Use the table to answer the question. Note: Round z-scores to the nearest hundredth and then find the required A values using the table.
The cholesterol levels of a group of young women at a university are normally distributed, with a mean of 188 and a standard deviation of 35. What percent of the young women have the following cholesterol levels? (Round your answers to one decimal place.)
(a) greater than 216
%
(b) between 190 and 220
%
To find the percentage of young women with cholesterol levels greater than 216, we first need to find the z-score for 216:
z = (216 - 188) / 35 = 0.80
Looking up a z-score of 0.80 in the table, we find that the area to the left of this z-score is 0.7881. So, the percentage of young women with cholesterol levels greater than 216 is:
100% - 78.81% = 21.19%
Therefore, 21.19% of young women have cholesterol levels greater than 216.
Next, to find the percentage of young women with cholesterol levels between 190 and 220, we need to find the z-scores for both values:
For 190:
z = (190 - 188) / 35 = 0.06
For 220:
z = (220 - 188) / 35 = 0.91
Looking up the z-scores in the table, we find the following areas:
- For z = 0.06, the area to the left is 0.5239
- For z = 0.91, the area to the left is 0.8186
Therefore, the percentage of young women with cholesterol levels between 190 and 220 is:
81.86% - 52.39% = 29.47%
So, 29.47% of young women have cholesterol levels between 190 and 220.