How to transform/shift the mean and standard deviation of a normal distribution?

Question

Given some Gaussian distribution with mean x and deviation s, how do I transform the distribution to have a new specific mean and specific deviation.

Say the distribution has a mean, $\bar x = 4$ and deviation, $s = 10$, and needs to be transformed so that the new mean and deviation are $\bar x = 0.50$ and $s = 2$.

My approach is to scale each element in the data set by $c = 0.20$, which will also scale the deviation to the desired $s = 2$, and will make the mean $\bar x = 0.80$. Finally I subtract 0.30 from each element to shift the mean to the desired $\bar x = 0.50$.

Answer 1

Let $X \sim N(\mu_1, \sigma_1^2)$, let $Y=aX+b$,

then we have $Y \sim N(a\mu_1+b, a^2\sigma_1^2)$.

So if you have a target normal distribution to map to, say $N(\mu_2, \sigma_2^2)$.

Solve for $a,b$ in

$$a\mu_1+b = \mu_2$$

$$a^2\sigma_1^2 = \sigma_2^2$$

Answer 2

Yes, your answer is correct.

Given $\bar{x}=E(X)=4, s_x=\sqrt{Var(X)}=10$, $\bar{y}=E(Y)=0.5, s_y=\sqrt{Var(Y)}=2$, you must be using the properties of the mean and variance: $$E(aX+b)=aE(x)+b;\\ Var(aX+b)=a^2Var(X).$$ Step 1: finding the coefficient: $$Var(Y)=Var(aX)=a^2Var(X) \iff 4=a^2\cdot 100 \Rightarrow a=0.2.$$ Step 2: finding the shift: $$E(Y)=E(0.2X+b)=0.2E(X)+b \iff 0.5=0.2\cdot 4+b \Rightarrow b=-0.3.$$ Hence: $$Y=aX+b=0.2X-0.3.$$