Let \( Z \sim N(0,1) \). Accurate to 4 decimal places, evaluate \( P(Z \leq-0.5 \) OR \( Z \geq 0.5) \) 0.6915 0.3829 0.6171 0.3085 None of the above.

Let \( Z \sim N(0,1) \). We need to evaluate \[ P\left(Z \leq -0.5 \text{ or } Z \geq 0.5\right) \] **Step 1: Use the symmetry of the normal distribution** Since the normal distribution is symmetric around 0, we have: \[ P\left(Z \leq -0.5 \text{ or } Z \geq 0.5\right) = P\left(|Z| \geq 0.5\right) \] This probability can be expressed as: \[ P\left(|Z| \geq 0.5\right) = 2P\left(Z \geq 0.5\right) \] **Step 2: Use the cumulative distribution function (CDF)** We use the standard normal CDF, \(\Phi(z)\), to find: \[ P\left(Z \geq 0.5\right) = 1 - \Phi(0.5) \] Given that \(\Phi(0.5) \approx 0.6915\), we have: \[ P\left(Z \geq 0.5\right) = 1 - 0.6915 = 0.3085 \] **Step 3: Calculate the final probability** Now, multiply by 2: \[ P\left(|Z| \geq 0.5\right) = 2 \times 0.3085 = 0.6170 \approx 0.6171 \] Thus, the probability \( P\left(Z \leq -0.5 \text{ or } Z \geq 0.5\right) \) is approximately \( 0.6171 \).