Let \( Z \sim N(0,1) \). Accurate to 4 decimal places, evaluate \( P(Z<-2.08 \) OR \( Z>0.18) \) 0.4286 0.4473 0.0188 0.5527 None of the above.

We are given that \( Z \sim N(0,1) \) and we need to compute \[ P(Z < -2.08 \text{ or } Z > 0.18). \] Since these two events are mutually exclusive, we have \[ P(Z < -2.08 \text{ or } Z > 0.18) = P(Z < -2.08) + P(Z > 0.18). \] **Step 1. Compute \( P(Z < -2.08) \):** By the symmetry of the normal distribution, \[ P(Z < -2.08) = P(Z > 2.08). \] Using a standard normal table or calculator, we find: \[ \Phi(2.08) \approx 0.9812, \] so \[ P(Z > 2.08) = 1 - \Phi(2.08) \approx 1 - 0.9812 = 0.0188. \] **Step 2. Compute \( P(Z > 0.18) \):** Similarly, using a standard normal table, \[ \Phi(0.18) \approx 0.5714, \] so \[ P(Z > 0.18) = 1 - \Phi(0.18) \approx 1 - 0.5714 = 0.4286. \] **Step 3. Sum the two probabilities:** \[ P(Z < -2.08 \text{ or } Z > 0.18) \approx 0.0188 + 0.4286 = 0.4474. \] Rounding to four decimal places, this is approximately \(0.4473\). Thus, the correct answer is: \[ 0.4473. \]