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In Statistics by Freedman et al. it is described how to construct a normal approximation for a histogram, as follows:

Calculate mean and SD for the histogram (in the image $63.5$ and $3$ inches).

Transform the scale in standard units using $\displaystyle\frac{x-\mu}{\delta}$ , changing the horizontal scale.

Fit the vertical scale passing from percent per inch to percent per standard unit, constructing the normal curve based on these scales.

This is the resulting image:

Even if I know that for a normal random variable $X$ we have $$\int_{\mu-k\delta}^{\mu+k\delta} f_X(\mu,\delta) = \int_{-k}^{k} f_Z(0,1) $$ I'm having problems to relate this notion to the graphs. I imagined constructing first $f_X$ and $f_Z$ together, then because of the equality I pointed out we know that for every $k$ the area under the curve of $f_X$ from $\mu-k\delta$ to $\mu+k\delta$ is equal to that from $-k$ to $k$ of $f_Z$. With the horizontal transformation then we are moving every $\pm k$ to $\pm(\mu+k\delta)$ , but I'm unsure why the vertical transformation is the "right" one to make the two graphs coincide.

I understand that I need to change the percent per inch to percent per standard unit, but I don't get why this is enough to pass from $f_Z$ to $f_X$. It has to be something related to changing the units of measure also on the vertical axes.

Can someone explain this to me please?

Edit :

even if I already accepted an answer I think that I'm still not able to understand really well what's going on : what I'd like to do is to trasform $f_Z$ in $f_X$ only by graphical transformations and make them coincide but I'm still not sure why the construction of the book works

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  • $\begingroup$ I think it would be clearer if you showed the integration variables in your equation as they, multiplied by the probability density, give a probability which is dimensionless. $\endgroup$ Commented Apr 27, 2021 at 7:51

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A lot of people don't understand the PDF. Some people think it's a probability, somehow. The term itself has a clue: a density. Think of a density of of a physical body. For instance, a density of gold in liquid form is 17.3 g/cc or0.63 lb/cubic inches. It doesn't surprise you that the density is different when we change the units of measure?

PDF is the probability per unit of measure of your variable. The same way as density of physical materials, when you change the units of measure of your variable, PDF changes. In your formula, the unit of measure is represented by that scaling factor $\delta$.

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  • $\begingroup$ Thank you @Aksakal, what I think I was misunderstanding was starting in my mind with $f_X$ and $f_Z$ on the same coordinate system; since the vertical axes-densities and so also horizontal axes-units are different It can't be done (at least if we care about units of measure), right? $\endgroup$ Commented Apr 26, 2021 at 22:23
  • $\begingroup$ $Z$ is a unitless variable, while $X$ has a unit of measure, so you can't really place them on the same coordinate system $\endgroup$ Commented Apr 26, 2021 at 22:39
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If $f$ is a probability density, what is conserved is the elementary probability $f\ dx$:

$$f_X dx = f_Y dy$$

This shows that the densities $f_X$ and $f_Y$ don't necessarily have the same dimensions.

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