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 \( P(X<7.99 \) OR \( X>8.1) \) 0.0062 0.4075 0.5987 0.5925 None of the above. ** Note: Do not round any z-score.

1. The lengths are normally distributed with mean \( \mu = 8 \) cm and standard deviation \( \sigma = 0.04 \) cm. 2. We need to compute \[ P(X < 7.99 \text{ or } X > 8.1). \] Since these two events are mutually exclusive, we have: \[ P(X < 7.99 \text{ or } X > 8.1) = P(X < 7.99) + P(X > 8.1). \] 3. **For \( X < 7.99 \):** Compute the \( z \)-score: \[ z = \frac{7.99 - 8}{0.04} = \frac{-0.01}{0.04} = -0.25. \] Thus, \[ P(X < 7.99) = \Phi(-0.25). \] By the symmetry of the normal distribution: \[ \Phi(-0.25) = 1 - \Phi(0.25). \] Typically, \( \Phi(0.25) \approx 0.5987 \), so: \[ \Phi(-0.25) \approx 1 - 0.5987 = 0.4013. \] 4. **For \( X > 8.1 \):** Compute the \( z \)-score: \[ z = \frac{8.1 - 8}{0.04} = \frac{0.1}{0.04} = 2.5. \] Thus, \[ P(X > 8.1) = 1 - \Phi(2.5). \] Typically, \( \Phi(2.5) \approx 0.9938 \), so: \[ P(X > 8.1) \approx 1 - 0.9938 = 0.0062. \] 5. **Combine the probabilities:** \[ P(X < 7.99 \text{ or } X > 8.1) \approx 0.4013 + 0.0062 = 0.4075. \] Thus, the correct answer is \( 0.4075 \).