\( X \) is a normally distributed random variable with mean 35 and standard deviation 11 . What is the probability that \( X \) is greater than 55 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is greater than 55 for a normally distributed random variable with mean \( \mu = 35 \) and standard deviation \( \sigma = 11 \), we can use the standard normal distribution (Z-score). ### Step 1: Calculate the Z-score The Z-score is calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \] Substituting the values: \[ Z = \frac{55 - 35}{11} = \frac{20}{11} \approx 1.8182 \] ### Step 2: Find the probability Next, we need to find the probability that \( Z \) is greater than 1.8182. This can be found using the standard normal distribution table or a calculator. \[ P(Z > 1.8182) = 1 - P(Z \leq 1.8182) \] ### Step 3: Look up the cumulative probability Using a standard normal distribution table or calculator, we find: \[ P(Z \leq 1.8182) \approx 0.9656 \] ### Step 4: Calculate the final probability Now we can calculate: \[ P(Z > 1.8182) = 1 - 0.9656 = 0.0344 \] ### Final Answer Thus, the probability that \( X \) is greater than 55 is approximately: \[ \boxed{0.034} \] This value is rounded to the nearest thousandth.