I have the following set of data:
Raw Scores x-x̄ (x-x̄)²
----------------------------- 7 -6 36 8 -5 25
10 -3 9
14 1 1
26 13 169
-----------------------------
65 0 240
-----------------------------From the raw scores we know that : $$ n=5 $$ and ∴ the mean, $$ x̄ = \frac{\sum}{n} = \frac{65}{5} = 13 $$
I then proceeded to complete the table above filling in $$ x-x̄ $$ & $$ (x-x̄ )² $$
I am now required to calculate the standard deviation & have the following eqation.
$$ \sigma = \sqrt{\frac{\sum(x_i-\overline{x})^2}{n}} = \sqrt{\frac{240}{5}} = 4 \sqrt{15} $$
However when I submit my data to wolfram and query standard deviatioin it returns the answer$$2\sqrt{15}$$
Maybe this is out of the scope of my learning but being curious I opened the step by step answer and I see that wolfram calculates the standard deviation by
$$ \sigma = \sqrt{\frac{\sum(x_i-\overline{x})^2}{n-1}} $$
My textbook accessible here and another textbook my teacher printed out have the eqation , I find it hard to believe that My teacher and 2 other textbooks can be wrong however Im pretty sure the developers at wolfram couldn't have got it wrong either.
What am I missing?
$\endgroup$1 Answer
$\begingroup$The difference is that you are calculating the true standard deviation, and Wolfram|Alpha is calculating the sample standard deviation.
Here's the difference, if you use the formula $$ \sigma = \sqrt{\frac{\sum(x_i-\overline{x})^2}{n}} $$ then you are answering the question "how much on average does my data differ from its mean?"
If you use the formula $$ \sigma_S = \sqrt{\frac{\sum(x_i-\overline{x})^2}{n-1}} $$ you are answering the question "if my data is a representative sample of a population, what is my best estimate of how much the population differs from its mean on average?"
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