I understood that naturally occurring phenomenon such a weight of all individuals form a bell curved normal distribution, with more frequency applied to the center. but I'm confused with the arithmetic operators used to derive the Z-score of the standard normal distribution.
Say, I found the mean weight of a population to around 60kg and to convert to standard normal distribution I first subtracted the individual weight/random measure by the mean i.e. with 60kg but then moving on why are we dividing it by (sigma) the standard deviation to find distance of how far the data point is away from the mean. Why did we use the arithmetic operation of division and why divide with sigma (the standard deviation) to find the Z-score? Can someone help out to conceptualize the idea.
There is a nice property of normal distributions, namely that if $Z$ is standard normal, then any linear function of $Z$ is normal:
$$Z\sim N(0,1)\implies X=\sigma Z+\mu\sim N(\mu,\sigma^2).$$
Reversing this tells us that if $X\sim N(\mu,\sigma^2)$, $$\frac{X-\mu}{\sigma}\sim N(0,1).$$
Thus, we can compute probabilities involving any normal random variable $X$ in terms of a standard normal just by standardizing $X$:
$$P(a<X<b)=P\left(\frac{a-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{b-\mu}{\sigma}\right)=P\left(\frac{a-\mu}{\sigma}<Z<\frac{b-\mu}{\sigma}\right)$$
where $Z\sim N(0,1).$