Nominal income refers to an income value that is not adjusted for inflation. Real income adjusts the nominal value for the rate of inflation. Real income per capita is considered a good measure of the buying power of an individual in a particular area because it adjusts for both the number of people in the area and the relative inflation of the area. According to Forbes magazine, the state with the highest annual real income per capita in 2020 is Connecticut with a real per capita income of \( \$ 68,533 \). Suppose that the annual real income of individuals in Connecticut follows a normal distribution with a mean of \( \$ 68,533 \) and standard deviation of \( \$ 23,480 \). (a) What is the probability that an individual in Connecticut has an annual real income of \( \$ 51,000 \) or more? (Round your answer to four decimal places.) (b) What is the probability that an individual in Connecticut has an annual real income of \( \$ 26,000 \) or less? (Round your answer to four decimal places.) (c) What is the probability that an individual in Connecticut has an annual real income between \( \$ 56,000 \) and \( \$ 76,000 \) ? (Round your answer to four decimal places.) (d) What is the annual real income of a person (in dollars) in the 99 th percentile of annual real income in Connecticut? (Round your answer to the nearest dollar.) \( \$ \$ \). (d)

**(a)** Let \( X \) be the annual real income, with \[ X \sim N(\mu=68533,\ \sigma=23480). \] We want \[ P(X \ge 51000). \] First, compute the \( z \)‐score: \[ z = \frac{51000 - 68533}{23480} \approx \frac{-17533}{23480} \approx -0.7466. \] Using the symmetry of the normal distribution, \[ P(X \ge 51000) = P(Z \ge -0.7466) = P(Z \le 0.7466). \] Looking up (or computing) the cumulative probability, \[ P(Z \le 0.7466) \approx 0.7734. \] **Answer (a):** \(0.7734\) --- **(b)** We need to find \[ P(X \le 26000). \] Compute the \( z \)-score: \[ z = \frac{26000 - 68533}{23480} \approx \frac{-42533}{23480} \approx -1.812. \] Thus, \[ P(X \le 26000) = P(Z \le -1.812). \] From standard normal tables or a calculator, \[ P(Z \le -1.812) \approx 0.0350. \] **Answer (b):** \(0.0350\) --- **(c)** We want the probability that \[ 56000 \le X \le 76000. \] First, compute the \( z \)-score for \(56000\): \[ z_1 = \frac{56000 - 68533}{23480} \approx \frac{-12533}{23480} \approx -0.5344. \] Next, compute the \( z \)-score for \(76000\): \[ z_2 = \frac{76000 - 68533}{23480} \approx \frac{7467}{23480} \approx 0.3180. \] The probability is then: \[ P(56000 \le X \le 76000) = P(z_1 \le Z \le z_2) = \Phi(0.3180) - \Phi(-0.5344). \] Using symmetry, note that \[ \Phi(-0.5344) = 1 - \Phi(0.5344). \] Assuming from tables/calculators: \[ \Phi(0.3180) \approx 0.6255 \quad \text{and} \quad \Phi(0.5344) \approx 0.7030, \] so that \[ \Phi(-0.5344) \approx 1 - 0.7030 = 0.2970. \] Thus, \[ P(56000 \le X \le 76000) \approx 0.6255 - 0.2970 = 0.3285. \] Rounded to four decimal places, this is \(0.3285\). **Answer (c):** \(0.3285\) --- **(d)** We require the income at the \(99^{\text{th}}\) percentile, i.e., find \( x \) such that \[ P(X \le x) = 0.99. \] Let \( z_{0.99} \) be the corresponding standard normal value. From tables, \[ z_{0.99} \approx 2.3263. \] Then, \[ x = \mu + z_{0.99}\cdot \sigma \approx 68533 + 2.3263 \times 23480. \] Compute the product: \[ 2.3263 \times 23480 \approx 54621.5. \] So, \[ x \approx 68533 + 54621.5 \approx 123154.5. \] Rounded to the nearest dollar, the \(99^{\text{th}}\) percentile income is \[ \$123155. \] **Answer (d):** \(\$123155\)