SOLUTION: compute the geometric and harmonic means for the following distribution of annual death rates: xi[3.95,4.95,5.95,6.95,7.95,8.95,9.95,10.95,11.95,12.95,13.95] Fi[1 ][ 4][5 ][1

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Question 174301: compute the geometric and harmonic means for the following distribution of annual death rates:
xi[3.95,4.95,5.95,6.95,7.95,8.95,9.95,10.95,11.95,12.95,13.95]
Fi[1 ][ 4][5 ][13 ][12] [19] [13] [10] [6 ] [4 ] [ 1 ]

Answer by Edwin McCravy(20081)   (Show Source): You can put this solution on YOUR website!
compute the geometric and harmonic means for the following distribution of annual death rates:
xi[3.95,4.95,5.95,6.95,7.95,8.95,9.95,10.95,11.95,12.95,13.95
You just have to learn some rules about the different kinds of averaging:

You didn't ask about arithmetic mean, but we have to use it
to find the harmonic mean.

How to find the arithmetic mean of some numbers:

1. Count them.
2. Add them all together.
3. Divide the result of 2 by the result of 1.

How to find the geometric mean of some numbers:

1. Count them.
2. Multiply them all together.
3. Take the root corresponding to 1 of the result of 1.

How to find the harmonic mean of some numbers:

1. Take the reciprocals of all the numbers
2. Take the arithmetic mean of the results of 1
3. Take the reciprocal of the result of 2.

3.95,4.95,5.95,6.95,7.95,8.95,9.95,10.95,11.95,12.95,13.95

To find the geometric mean:

1. Count them.

There are 11 of them

2. Multiply them all together.

(3.95)(4.95)(5.95)(6.95)(7.95)(8.95)(9.95)(10.95)(11.95)(12.95)(13.95)
= 1.353141797 × 1010.

3. Take the root corresponding to 1 of the result of 1.

Since we counted 11 in the first step, we take the 11th root:



----

To find the harmonic means:

1. Take the reciprocals of all the numbers:

,,,,,,,,,,

2. Take the arithmetic mean of the results of 1

To find the arithmetic mean of those numbers:

      1a. Count them.  There are 11
      2a. Add them all together. 1.19694482
      3a. Divide the result of 2 by the result of 1a.
           

3. Take the reciprocal of the result of 2.

       

Edwin

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