8. In a randomly selected sample of women ages 20-34, the mean total cholesterol level is $190 \mathrm{mg}$ per deciliter with a standard deviation of $40.1 \mathrm{mg} / \mathrm{dl}$. Assume the total cholesterol levels are normally distributed. Find the cut off cholesterol level a woman in this 20-34 age group can have to be in the bottom $10 \%$.

Step 1 ：Given that the mean total cholesterol level is \(190 \, \mathrm{mg/dl}\) and the standard deviation is \(40.1 \, \mathrm{mg/dl}\). We are asked to find the cholesterol level that a woman in the 20-34 age group can have to be in the bottom 10%.

Step 2 ：We know that the cholesterol levels are normally distributed. In a normal distribution, the z-score can be used to find the value that corresponds to a certain percentile. The z-score is a measure of how many standard deviations a value is from the mean.

Step 3 ：The formula for z-score is: \(Z = \frac{(X - \mu)}{\sigma}\), where \(Z\) is the z-score, \(X\) is the value from the dataset (cholesterol level in this case), \(\mu\) is the mean of the dataset, and \(\sigma\) is the standard deviation of the dataset.

Step 4 ：We need to find the z-score that corresponds to the bottom 10% of the population. This can be found using a z-table or a statistical function, which gives a z-score of approximately -1.2816.

Step 5 ：We can rearrange the z-score formula to solve for \(X\), the cholesterol level: \(X = Z\sigma + \mu\).

Step 6 ：Substituting the z-score, the mean, and the standard deviation into the formula, we get: \(X = -1.2816 \times 40.1 + 190\), which simplifies to approximately 138.6.

Step 7 ：Final Answer: The cut off cholesterol level a woman in this 20-34 age group can have to be in the bottom 10% is approximately \(\boxed{138.6 \, \mathrm{mg/dl}}\).