The length of similar components produced by a company are approximated by a normal distribution model with a mean of 8 cm and a standard deviation of 0.04 cm . Let the random variable \( X \) represent the length of one component chosen at random. Accurate to 4 decimal places, evaluate \( P(X>8.08) \) 0.9772 0.0455 0.9545 0.0228 None of the above. ** Note: Do not round any z-score.

To find \( P(X > 8.08) \), let's first determine the z-score using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] Here, \( X = 8.08 \), \( \mu = 8 \), and \( \sigma = 0.04 \): \[ z = \frac{(8.08 - 8)}{0.04} = \frac{0.08}{0.04} = 2 \] Next, we check the z-table (or standard normal distribution table) for \( P(Z > 2) \). The value \( P(Z < 2) \) typically is about 0.9772, so: \[ P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228 \] Final answer: \( P(X > 8.08) \approx 0.0228 \). Did your math brain skip a beat? No worries! Misreading the question or skipping the z-score step is a common oopsie! Brush up on those z-score calculations, and double-check the tables – remember, your z-scores are valuable little tools, guiding you through the wild world of probabilities!