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:

$z=-2.33$

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)

$P(X$   (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:

$P(X$  

$P(z$

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

$z=-2.33$

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