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We have a function which returns a value d with a standard deviation of s.

Afterwards, let us plug d into the following formula:

y = 1.12-0.01*d

Would y still have the same standard deviation s? If not, how would it change?

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    $\begingroup$ The new standard deviation would be $0.01 s$. Adding a constant does not change the standard deviation. Multiplying by a constant $c$ scales the standard deviation by $|c|$. $\endgroup$
    – user169852
    Commented May 23, 2018 at 17:04
  • $\begingroup$ Would this scale indefinitely? Meaning, if we now multiply y with constant k, let's say, 0.7, would s scale again by 0.7? $\endgroup$ Commented May 23, 2018 at 17:11
  • $\begingroup$ Sure, $z = 0.7y = 0.7(1.12 - 0.01d) = 0.784 - 0.07d$, so the standard deviation of $z$ is $0.07s$. $\endgroup$
    – user169852
    Commented May 23, 2018 at 17:13
  • $\begingroup$ Perfect - Thanks! Would you like to write it as a formal answer so I can accept it? $\endgroup$ Commented May 23, 2018 at 17:17

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The standard deviation is a measure of "spread", i.e. how far values vary from the mean. Adding the same fixed number to each output changes the "location" of each data point, but it doesn't change the spread. Multiplying each number by a constant doesn't change the location, but it changes the spread: multiplying by $2$ changes a gap of $7$ to a gap of $14$.

There are standard formulae:

The mean, or expected value, written $\mathrm E[X]$, has the property that $$\mathrm E[aX+b]=a\mathrm E[X]+b$$ If the mean of $X$ is $\mu$, then the mean of $aX+b$ is $a\mu+b$.

The variance, or standard deviation squared, written $\mathrm{Var}[X]$, has the property that $$ \mathrm{Var}[aX+b] = a^2\mathrm{Var[X]} $$ If the standard deviation of $X$ is $\sigma$, then the standard deviation of $aX+b$ is $|a|\sigma$.

(Note: $\sqrt{a^2} = |a|$ for all real $a$.)

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In general: $$\text{Var}(aX+b)=\mathbb E(aX+b-\mathbb Ea(X+b))^2=a^2\mathbb E(X-\mathbb EX)^2=a^2\text{Var}X$$

so that:$$\sigma(aX+b)=(\text{Var}(aX+b))^\frac12=(a^2\text{Var}X)^{\frac12}=|a|\sigma(X)$$

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As Bungo says, adding a constant will not change the standard deviation. Multiplying by a constant will; it will multiply the standard deviation by its absolute value.

To see this, calculate a few simple cases. E.g. calculate the mean and standard deviation of a standard fair six sided die. Now do the same for a few non-standard dice. Still six sided and fair but with non-standard labels.

$2, 3, 4, 5, 6, 7$

$2, 4, 6, 8, 10, 12$

$-1, -2, -3, -4, -5, -6$

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One should be clear about what is multiplied by a constant. If the question is to make sense, the thing that is multiplied by a constant should be the thing whose standard deviation is taken. What happens to the standard deviation when the standard deviation itself is multiplied by a constant is a simpler question.

Suppose the thing whose standard deviation is to be found is multiplied by $c.$

Then the variance is multiplied by $c^2$ and the standard deviation by $|c|.$

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