The mean serum cholesterol of a large population of overweight adults is \( 210 \mathrm{mg} / \mathrm{dl} \) and the standard deviation of \( 19 \mathrm{mg} / \mathrm{dl} \). If the sample of 40 adults is selected, find the probability that the mean will be greater than \( 210 \mathrm{mg} / \mathrm{dl} \).

To find the probability that the sample mean of cholesterol will be greater than \( 210 \mathrm{mg}/\mathrm{dl} \), we can use the Central Limit Theorem. For a sample size of \( n = 40 \), the standard error (SE) is calculated as: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{19}{\sqrt{40}} ≈ 3.0 \mathrm{mg}/\mathrm{dl}. \] Since the sample mean's distribution is normal (thanks to the Central Limit Theorem), we can standardize the sample mean of \( 210 \mathrm{mg}/\mathrm{dl} \) to find our z-score: \[ z = \frac{\bar{x} - \mu}{SE} = \frac{210 - 210}{3.0} = 0. \] Now, we look up the z-score of \( 0 \) in the standard normal distribution, which gives us a probability of \( 0.5 \). This means there is a \( 50\% \) chance that the mean of a sample of \( 40 \) adults will be greater than \( 210 \mathrm{mg}/\mathrm{dl} \).