[Math] How to determine if Standard Deviation is high/low

standard deviationstatistics

I have derived the following response time data for a performance test I am running:

Min – 8sec
Max – 284sec
Average – 28sec
Standard Deviation – 27sec

What does the standard deviation say about the response time data distribution? When you say low/high standard deviation, what does this actually mean? Is this in comparison to the Average/Min/Max?

I know what standard deviation is and how it's computed. I'm just not sure how to tell if it is high or low.

Best Answer

If you take your cues from the financial industry, you can use the coefficient of variation (CV), which is the standard deviation / mean. This formula is used to normalize the standard deviation so that it can be compared across various mean scales.

As a rule of thumb, a CV >= 1 indicates a relatively high variation, while a CV < 1 can be considered low.

Some references to usage as "rule of thumb"

http://www.readyratios.com/reference/analysis/coefficient_of_variation.html

http://www.mhnocc.org/forum/index.php?t=msg&goto=234&

Related Solutions

[Math] Calculating mean and standard deviation of very large sample sizes

Posting as an answer in response to comments.

Here's a way to compute the mean and standard deviation in one pass over the file. (Pseudocode.)

n = r1 = r2 = 0;
while (more_samples()) {
    s = next_sample();
    n += 1;
    r1 += s;
    r2 += s*s;
}
mean = r1 / n;
stddev = sqrt(r2/n - (mean * mean));

Essentially, you keep a running total of the sum of the samples and the sum of their squares. This lets you easily compute the standard deviation at the end.

[Math] Average, standard deviation and min/max values

Suppose we have $n$ numbers $x_1, x_2, \ldots, x_n$ ranging from $a$ to $b$, meaning that $x_i = a$ for at least one $i$, $1 \leq i \leq n$, and $x_j = b$ for at least one $j$, $1 \leq j \leq n$. Duplicates are allowed, meaning that two or more of the $n$ numbers could possibly have the same value in $[a, b]$. Define the mean $\bar{x}$ and the variance $\sigma_x^2$ of the set of $n$ numbers as $$\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i, ~~ \sigma_x^2 = \frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})^2 = \left(\frac{1}{n}\sum_{i=1}^n x_i^2\right) - \bar{x}^2.$$ For convenience, let us change the set to $y_1, y_2, \ldots, y_n$ where $y_i = x_i - a$, so that the new set has values ranging from $0$ to $b-a$. The mean $\bar{y}$ is just $\bar{x}-a$, while the variance is unchanged: $\sigma_y^2 = \sigma_x^2$. Now, it is shown in the answers to this question on stats.SE that the ratio $\sigma_y/\bar{y} = \sigma_x/(\bar{x}-a)$ can be no larger than $\sqrt{n-1}$. Note that the value does not depend on $b$ at all You don't say what the value of $n$ is, but given the numbers in your answer, the upper bound on the standard deviation is $$\sigma_x = 554 \leq (464-100)\sqrt{n-1}$$ which is certainly satisfied except in the unusual circumstance that there are only $3$ peers in the experiment and thus only $3$ numbers $x_1$, $x_2$, and $x_3$ are being described in terms of mean/standard-deviation/min-max: certainly overkill!

In summary, there are no obvious problems with the standard deviation being larger than the mean.

Best Answer

Related Solutions

[Math] Calculating mean and standard deviation of very large sample sizes

[Math] Average, standard deviation and min/max values

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