Answer: 0.1056

## Explanation Given: - X ~ N(3000, 400²) - Normal distribution with mean μ = 3000 and standard deviation σ = 400 - We need to find P(X > 3500) **Step 1: Standardize the variable** \[Z = \frac{X - \mu}{\sigma} = \frac{3500 - 3000}{400} = \frac{500}{400} = 1.25\] **Step 2: Find the probability** \[P(X > 3500) = P(Z > 1.25)\] **Step 3: Use the standard normal table** From the standard normal table: - P(Z < 1.25) = 0.8944 - Therefore: P(Z > 1.25) = 1 - P(Z < 1.25) = 1 - 0.8944 = 0.1056 **Verification:** - The standard normal distribution is symmetric - About 68% of values lie within ±1 standard deviation - About 95% of values lie within ±2 standard deviations - Since 3500 is 1.25 standard deviations above the mean, the probability of exceeding this value should be relatively small, which matches our calculated result of 0.1056 (about 10.56%).

Author: Tanishq Prabhu