Let’s consider the z-score calculation of data $x_i$ with mean $\bar{x}$ and standard deviation $s_x$.

$$z_i = \dfrac{x_i-\bar{x}}{s_x}$$

This means that, given some data $(x_i)$, we can transform to data with a mean of $0$ and standard deviation of $1$.

Rearranging, we get:

$$x_i = z_i s_x+ \bar{x}$$

This gives us back our original data with the original mean $\bar{x}$ and standard deviation $s_x$. But we could’ve gone to data $y_i$ with any mean $\bar{y}$ and standard deviation $s_y$.

$$y_i = z_i s_y +\bar{y}$$

Now combine the two transformations, first to $z_i$ and then to $y_i$.

$$y_i = \dfrac{x_i-\bar{x}}{s_x}s_y + \bar{y}$$

This is the same as what Henry posted, but I do think it is helpful to see that we get there by first going to standardized data and the transforming to data with the mean and standard deviation values we desire.

Let’s consider the z-score calculation of data $x_i$ with mean $\bar{x}$ and standard deviation $s_x$.

$$z_i = \dfrac{x_i-\bar{x}}{s_x}$$

This means that, given some data $(x_i)$, we can transform to data with a mean of $0$ and standard deviation of $1$.

Rearranging, we get:

$$x_i = z_i s_x+ \bar{x}$$

This gives us back our original data with the original mean $\bar{x}$ and standard deviation $s_x$. But we could’ve gone to data $y_i$ with any mean $\bar{y}$ and standard deviation $s_y$.

$$y_i = z_i s_y +\bar{y}$$

Now combine the two transformations, first to $z_i$ and then to $y_i$.

$$y_i = \dfrac{x_i-\bar{x}}{s_x}s_y + \bar{y}$$

This is the same as what Henry posted, but I do think it is helpful to see that we get there by first going to standardized data and then transforming to data with the mean and standard deviation values we desire.