1. **Stating the problem:** We want to find the probability that the sample mean cholesterol level is between 187 and 193. 2. **Given information:** Assume the population mean $\mu$ and population standard deviation $\sigma$ are known, and the sample size is $n$. 3. **Formula used:** The sampling distribution of the sample mean $\bar{x}$ is approximately normal with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$. 4. **Calculate the z-scores:** $$z_1 = \frac{187 - \mu}{\sigma / \sqrt{n}}, \quad z_2 = \frac{193 - \mu}{\sigma / \sqrt{n}}$$ 5. **Find the probability:** $$P(187 < \bar{x} < 193) = P(z_1 < Z < z_2) = \Phi(z_2) - \Phi(z_1)$$ where $\Phi$ is the standard normal cumulative distribution function. 6. **Interpretation:** This probability represents the likelihood that the sample mean cholesterol level falls between 187 and 193. *Note:* Since the problem does not provide $\mu$, $\sigma$, or $n$, the exact numerical answer cannot be computed here.