - \mu}{\sigma} = \frac{21 - 33}{6.51} \approx -1.84 \] Now, we find \( P(X > 21) = 1 - P(Z \leq -1.84) \). Looking up the z-value in the standard normal distribution table gives us \( P(Z \leq -1.84) \approx 0.0329 \). Therefore: \[ P(X > 21) = 1 - 0.0329 = 0.9671 \] So, rounding to four decimal places, \( P(X > 21) \approx 0.9671 \). The final answer is: \[ P(X > 21) \approx 0.9671 \]