1. **Problem statement:** We have an aircraft's fuel consumption rate that follows a normal distribution with mean $\mu = 8000$ kg and standard deviation $\sigma = 500$ kg. We want to find: a. The probability that a randomly chosen flight consumes more than 8500 kg of fuel. b. The fuel consumption value below which 90% of flights fall. --- 2. **Key formulas and concepts:** - For a normal distribution $X \sim N(\mu, \sigma^2)$, the standardized variable (z-score) is: $$ z = \frac{X - \mu}{\sigma} $$ - The probability that $X$ is greater than a value $x$ is: $$ P(X > x) = 1 - P(Z \leq z) $$ where $P(Z \leq z)$ is the cumulative distribution function (CDF) of the standard normal distribution. - To find the value $x_p$ below which $p$ percent of data fall, use the inverse CDF (quantile function): $$ x_p = \mu + z_p \sigma $$ where $z_p$ is the z-score corresponding to cumulative probability $p$. --- 3. **Part a: Probability that consumption > 8500 kg** Calculate the z-score: $$ z = \frac{8500 - 8000}{500} = \frac{500}{500} = 1 $$ Using standard normal tables or a calculator, the CDF at $z=1$ is approximately 0.8413. Therefore, $$ P(X > 8500) = 1 - 0.8413 = 0.1587 $$ So, there is about a 15.87% chance that a flight consumes more than 8500 kg. --- 4. **Part b: Fuel consumption value below which 90% of flights fall** We want $x$ such that: $$ P(X \leq x) = 0.90 $$ From standard normal tables, the z-score for 0.90 cumulative probability is approximately $z = 1.2816$. Calculate $x$: $$ x = \mu + z \sigma = 8000 + 1.2816 \times 500 = 8000 + 640.8 = 8640.8 $$ So, 90% of flights consume less than approximately 8641 kg of fuel. --- **Final answers:** - a. $P(X > 8500) \approx 0.1587$ - b. Fuel consumption value below which 90% fall is approximately 8641 kg.