Question

4.10 Find the SD. Cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 185 milligrams per deciliter (mg/dl). Women with cholesterol levels above 220 mg/dl are con- sidered to have high cholesterol and about 18.5% of women fall into this category. What is the standard deviation of the distribution of cholesterol levels for women aged 20 to 34?

Answer 1

Answer:

Standard deviation = 38.89 ≈ 39 mg/dl

Step-by-step explanation:

Data provided in the question:

Mean of the normal distribution = 185 milligrams per deciliter

Let X denote the cholesterol level of women aged 20 to 34 (in mg/dl)

Therefore,

P( X > 220 ) = 18.5% = 0.185

Now,

P( X > 220 ) = $P(Z > \frac{220-mean}{\textup{Standard deviation}})$

Now,

From the standard Z table $P(Z > \frac{220-mean}{\textup{Standard deviation}})$ = 0.185

we have Z = 0.90

Thus,

$\frac{220-mean}{\textup{Standard deviation}}$ = 0.90

or

$\frac{220-185}{\textup{Standard deviation}}$ = 0.90

or

Standard deviation = $\frac{35}{0.90}$

or

Standard deviation = 38.89 ≈ 39 mg/dl

Answer 2

Using the normal distribution, it is found that the standard deviation is of 39 mg/dl.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It 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, which is the percentile of X.

In this problem:

• Mean of 185 mg/dl, thus, $\mu = 185$.
• 18.5% of women have cholesterol levels above 220 mg/dl, which means that when X = 220, Z has a p-value of 1 - 0.185 = 0.815. Thus, when X = 220, Z = 0.895.

Then, to find $\sigma$:

$Z = \frac{X - \mu}{\sigma}$

$0.895 = \frac{220 - 185}{\sigma}$

$0.895\sigma = 35$

$\sigma = \frac{35}{0.895}$

$\sigma = 39$

The standard deviation is of 39 mg/dl.

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