SOLUTION: how it solves this problem? 5. Diastolic blood pressures are assumed to follow a normal distribution with a mean of 85 and a standard deviation of 12 a. What proportion of p

Question 304692: how it solves this problem?
5. Diastolic blood pressures are assumed to follow a normal distribution with a mean of 85 and a standard deviation of 12
a. What proportion of people have disastolic blood pressure less than 90% ?

b. What proportion have diatolic blood pressure between 80 and 90% ?

c. If someone has a diatolic blood pressure of 100. what is percentile does represent?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
You can use this tool to help you calculate the answer to this problem.

http://davidmlane.com/hyperstat/z_table.html

In order to use the tool, you need to have java enabled on you computer.

To answer these questions, you would use the top graph and enter a mean of 85 and a standard deviation of 12 where indicated.

I believe your question should be worded as follows:

My answers are below each of the parts of the question.

5. Diastolic blood pressures are assumed to follow a normal distribution with a mean of 85 and a standard deviation of 12.

a. What proportion of people have diastolic blood pressure less than 90?

The proportion of the people with diastolic blood pressure less than 90 would be .661539 or 66.13539%.

Using the tool, you would select "below" and enter a 90 in the box.

b. What proportion have diastolic blood pressure between 80 and 90?

The proportion of the people with diastolic blood pressure between 80 and 90 would be .323078 or 32.3078%

Using the tool, you would select "between" and enter 80 in the first box and 90 in the second box.

c. If someone has a diastolic blood pressure of 100. what percentile does that represent?

The proportion of the people with diastolic blood pressure of 100 and above would be .105650 or 10.5650%

Using the tool, you would select "above" and enter 90 in the box.

A picture of the three graphs is shown below:

***** picture not found *****

Without the use of this tool, or a tool like it, you would need to access the Z-Tables and manually calculate the proportions required.

This is considerably more cumbersome and definitely not recommended unless you absolutely have to use the tables.

In that case, you may want to look at the lesson referenced below:

http://www.algebra.com/algebra/homework/Probability-and-statistics/change-this-name28108.lesson
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