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 \( P(7.88 \leq X \leq 7.99) \) 0.0013 0.3999 0.4013 0.6001 None of the above. ? Note: Do not round any z-score.

Let \( X \) be the length of one component, where \[ X \sim N(8,\;0.04^2). \] We need to evaluate \[ P(7.88 \leq X \leq 7.99). \] **Step 1. Standardize the variable** Define \[ Z = \frac{X - 8}{0.04}. \] Then, \[ P(7.88 \leq X \leq 7.99) = P\left(\frac{7.88-8}{0.04} \leq Z \leq \frac{7.99-8}{0.04}\right). \] **Step 2. Compute the z-scores** For \( X = 7.88 \): \[ z_{\text{low}} = \frac{7.88 - 8}{0.04} = \frac{-0.12}{0.04} = -3. \] For \( X = 7.99 \): \[ z_{\text{high}} = \frac{7.99 - 8}{0.04} = \frac{-0.01}{0.04} = -0.25. \] Thus, \[ P(7.88 \leq X \leq 7.99) = P(-3 \leq Z \leq -0.25). \] **Step 3. Express using the standard normal CDF** This probability can be written as: \[ P(-3 \leq Z \leq -0.25) = \Phi(-0.25) - \Phi(-3), \] where \(\Phi(z)\) is the cumulative distribution function (CDF) of the standard normal distribution. **Step 4. Use symmetry properties of the standard normal** Recall that: \[ \Phi(-a) = 1 - \Phi(a). \] So, \[ \Phi(-0.25) = 1 - \Phi(0.25) \quad \text{and} \quad \Phi(-3) = 1 - \Phi(3). \] From standard normal tables: \[ \Phi(0.25) \approx 0.5987 \quad \text{and} \quad \Phi(3) \approx 0.9987. \] Thus, \[ \Phi(-0.25) \approx 1 - 0.5987 = 0.4013, \] \[ \Phi(-3) \approx 1 - 0.9987 = 0.0013. \] **Step 5. Calculate the probability** Substitute these values into the probability expression: \[ P(-3 \leq Z \leq -0.25) \approx 0.4013 - 0.0013 = 0.4000 - 0.0010 = 0.3999. \] **Final Answer** \[ P(7.88 \leq X \leq 7.99) \approx 0.3999. \] Thus, the correct answer from the options is \(0.3999\).