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. In centimeters (cm) and accurate to at most 4 decimal places, the \( 59^{\text {th }} \) percentile of all lengths has value 8.00455 8.0091 7.9909 7.99545 None of the above. Note: Round each of your z-scores to 3 decimal places.

We model the length \( X \) as \[ X \sim N(8, 0.04^2). \] We seek the \( 59^{\text{th}} \) percentile, i.e., the value \( x_{0.59} \) that satisfies \[ P(X \leq x_{0.59}) = 0.59. \] 1. Standardize the variable: \[ Z = \frac{X - 8}{0.04} \quad \Rightarrow \quad P\left(Z \leq \frac{x_{0.59} - 8}{0.04}\right) = 0.59. \] Let \( z_{0.59} = \frac{x_{0.59} - 8}{0.04} \). 2. Find \( z_{0.59} \) using a standard normal table or calculator. We solve: \[ \Phi(z_{0.59}) = 0.59. \] Referring to a standard normal table (or using a calculator) we find: \[ z_{0.59} \approx 0.227. \] 3. Convert back to the original variable: \[ x_{0.59} = 8 + 0.04 \times 0.227 = 8 + 0.00908 \approx 8.0091. \] Thus, the \( 59^{\text{th}} \) percentile of all lengths is approximately \( 8.0091\,\text{cm} \).