1. The problem asks us to find the area under the standard normal curve for various Z-value ranges. 2. We use the provided Z-table, where each entry represents the area from the mean (Z = 0) to the positive Z value. --- **1. Area from Z = 0 to Z = 1.35:** - Locate Z = 1.3 in the left column and 0.05 in the top row. - From the table, area = 0.4115. **2. P(-2.05 \u2264 Z \u2264 1.96):** - Find area from mean to 2.05: - Z = 2.0 and 0.05: area = 0.4798. - Find area from mean to 1.96: - Z = 1.9 and 0.06: area = 0.4750. - Find area from mean to 2.05 negative side, same as positive side area: 0.4798. - Probability = area from -2.05 to 0 plus area from 0 to 1.96 = 0.4798 + 0.4750 = 0.9548. **3. P(Z \u2264 -1.47):** - Area to left of negative Z = area to the right of positive 1.47 (by symmetry). - Find area mean to 1.4 and subtract mean to 0.03 for Z=1.47 approx: - Z=1.4 and 0.00:0.4192 - Z=0.0 and 0.03:0.0120 - Area mean to 1.47 = 0.4192 + 0.0120 (since 1.47 is between 1.4 and 1.5 approx) ~0.4312 - Left tail = 0.5 - 0.4312 = 0.0688. **4. Area to left of Z = 2.01:** - Z=2.0 and 0.01: 0.4778 (area from mean to 2.01) - Area left = 0.5 + 0.4778 = 0.9778. **5. P(-3.1 \u2264 Z \u2264 -1.27):** - Find area mean to 3.1: approximately 0.4990 - Find area mean to 1.27 approximately: - Z=1.2 and 0.07:0.3980 - Z=1.3 and 0.00:0.4032 - Interpolate for 1.27 ~ 0.4006 - Area between -3.1 to -1.27 = (0.5 - 0.4006) - (0.5 - 0.4990) = 0.0994 - 0.0010 = 0.0984 --- **Final Answers:** 1. 0.4115 2. 0.9548 3. 0.0688 4. 0.9778 5. 0.0984