Question

A pharmaceutical company wants to test a new cholesterol drug. The average cholesterol of the target population is 200 mg and they have a standard deviation of 25 mg. If a member of the target population has a probability of 93.06% of having a population of Zmg cholesterol. What is the target member’s cholesterol, Z?

Answer 1

To find the cholesterol level \( Z \) for which a member of the target population has a probability of 93.06% of having a cholesterol level less than \( Z \), we can use the concepts of the normal distribution.

Given that:

- The average cholesterol level (mean, \( \mu \)) is 200 mg.
- The standard deviation (\( \sigma \)) is 25 mg.

We know that the Z-score formula for a normal distribution is given by:

\[
Z = \frac{X - \mu}{\sigma}
\]

where:
- \( X \) is the value we want to find (the cholesterol level, in this case),
- \( \mu \) is the mean, and
- \( \sigma \) is the standard deviation.

We're looking for \( Z \) such that:

\[
P(X < Z) = 0.9306
\]

To find \( Z \), we first need to find the corresponding Z-score from the standard normal distribution for the cumulative probability of 0.9306. You may look this up in Z-tables or use a statistical software/tool.

The Z-score that corresponds to a cumulative probability of 0.9306 is approximately:

\[
Z \approx 1.48
\]

Now we can convert this Z-score back into the cholesterol level using the Z-score formula by rearranging it:

\[
Z = \mu + Z \cdot \sigma
\]

Substituting our values:

\[
Z = 200 + 1.48 \cdot 25
\]

Calculating the right side gives:

\[
Z = 200 + 37
\]
\[
Z = 237
\]

So, the target member's cholesterol level \( Z \) is approximately **237 mg**.