Respuesta :
Answer:
0.2514 = 25.14% of women have HDL below 45 mg/dl or less.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean HDL cholesterol levels of 55mg/dl with a standard deviation of 15 mg/dl.
This means that [tex]\mu = 55, \sigma = 15[/tex]
What proportion of women have HDL below 45 mg/dl or less?
This is the p-value of Z when X = 45. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{45 - 55}{15}[/tex]
[tex]Z = -0.67[/tex]
[tex]Z = -0.67[/tex] has a p-value of 0.2514
0.2514 = 25.14% of women have HDL below 45 mg/dl or less.
