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, the cutoff of the bottom \( 41 \% \) has value: 7.99545 7.9909 8.0091 8.00455 None of the above. Note: Round each of your z-scores to 3 decimal places.

We are given that the lengths \( X \) follow a normal distribution with \[ \mu = 8 \quad \text{and} \quad \sigma = 0.04. \] We need to find the cutoff value \( c \) such that \[ P(X \le c) = 0.41. \] **Step 1. Standardize the variable** Let \[ Z = \frac{X - \mu}{\sigma}. \] Then \[ P(X \le c) = P\left(Z \le \frac{c-8}{0.04}\right) = 0.41. \] **Step 2. Find the corresponding \( z \)-score** We have \[ P(Z \le z_c) = 0.41. \] Using a standard normal table or calculator (rounding the \( z \)-score to 3 decimal places), we typically find \[ z_c \approx -0.227. \] **Step 3. Solve for \( c \)** Now, convert back to the original variable: \[ z_c = \frac{c-8}{0.04} \quad \Rightarrow \quad c = 8 + 0.04 \cdot z_c. \] Substitute \( z_c \approx -0.227 \): \[ c = 8 + 0.04(-0.227) \approx 8 - 0.00908 = 7.99092. \] Rounding accurate to 4 decimal places, we get \[ c \approx 7.9909. \] Thus, the cutoff value is \[ \boxed{7.9909}. \]