1. The problem asks to find the probability that the standard normal variable $Z$ lies between 1.02 and 2.03, i.e., $P(1.02 \leq Z \leq 2.03)$. 2. Recall that for a standard normal distribution, probabilities correspond to areas under the curve and can be found using the cumulative distribution function (CDF) $\Phi(z)$. 3. The probability $P(a \leq Z \leq b)$ is given by $\Phi(b) - \Phi(a)$. 4. Using standard normal tables or a calculator, find $\Phi(2.03)$ and $\Phi(1.02)$. 5. From tables, $\Phi(2.03) \approx 0.9788$ and $\Phi(1.02) \approx 0.8461$. 6. Therefore, $P(1.02 \leq Z \leq 2.03) = 0.9788 - 0.8461 = 0.1327$. 7. The final answer is $\boxed{0.1327}$, meaning there is approximately a 13.27% chance that $Z$ falls between 1.02 and 2.03.