Answer:
- <em>About 34% of women will have levels between:</em>
<em> </em><u><em> 171 mg/dl and 207 mg/dl </em></u>
Explanation:
You might find several intervals of <em>cholesterol levels</em> that contain about 34% of the women.
The easiest way is to find a symmetric interval. This is, with the same number of women below and above the mean.
Then, if <em>34% of the women</em> are within the interval, 100% - 34% = 66% are out of the interval.
For a symmetric interval, half of 66% would be above the median and half-below the median.
Thus, 33% above and 33% below the median.
Now, you can look up the Z-score in a standard normal distribution table.
There are two types of standard distribution tables: tables that show values that represent the AREA to the LEFT of the Z-score, and tables that show values that represent the AREA to the RIGHT of the Z-score.
Using the second, find the Z-score for a probability of 33%, i.e. 0.33, it is Z-score = 0.44.
That means that the interval must be - 0.44 < Z-score < 0.44
Now that you have the Z-score you can find the cholesterol levels:
For the upper level:




For the lower level:


Rounding to whole numbers the interval would be between 171 mg/dl and 207 mg/dl.