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 \( P(X<7.98 \) OR \( X>7.99) \) 0.9072 0.0928 0.6915 0.5987 None of the above. ** Note: Do not round any z -score.

To find \( P(X < 7.98 \) OR \( X > 7.99) \), we can break this down into two parts and use the z-score formula. 1. For \( P(X < 7.98) \): \[ z = \frac{7.98 - 8}{0.04} = \frac{-0.02}{0.04} = -0.5 \] Looking up \( z = -0.5 \) in the standard normal distribution table gives approximately \( P(Z < -0.5) \approx 0.3085 \). 2. For \( P(X > 7.99) \): \[ z = \frac{7.99 - 8}{0.04} = \frac{-0.01}{0.04} = -0.25 \] Looking up \( z = -0.25 \) gives approximately \( P(Z < -0.25) \approx 0.4013 \), so \( P(Z > -0.25) = 1 - 0.4013 = 0.5987 \). Now we combine the two parts: \[ P(X < 7.98 \text{ OR } X > 7.99) = P(X < 7.98) + P(X > 7.99) = 0.3085 + 0.5987 = 0.9072. \] So the correct answer is **0.9072**. Now onto some additional information: Ever wonder why we rely on the normal distribution? It’s fascinating! This distribution is central to the field of statistics due to the Central Limit Theorem, which tells us that in many situations, the average of a sufficiently large number of observations will be normally distributed, regardless of the original distribution of the data. This helps in making predictions and performance evaluations in various fields! In practical terms, understanding the normal distribution is crucial when assessing quality control processes in manufacturing. For instance, a company can use it to identify when a batch of products has variations that fall outside of acceptable limits. By monitoring the mean and standard deviation, businesses can proactively address production issues before they result in significant losses or customer complaints. So, keep an eye on your z-scores; they might save your day!