Question

In a randomly selected sample of women ages 20–34, the mean total cholesterol level is 188 milligrams per deciliter with a standard deviation of 41.3 milligrams per deciliter. Assume the total cholesterol levels are normally distributed. Find the highest total cholesterol level a woman in this 20–34 age group can have and still be in the bottom 1%. (Round your answers to two decimal places)

Answer 1

Answer:

$z=-2.33<\frac{a-188}{41.3}$

And if we solve for a we got

$a=188 -2.33*41.3=91.771$

The highest total cholesterol level a woman in this 20–34 age group can have and still be in the bottom 1% is 91.771 mg per deciliter

Step-by-step explanation:

Let X the random variable that represent the cholesterol level of a population of women between 20-34, and for this case we know the distribution for X is given by:

$X \sim N(188,41.3)$  

Where $\mu=188$ and $\sigma=41.3$

We want to find the highest value for the bottom 1% in the distribution, so we need to find a value a who satisfy the following conditions:

$P(X>a)=0.99$   (a)

$P(X<a)=0.01$   (b)

We can find a z value that satisfy the condition with 0.01 of the area on the left and 0.99 of the area on the right it's z=-2.33. And we can verify that on this case P(Z<-2.33)=0.01 and P(z>-2.33)=0.099

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.01$  

$P(z<\frac{a-\mu}{\sigma})=0.01$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-2.33<\frac{a-188}{41.3}$

And if we solve for a we got

$a=188 -2.33*41.3=91.771$

The highest total cholesterol level a woman in this 20–34 age group can have and still be in the bottom 1% is 91.771 mg per deciliter

Answer 2

Using the normal distribution, it is found that the highest total cholesterol level a woman in this 20–34 age group can have and still be in the bottom 1% is of 91.89 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 188 mg/dl, hence $\mu = 188$
• Standard deviation of 41.3 mg/dl, hence $\sigma = 41.3$

The highest total cholesterol level a woman in this 20–34 age group can have and still be in the bottom 1% is the 1st percentile, which is X when Z has a p-value of 0.99, so X when Z = -2.327.

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

$-2.327 = \frac{X - 188}{41.3}$

$X - 188 = -2.327(41.3)$

$X = 91.89$

The highest total cholesterol level a woman in this 20–34 age group can have and still be in the bottom 1% is of 91.89 mg/dl.

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