1. **State the problem:** We want to find the probability that a newborn giraffe's height $X$ is between 82 and 90 inches, given that $X$ is normally distributed with mean $\mu=88$ inches and standard deviation $\sigma=4$ inches. 2. **Formula and rules:** For a normal distribution, the probability that $X$ lies between $a$ and $b$ is given by $$P(a \leq X \leq b) = P\left(\frac{a-\mu}{\sigma} \leq Z \leq \frac{b-\mu}{\sigma}\right) = \Phi\left(\frac{b-\mu}{\sigma}\right) - \Phi\left(\frac{a-\mu}{\sigma}\right)$$ where $Z$ is the standard normal variable and $\Phi(z)$ is the cumulative distribution function (CDF) of the standard normal distribution. 3. **Calculate the z-scores:** $$z_1 = \frac{82 - 88}{4} = \frac{-6}{4} = -1.5$$ $$z_2 = \frac{90 - 88}{4} = \frac{2}{4} = 0.5$$ 4. **Find the probabilities from the standard normal table:** From the table, $$\Phi(-1.5) = 0.0668$$ $$\Phi(0.5) = 0.6915$$ 5. **Calculate the probability between 82 and 90 inches:** $$P(82 \leq X \leq 90) = \Phi(0.5) - \Phi(-1.5) = 0.6915 - 0.0668 = 0.6247$$ 6. **Interpretation:** There is approximately a 62.47% chance that a newborn giraffe's height will be between 82 and 90 inches. **Final answer:** $$\boxed{0.6247}$$