1. The problem asks to find the proportion of the area under the normal curve for a specified region and to sketch the normal curve with that region shaded. 2. The normal distribution is given by the probability density function: $$f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$ where $\mu$ is the mean and $\sigma$ is the standard deviation. 3. The area under the normal curve between two points corresponds to the probability that a normally distributed random variable falls within that range. 4. To find the proportion of the area, we use the standard normal distribution $Z = \frac{X-\mu}{\sigma}$ and look up the cumulative probabilities in the standard normal table or use technology. 5. Since the problem does not specify the exact region or parameters, assume the standard normal curve ($\mu=0$, $\sigma=1$) and a region, for example, $Z \leq 1$. 6. The proportion of area to the left of $Z=1$ is: $$P(Z \leq 1) = \Phi(1) \approx 0.8413$$ 7. This means about 84.13% of the area under the curve lies to the left of $Z=1$. 8. The sketch would show the bell curve centered at 0 with the area to the left of 1 shaded. Final answer: The proportion of area under the standard normal curve to the left of $Z=1$ is approximately 0.8413.