The national average of car insurance is \( \$ 1,427 \) per year and standard deviation is \( \$ 354 \). Alex paid \( \$ 1,781 \) a year for his car insurance. What percentage of people paid more than Alex did on their car insurance? (Just put a number between 0 and 100 without \( \% \) unit)

To find the percentage of people who paid more than Alex for car insurance, we can use the properties of the normal distribution. We will first calculate the z-score for Alex's car insurance payment. The z-score is calculated using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where: - \( X \) is the value we are interested in (Alex's payment), - \( \mu \) is the mean (national average), - \( \sigma \) is the standard deviation. Given: - \( X = 1781 \) - \( \mu = 1427 \) - \( \sigma = 354 \) Substituting the values into the formula: \[ z = \frac{(1781 - 1427)}{354} = \frac{354}{354} = 1 \] Now, we need to find the percentage of people who paid more than Alex. This corresponds to finding the area to the right of the z-score of 1 in the standard normal distribution. Using a standard normal distribution table or calculator, we find the area to the left of \( z = 1 \): \[ P(Z < 1) \approx 0.8413 \] This means that approximately 84.13% of people paid less than Alex. To find the percentage of people who paid more than Alex, we subtract this value from 1: \[ P(Z > 1) = 1 - P(Z < 1) \approx 1 - 0.8413 = 0.1587 \] To convert this to a percentage, we multiply by 100: \[ 0.1587 \times 100 \approx 15.87 \] Thus, the percentage of people who paid more than Alex is approximately: \[ \boxed{15.87} \]