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 \( P(7.93 \leq X \leq 7.95) \) 0.0401 0.1056 0.0656 0.9344 None of the above. **Note: Do not round any z-score.

We are given that the component lengths are normally distributed as \[ X \sim N \left(8,\, 0.04\right) \] with mean \(\mu = 8\) and standard deviation \(\sigma = 0.04\). **Step 1: Standardize the variable** The \(z\)-score transformation is given by \[ z = \frac{x - \mu}{\sigma} \] For the lower bound \(x = 7.93\): \[ z_1 = \frac{7.93 - 8}{0.04} = \frac{-0.07}{0.04} = -1.75 \] For the upper bound \(x = 7.95\): \[ z_2 = \frac{7.95 - 8}{0.04} = \frac{-0.05}{0.04} = -1.25 \] **Step 2: Express the required probability in terms of \(z\)** We want to find: \[ P(7.93 \leq X \leq 7.95) = P(-1.75 \leq Z \leq -1.25) \] This can be written as: \[ P(-1.75 \leq Z \leq -1.25) = \Phi(-1.25) - \Phi(-1.75) \] where \(\Phi(z)\) is the cumulative distribution function (CDF) of the standard normal distribution. **Step 3: Use standard normal distribution values** From the standard normal table (accurate to 4 decimal places): \[ \Phi(-1.25) \approx 0.1056 \] \[ \Phi(-1.75) \approx 0.0401 \] **Step 4: Compute the final probability** Subtract the CDF values: \[ P(-1.75 \leq Z \leq -1.25) = 0.1056 - 0.0401 = 0.0655 \] Rounded to 4 decimal places, the probability is: \[ 0.0656 \] Thus, the correct answer is \(\boxed{0.0656}\).