Quadratic Mean
The quadratic mean is the square root of the arithmetic mean of the squares of the terms. $$ \mu_q = \sqrt{ \frac{x_1^2+x_2^2+x_3^2+...+x_n^2}{n}}$$
Purpose
The quadratic mean is particularly sensitive to the variations among the terms x1, x2, ..., xn in a distribution and their deviation from the average value μ.
For instance, it is useful for measuring deviation, variance, or error in a dataset relative to the arithmetic mean, as it ignores the direction of deviations (positive or negative) and focuses solely on their magnitude.
In contrast, the arithmetic mean can neutralize opposing values due to their signs.
Note. One way to estimate the dispersion of a distribution around the mean is to compute the difference between the quadratic mean μq and the arithmetic mean μ: $$ \mu_q - \mu $$ or $$ \frac{\mu_q - \mu}{\mu_q} $$. In the latter case, the quadratic mean must never be zero, as this would result in an undefined division by zero.
A Practical Example
Consider the following set of values:
$$ X = \{ 1,5,7,3,6,8 \} $$
This distribution contains n=6 terms, and the arithmetic mean is μ=5.
Note. The arithmetic mean of this dataset is μ=5: $$ \mu= \frac{1+5+7+3+6+8}{6} = \frac{30}{6} = 5 $$
Next, let's calculate the quadratic mean for this distribution:
$$ \mu_q = \sqrt{ \frac{1^2+5^2+7^2+3^2+6^2+8^2}{6} } $$
$$ \mu_q = \sqrt{ \frac{1+25+49+9+36+64}{6} } $$
$$ \mu_q = \sqrt{ \frac{184}{6} } $$
$$ \mu_q = 5.54 $$
The quadratic mean is μq = 5.54.
This value is higher than the arithmetic mean (μ=5), indicating some variation between the data points and the average.
$$ \frac{\mu_q - \mu}{\mu_q} = \frac{5.54 - 5}{5.54} = 0.09 $$
Example 2
Now, let's examine this distribution:
$$ X = \{ 1,1,1,9,9,9 \} $$
This set also has n=6 terms, and the arithmetic mean remains μ=5.
Note. The arithmetic mean of this dataset is 5: $$ \mu= \frac{1+1+1+9+9+9}{6} = \frac{30}{6} = 5 $$
However, in this distribution, the terms are much more spread out from the mean value (μ=5).
Let's find the quadratic mean:
$$ \mu_q = \sqrt{ \frac{1^2+1^2+1^2+9^2+9^2+9^2}{6} } $$
$$ \mu_q = \sqrt{ \frac{1+1+1+81+81+81}{6} } $$
$$ \mu_q = \sqrt{ \frac{246}{6} } $$
$$ \mu_q = 6.4 $$
The quadratic mean is μq = 6.4.
This value is significantly larger than in the previous example, reflecting a greater spread between the terms x1, x2, ..., xn and the arithmetic mean (μ=5).
$$ \frac{\mu_q - \mu}{\mu_q} = \frac{6.4 - 5}{6.4} = 0.21 $$
Example 3
Consider this distribution:
$$ X = \{ 4,6,4,6,4,6 \} $$
This set also has n=6 numbers, and the arithmetic mean is still μ=5.
Note. The arithmetic mean of this dataset is 5: $$ \mu= \frac{4+6+4+6+4+6}{6} = \frac{30}{6} = 5 $$
In this case, the terms of the distribution are very close to the arithmetic mean (μ=5).
Now, let's calculate the quadratic mean:
$$ \mu_q = \sqrt{ \frac{4^2+6^2+4^2+6^2+4^2+6^2}{6} } $$
$$ \mu_q = \sqrt{ \frac{16+36+16+36+16+36}{6} } $$
$$ \mu_q = \sqrt{ \frac{156}{6} } $$
$$ \mu_q = 5.01 $$
The quadratic mean is μq = 5.01.
In this example, the quadratic mean is almost identical to the arithmetic mean (μ=5) because there is minimal variation between the individual values and the mean.
$$ \frac{\mu_q - \mu}{\mu_q} = \frac{5.01 - 5}{5.01} = 0.01 $$
Observations
Here are some key observations about the quadratic mean:
- The quadratic mean equals the arithmetic mean when all terms in the distribution are constant.
Example. Consider this distribution with constant values: $$ X = \{ 5,5,5,5,5,5 \} $$. The arithmetic mean is μ=5: $$ \mu = \frac{5+5+5+5+5+5}{6} = \frac{30}{6} = 5 $$. The quadratic mean is also μq = 5: $$ \mu_q = \sqrt{ \frac{5^2+5^2+5^2+5^2+5^2+5^2}{6} } $$ $$ \mu_q = \sqrt{ \frac{25+25+25+25+25+25}{6} } $$ $$ \mu_q = \sqrt{ \frac{150}{6} } = \sqrt{ 25 } = 5 $$. In this scenario, the data dispersion around the mean is zero: $$ \frac{\mu_q - \mu}{\mu_q} = \frac{5-5}{5} = 0 $$
- The quadratic mean (μq) is always greater than or equal to the arithmetic mean (μ). Similarly, the arithmetic mean is greater than or equal to the geometric mean (μg) and the harmonic mean (μh): $$ \mu_h \le \mu_g \le \mu \le \mu_q $$
- Weighted Quadratic Mean
The weighted quadratic mean is an alternative form of the quadratic mean where each term is assigned a specific weight (wi): $$ \mu = \sqrt{ \frac{ \sum x^2_i \cdot w_i}{ \sum w_i} } $$
And so on.
