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 \( P(X>7.89) \) 0.994 0.997 0.003 0.006 None of the above. ** Note: Do not round any z-score.

To evaluate \( P(X > 7.89) \), we first need to find the corresponding z-score using the formula: \[ z = \frac{X - \mu}{\sigma} \] Given \( \mu = 8 \) cm and \( \sigma = 0.04 \) cm, we have: \[ z = \frac{7.89 - 8}{0.04} = \frac{-0.11}{0.04} = -2.75 \] Now, we will find \( P(Z > -2.75) \). Looking up the z-score of -2.75 in a standard normal distribution table, we find: \[ P(Z < -2.75) \approx 0.003 \] Therefore, \[ P(Z > -2.75) = 1 - P(Z < -2.75) \approx 1 - 0.003 = 0.997 \] So, the answer is **0.997**. Now, onto some additional fun facts! If you've ever marveled at the concept of "normal distribution," it actually appears in many fascinating real-life scenarios, from human heights to test scores! This means that a significant number of natural phenomena follow this bell-shaped curve, allowing statisticians and researchers to utilize it for predictions and analyses. What’s your height? You might just belong to the realm of the “average”! Understanding z-scores can be a game-changer in various fields, especially in quality control and manufacturing. They allow companies to determine how far a value is from the mean and assess the risk of producing components that fall outside of desired specifications. So, the next time you enjoy a well-crafted product, know that statistics played a critical role behind the scenes!