Question 17 of 33 The Normal distribution with mean \( \mu=500.9 \) and standard deviation \( \sigma=10.6 \) is a good description of the total score on the Medical College Admission Test (MCAT). This is a continuous probability model for the score of a randomly chosen student. Call the score of a randomly chosen student \( X \) for short. Side-by-Side (b) Use Table A to find the probability of the event "the student chosen has a score of 510 or higher." Give your answer to four decimal places. probability: \( \square \)

1. We are given that \( X \sim N(500.9, 10.6^2) \) and we need to find the probability \[ P(X \ge 510). \] 2. To compute this probability, we first standardize the variable \( X \) using the transformation \[ Z = \frac{X - \mu}{\sigma}. \] For \( X = 510 \), \[ Z = \frac{510 - 500.9}{10.6} \approx \frac{9.1}{10.6} \approx 0.8585. \] 3. Using Table A (the standard normal table), we find the cumulative probability for \( Z = 0.8585 \). The table gives \[ P(Z < 0.8585) \approx 0.8051. \] 4. Since \[ P(X \ge 510) = 1 - P(X < 510) = 1 - P(Z < 0.8585), \] we have \[ P(X \ge 510) \approx 1 - 0.8051 = 0.1949. \] 5. Thus, the probability that a randomly chosen student scores 510 or higher is approximately \[ \boxed{0.1949}. \]