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. Without using the normalcdf(...) function, evaluate \( P(7.88 \leq X \leq 8.04) \) 1.677 0.8385 0.1615 0.3385 0.677

We start with the given normal distribution for \( X \) having mean \(\mu = 8\) cm and standard deviation \(\sigma = 0.04\) cm. We are asked to find \[ P(7.88 \leq X \leq 8.04). \] **Step 1. Standardize the variable** Convert the \( X \)-values to \( Z \)-scores using \[ Z = \frac{X-\mu}{\sigma}. \] For \( X = 7.88 \): \[ Z = \frac{7.88 - 8}{0.04} = \frac{-0.12}{0.04} = -3. \] For \( X = 8.04 \): \[ Z = \frac{8.04 - 8}{0.04} = \frac{0.04}{0.04} = 1. \] Thus, the probability becomes \[ P(7.88 \leq X \leq 8.04) = P(-3 \leq Z \leq 1). \] **Step 2. Use the Standard Normal Distribution** The probability \(P(-3 \leq Z \leq 1)\) can be expressed as: \[ P(-3 \leq Z \leq 1) = P(Z \leq 1) - P(Z \leq -3). \] From standard normal tables: - \(P(Z \leq 1) \approx 0.8413\), - \(P(Z \leq -3) \approx 0.00135\). Thus, \[ P(-3 \leq Z \leq 1) \approx 0.8413 - 0.00135 = 0.83995 \approx 0.84. \] **Step 3. Match with Given Options** The option that is closest to \(0.84\) is \[ 0.8385. \]