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On standardization of normal distribution

  1. Feb 14, 2015 #1
    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. jcsd
  3. Feb 14, 2015 #2
    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) .
  4. Feb 14, 2015 #3


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    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?
  5. Feb 15, 2015 #4
    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
  6. Feb 15, 2015 #5


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    Standardization means what you think it means: when you standardize data in this setting you subtract the mean and divide that difference by the standard deviation. The fact that this operation shifts the work from an arbitrary normal distribution to the standard normal distribution is a mathematical result that needs to be demonstrated: simply making the substitution in the density function isn't enough.

    Is your course calculus based?
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