Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
1. Measures of Center and Variation
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Exercise 38 Page 592

Consider an example data set and find the two means. Which one is greater?

The arithmetic mean is always greater than or equal to the geometric mean.

Practice makes perfect
The most common type of mean, the arithmetic mean x_a, is given by the sum of the data values divided by the number of data values. x_a=Sum of Values/Number of Values However, there are other kinds of means. One of these is the geometric mean x_g. It is the n^\text{th} root of the product of the n data values. x_g=sqrt(a_1* a_2* ... * a_n) To compare these two means, let's consider an example data set. Keep in mind that there are many possible examples.

{3,5,1,5,1,1,2,3,15}

Comparing the Means

Let's find the arithmetic mean of the data set.
x_a=Sum of Values/Number of Values
x_a=3+5+1+5+1+1+2+3+15/9
Evaluate right-hand side
x_a=36/9
x_a=4
Now, let's calculate the geometric mean. In this case, n is 9.
x_g=sqrt(a_1* a_2*...* a_n)
x_g=sqrt(3*5*1*5*1*1*2*3*15)
Evaluate right-hand side
x_g=sqrt(6750)
x_g=2.663656...
x_g≈2.66
In this example, the arithmetic mean 4 is greater than the geometric mean 2.66.

Another Example

To see if this is always true, let's find these means for a data set where all of its values are the same. {5,5,5,5,5,5} Let's find the arithmetic mean of the data set.
x_a=Sum of Values/Number of Values
x_a=5+5+5+5+5+5/6
Evaluate right-hand side
x_a=30/6
x_a=5
Again, let's calculate the geometric mean of the data. In this case, n=6.
x_g=sqrt(a_1* a_2*...* a_n)
x_g=sqrt(5*5*5*5*5*5)
Evaluate right-hand side
x_g=sqrt(15625)
x_g=5
We have found a case where the arithmetic mean is equal to the geometric mean.

Conclusion

It seems that the arithmetic mean is always greater than or equal to the geometric mean. Let's look at some random examples to confirm this fact.
Geometric Mean and Arithmetic Mean
From the above examples, we can conclude that the arithmetic mean is always greater than or equal to the geometric mean. x_a≥x_g

Extra

Data Set With Negative Values
What if the data set contained negative values? Would the inequality remain true? Let's consider an example data set. {-1, -2, -3, -4} Now, let's calculate the mean of the data.
x_a=Sum of Values/Number of Values
x_a=(-1)+(-2)+(-3)+(-4)/4
Evaluate right-hand side
x_a=-1-2-3-4/4
x_a=-10/4
x_a=-10/4
x_a=-2.5
Now, let's calculate the geometric mean. In this case, n is 4.
x_g=sqrt(a_1* a_2*...* a_n)
x_g=sqrt((-1)*(-2)*(-3)*(-4))
Evaluate right-hand side
x_g=sqrt(2*12)
x_g=sqrt(24)
x_g=4.898979...
x_g≈4.90
This time, the geometric mean is greater than the arithmetic mean. It is because the inequality is only valid when the data set values are nonnegative. For now, the proof of the inequality is beyond our knowledge.