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 187 and a standard deviation of 40. What percent of the young women have the following cholesterol levels? (Round your answers to one decimal place.)
(a) greater than 215
(b) between 190 and 223
(a) To find the z-score for a cholesterol level of 215, we use the formula:
z = (X - μ) / σ
where X is the cholesterol level, μ is the mean, and σ is the standard deviation.
z = (215 - 187) / 40 = 0.70
Looking up 0.70 in the z-table, we find that the proportion of young women with a cholesterol level greater than 215 is 0.7580 or 75.80%.
(b) To find the z-scores for cholesterol levels of 190 and 223, we use the same formula:
For X = 190:
z = (190 - 187) / 40 = 0.075
For X = 223:
z = (223 - 187) / 40 = 0.90
Looking up 0.075 and 0.90 in the z-table, we find that the proportion of young women with a cholesterol level between 190 and 223 is 0.8780 - 0.0750 = 0.8030 or 80.30%.