Solved – How to calculate standard deviation of function with two variables

standard deviation

I have the following:

x is mean of X={4,6} and equals to 5 and standard deviation of X equals to 1.41
y is mean of Y={3,9} and equals to 6 and standard deviation of Y equals to 4.24

How can I calculate the standard deviation of z which is equal to:

$$ z=\frac{y-x}{x}100 = \frac{6-5}{5}100=20$$

Best Answer

Suppose you have quantities $a,b$ and their standard deviations are $\Delta a, \Delta b$. Then if $s=a+b$ or $s=a-b$ then $\Delta s = \sqrt{(\Delta a)^2+(\Delta b)^2}$.

And if $q = a/b$ then $\dfrac{\Delta q}{q} = \sqrt{\left(\dfrac{\Delta a}{a}\right)^2+\left(\dfrac{\Delta b}{b}\right)^2}$

Your problem involves only subtraction and division, so you should be able to apply these easily. Look up "propagation of uncertainty" for more information.

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Clustering – How to Use Standard Deviation to Place Items in Categories?

My idea would be this:

subtract its mean from the set
categorise into bands of multiples of the standard deviation

In the programming language Python this would look something like this:

# load array- and absolute-value function from numerical package
from numpy import array, abs

# have data available in an array
s = array([1,1,15,22,9,99,100,240,3,7])

# calculate moments
mu = s.mean()
sigma = s.std()  # standard deviation

# standardise and categorise
z = s - mu  # subtract mu from each element

s1 = s[abs(z)<0.4*sigma]  # get elements with abs value < 0.4*sigma
s2 = s[(abs(z)>0.4*sigma) & (abs(z)<0.5*sigma)]
s3 = s[(abs(z)>0.5*sigma) & (abs(z)<0.6*sigma)]
s4 = s[(abs(z)>0.6*sigma) & (abs(z)<0.7*sigma)]
s5 = s[abs(z)>0.7*sigma]

This results in:

In [36]: s1
Out[36]: array([22])

In [37]: s2
Out[37]: array([15])

In [38]: s3
Out[38]: array([9, 7])

In [39]: s4
Out[39]: array([  1,   1,  99, 100,   3])

In [40]: s5
Out[40]: array([240])

s1-s5 correspond to the categories. Maybe this is an inspiration for something you like.

Solved – Calculate standard deviation given tabled number of occurrences

Your formula for the mean is correct.

Presumably you can work out the mean from the variance.

Have you seen the formula for the population variance $\text{Var}(X)=E(X^2)-E(X)^2$?

The usual sample variance is the same, with Bessel's correction -
$\text{Var}(x)=\frac{n}{n-1} [\overline{x^2}-\overline{x}^2]$.

(Where $\overline{x^2}$ means 'take the mean of the squared observations')

From that you can use exactly the same method you did for the mean to work out the mean of the $x^2$ values and hence obtain the variance and then the standard deviation.

Best Answer

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Clustering – How to Use Standard Deviation to Place Items in Categories?

Solved – Calculate standard deviation given tabled number of occurrences

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