The cholesterol levels of young women aged 20 to 34 vary approximately normal with mean 185 milligrams per deciliter and standard deviation 39 mg/dl. Cholesterol levels for men vary with mean 222mg/dl and standard deviation 37 mg/dl Sandy's cholesterol level is 220 her fathers is level is 250 who's is higher
Z = (score-mean)/SD
Compare Z scores.
To determine whose cholesterol level is higher, we need to compare Sandy's cholesterol level to her father's cholesterol level.
Given:
For young women aged 20 to 34:
Mean (μ) = 185 mg/dl
Standard deviation (σ) = 39 mg/dl
For men:
Mean (μ) = 222 mg/dl
Standard deviation (σ) = 37 mg/dl
Sandy's cholesterol level = 220 mg/dl
Father's cholesterol level = 250 mg/dl
To compare their cholesterol levels, we can calculate the z-scores for both Sandy and her father, and compare them. The z-score indicates how many standard deviations away from the mean a particular data point is.
For Sandy:
z-score = (x - μ) / σ
z-score = (220 - 185) / 39
z-score = 35 / 39
z-score ≈ 0.897
For her father:
z-score = (x - μ) / σ
z-score = (250 - 222) / 37
z-score = 28 / 37
z-score ≈ 0.757
Since Sandy's z-score is higher (0.897 > 0.757), we can conclude that her cholesterol level is relatively higher than her father's cholesterol level.
To determine whose cholesterol level is higher between Sandy and her father, we need to compare their levels to the respective means of their age and gender groups.
Given that Sandy's cholesterol level is 220 mg/dl and her father's level is 250 mg/dl, we can compare them to the mean cholesterol levels for women and men, respectively.
For Sandy:
- Mean cholesterol level for young women (20-34 years old) = 185 mg/dl
For her father:
- Mean cholesterol level for men = 222 mg/dl
To determine who has a higher cholesterol level, we need to calculate the number of standard deviations each value is from its respective mean.
For Sandy:
Standard deviation for young women = 39 mg/dl
Z-score = (x - mean) / standard deviation = (220 - 185) / 39 ≈ 0.897
For her father:
Standard deviation for men = 37 mg/dl
Z-score = (x - mean) / standard deviation = (250 - 222) / 37 ≈ 0.757
Comparing the Z-scores, we see that Sandy's Z-score is higher (0.897) than her father's Z-score (0.757). This indicates that Sandy's cholesterol level is farther away from the mean for her age group (young women) compared to her father's cholesterol level from the mean for his gender group (men).
Therefore, Sandy's cholesterol level of 220 mg/dl is higher (relatively speaking) compared to her father's cholesterol level of 250 mg/dl.