6. To the left of \( z=0.72 \) \( \begin{array}{ll}\text { 8. To the right of } z=-0.84 & \text { 7. To the left of } z=-3.08 \\ \text { 10. Between } z=-1.22 \text { and } & \text { 11. To the left of } z=0.12 \text { or to } \\ z=-0.26 & \text { the right of } z=1.72\end{array} \)

**6. To the left of \( z=0.72 \)** We need to find \[ P(Z < 0.72) = \Phi(0.72). \] Reading the standard normal table, \[ \Phi(0.72) \approx 0.7642. \] --- **7. To the left of \( z=-3.08 \)** We find \[ P(Z < -3.08) = \Phi(-3.08). \] Because \(-3.08\) is far in the left tail, the probability is very small. The table gives approximately \[ \Phi(-3.08) \approx 0.0010. \] --- **8. To the right of \( z=-0.84 \)** We need \[ P(Z > -0.84) = 1 - \Phi(-0.84). \] Using the symmetry of the normal distribution, note that \[ \Phi(-0.84) = 1 - \Phi(0.84). \] For \( z=0.84 \), the table gives \( \Phi(0.84) \approx 0.7995 \). Therefore, \[ \Phi(-0.84) \approx 1 - 0.7995 = 0.2005, \] and hence, \[ P(Z > -0.84) \approx 1 - 0.2005 = 0.7995. \] --- **10. Between \( z=-1.22 \) and \( z=-0.26 \)** We calculate \[ P(-1.22 < Z < -0.26) = \Phi(-0.26) - \Phi(-1.22). \] Using the symmetry property, we write \[ \Phi(-a) = 1 - \Phi(a). \] For \( z=0.26 \), suppose \( \Phi(0.26) \approx 0.6026 \); then \[ \Phi(-0.26) \approx 1 - 0.6026 = 0.3974. \] For \( z=1.22 \), if \( \Phi(1.22) \approx 0.8888 \); then \[ \Phi(-1.22) \approx 1 - 0.8888 = 0.1112. \] Thus, \[ P(-1.22 < Z < -0.26) \approx 0.3974 - 0.1112 = 0.2862. \] --- **11. To the left of \( z=0.12 \) or to the right of \( z=1.72 \)** This probability is given by \[ P(Z < 0.12 \text{ or } Z > 1.72) = P(Z < 0.12) + P(Z > 1.72). \] First, \[ P(Z < 0.12) = \Phi(0.12). \] Assume \( \Phi(0.12) \approx 0.5478 \). Next, \[ P(Z > 1.72) = 1 - \Phi(1.72). \] For \( z=1.72 \), suppose \( \Phi(1.72) \approx 0.9573 \); then \[ P(Z > 1.72) \approx 1 - 0.9573 = 0.0427. \] Thus, \[ P(Z < 0.12 \text{ or } Z > 1.72) \approx 0.5478 + 0.0427 = 0.5905. \]