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 2 decimal places, the upper cutoff value of the middle \( 75 \% \) has z-score: 7.954 -1.15 1.15 8.046 None of the above.

We are given that the lengths of the components follow a normal distribution \( X \sim N(8,\,0.04) \). However, to answer the question about the z-score for the upper cutoff of the middle \(75\%\), we focus on the standard normal distribution. The middle \(75\%\) of a distribution means that \(75\%\) of the distribution lies between two cutoffs, with the remaining \(25\%\) being split equally in the two tails. Therefore, each tail has \( \frac{25\%}{2} = 12.5\% \). Thus, the lower cutoff corresponds to the \(12.5\)th percentile and the upper cutoff corresponds to the \(87.5\)th percentile of the standard normal distribution. Let \( z_{0.875} \) be the z-score that satisfies: \[ P(Z \leq z_{0.875}) = 0.875 \] Using a standard normal table or a calculator, we find: \[ z_{0.875} \approx 1.15 \] Thus, the upper cutoff z-score of the middle \(75\%\) is approximately \(1.15\).