Given that \( z \) is a standard normal random variable, compute the following probabilities. (Round your answers to decimal places.) (a) \( P(0 \leq z \leq 0.84) \) (b) \( P(-1.59 \leq z \leq 0) \) (c) \( P(z>0.42) \) (d) \( P(z \geq-0.22) \) (e) \( P(z<1.60) \) (f) \( P(z \leq-0.74) \)

### (a) \( P(0 \leq z \leq 0.84) \) 1. Recall that for the standard normal variable, \[ P(0 \leq z \leq 0.84) = \Phi(0.84) - \Phi(0) \] 2. Since \(\Phi(0) = 0.5\) and from the table \(\Phi(0.84) \approx 0.7995\), \[ P(0 \leq z \leq 0.84) \approx 0.7995 - 0.5 = 0.2995 \] 3. Rounded to two decimal places, the answer is approximately \(0.30\). --- ### (b) \( P(-1.59 \leq z \leq 0) \) 1. Write the probability as: \[ P(-1.59 \leq z \leq 0) = \Phi(0) - \Phi(-1.59) \] 2. Since \(\Phi(0) = 0.5\) and using the symmetry of the normal distribution, \[ \Phi(-1.59) = 1 - \Phi(1.59) \] 3. From tables, \(\Phi(1.59) \approx 0.9441\). Thus, \[ \Phi(-1.59) \approx 1 - 0.9441 = 0.0559 \] 4. Therefore, \[ P(-1.59 \leq z \leq 0) \approx 0.5 - 0.0559 = 0.4441 \] 5. Rounded to two decimal places, the answer is approximately \(0.44\). --- ### (c) \( P(z > 0.42) \) 1. The probability is given by: \[ P(z > 0.42) = 1 - \Phi(0.42) \] 2. From the table, \(\Phi(0.42) \approx 0.6628\), so: \[ P(z > 0.42) \approx 1 - 0.6628 = 0.3372 \] 3. Rounded to two decimal places, the answer is approximately \(0.34\). --- ### (d) \( P(z \geq -0.22) \) 1. For a continuous variable, \[ P(z \geq -0.22) = 1 - P(z < -0.22) = 1 - \Phi(-0.22) \] 2. Using the symmetry property: \[ \Phi(-0.22) = 1 - \Phi(0.22) \] 3. From tables, \(\Phi(0.22) \approx 0.5871\). Thus, \[ \Phi(-0.22) \approx 1 - 0.5871 = 0.4129 \] 4. Therefore, \[ P(z \geq -0.22) \approx 1 - 0.4129 = 0.5871 \] 5. Rounded to two decimal places, the answer is approximately \(0.59\). --- ### (e) \( P(z < 1.60) \) 1. This probability is simply: \[ P(z < 1.60) = \Phi(1.60) \] 2. From the standard normal table, \(\Phi(1.60) \approx 0.9452\). 3. Rounded to two decimal places, the answer is approximately \(0.95\). --- ### (f) \( P(z \leq -0.74) \) 1. Express the probability as: \[ P(z \leq -0.74) = \Phi(-0.74) \] 2. Using the symmetry of the normal distribution: \[ \Phi(-0.74) = 1 - \Phi(0.74) \] 3. From tables, \(\Phi(0.74) \approx 0.7704\). Thus, \[ \Phi(-0.74) \approx 1 - 0.7704 = 0.2296 \] 4. Rounded to two decimal places, the answer is approximately \(0.23\).