Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

To standardize a random variable that is normally distributed, it makes absolute sense to subtract the expected value $\mu$ , from each value that the random variable can assume--it shifts all of the values such that the expected value is centered at the origin. But how does dividing by the standard deviation play a role in the standardization of a random variable? That part is not as intuitive to me as is subtracting $\mu$.

share|cite|improve this question
up vote 8 down vote accepted

Let $X \sim N(\mu,\sigma^2)$.

Let $Y = \large\frac{X-\mu}{\sigma}$.

$E[Y] = \large\frac{E[X] - \mu}{\sigma} = \large\frac{\mu-\mu}{\sigma} = 0$

$\text{Var}(Y) = \large\frac{1}{\sigma^2}\text{Var}(X) = \large\frac{1}{\sigma^2}\sigma^2 = 1$.

So that $Y \sim N(0,1)$.

This is precisely why we subtract the mean and divide by the standard deviation.

share|cite|improve this answer

Recall that $\operatorname{Var}{X}=E(X-\mu)^2$, where $\mu$ is the mean of $X$. So $\operatorname{Var}(X)$ is an average of squares. Thus if we scale $X$ by a factor $\rho$, then the variance gets multiplied by $\rho^2$.

To get a feeling for this, recall that if we scale a geometric figure, such as a square or a triangle, by the factor $\rho$, then area gets scaled by a factor of $\rho^2$.

Now suppose that the variance of $X$ is $\sigma^2$. Then we must scale $X$ by the factor $\dfrac{1}{\sigma}$ to bring the variance to $1$.

share|cite|improve this answer

Dividing by the standard deviation lets the variance of your new random variable be 1. That is what the standardizing a random variable means. You simply let the mean and variance of your random variable be 0 and 1, respectively

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.