1. The problem asks to find the values of $z_1$ and $z_2$ such that the area under the standard normal curve between $z_1$ and $z_2$ is 0.8812. 2. The standard normal distribution is symmetric about zero, so if $P(z_1 < Z < z_2) = 0.8812$, then $z_1 = -z_2$. 3. The total area under the curve is 1, so the area outside the interval is $1 - 0.8812 = 0.1188$. 4. Since the distribution is symmetric, the area in each tail is $0.1188 / 2 = 0.0594$. 5. We look for the $z$-value such that the cumulative area to the left is $0.0594$ in the lower tail, which corresponds to $z_1$, and the cumulative area to the left is $1 - 0.0594 = 0.9406$ for $z_2$. 6. Using standard normal distribution tables or a calculator, the $z$-value for cumulative area $0.0594$ is approximately $-1.56$, and for $0.9406$ is approximately $1.56$. 7. Therefore, the values are $z_1 = -1.56$ and $z_2 = 1.56$. Final answer: $z_1 = -1.56$ and $z_2 = 1.56$.