# On standardization of normal distribution

1. Feb 14, 2015

### jwqwerty

Let X be random variable and X~N(u,σ^2)
Thus, normal distribution of x is
f(x) = (1/σ*sqrt(2π))(e^(-(x-u)^2)/(2σ^2)))

If we want to standardize x, we let z=(x-u)/σ
Then the normal distribution of z becomes
z(x) = (1/σ*sqrt(2π))(e^(-(x^2)/(2))

and we usually write Z~N(0,1)

But as you can see, sigma in z(x) does not disappear. Thus, in my opinion Z~N(0,1) should be actually written as Z~(1/σ)N(0,1). So here goes my question :
why does every textbook use the notation Z~N(0,1), not Z~(1/σ)N(0,1)

Last edited: Feb 14, 2015
2. Feb 14, 2015

### PWiz

By subtracting μ from $x$, f(x) will become centered at 0 instead of the mean, because all values have been reduced by μ. After this, dividing by σ has the effect of altering the spread of the data, since where x was originally σ, it has now become 1. Similarly, 2σ becomes 2 and so on, and now the curve has standard deviation (and variance) of 1. Hence, if $Z=\frac{X-μ}{σ}$, then Z ~ N(0,1) .

3. Feb 14, 2015

You can't do the standardization the way you did (simply by substituting the expression for z in the density). What you are doing is attempting to find the density of a
new random variable Z given an existing density and a transformation. Have you studied that technique?

4. Feb 15, 2015

### jwqwerty

Then what does standardization mean? How can we standardize?
Sorry i have just started studying statistics and i need your help, statdad!

Last edited: Feb 15, 2015
5. Feb 15, 2015