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    Hi,

    I'm a bit confused about how the normal standardisation formula is derived. We know if:

     X \sim N(\mu,\sigma^2)

    The pdf f(X) is:

     f(X) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(X-\mu)^2}{2\sigma^2}}

    and:

     if \ Z = \frac{X-\mu}{\sigma}


     f(Z) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{Z^2}{2}}

    My question is why is there still a sigma on the bottom of the fraction? Have I done something wrong with the algebra? My S2 book insists that it is a simple C1 transformation which makes me feel a bit stupid not understanding where this standardisation formula comes from! Can somebody please explain it for me? Thanks
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    Huh what exam board im.on.wjec and dont follow this could you explain in differentvway might have been taught alternative notatoins

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    (Original post by jadpan)
    Huh what exam board im.on.wjec and dont follow this could you explain in differentvway might have been taught alternative notatoins

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    Hi, just want to know where the standardising formula:

     Z = \frac{X-\mu}{\sigma}

    comes from. I tried the method in my original post to try to persuade myself that it works. It also confused me when it said if you have f(X), it is symmetrical about X = mu, but if we want it to be symmetrical about X = 0, you make the transformation f(X) to f(X-mu), but shouldn't you make the transformation f(X) to f(X+mu) ?
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    (Original post by metaltron)
    Hi, just want to know where the standardising formula:

     Z = \frac{X-\mu}{\sigma}

    comes from. I tried the method in my original post to try to persuade myself that it works. It also confused me when it said if you have f(X), it is symmetrical about X = mu, but if we want it to be symmetrical about X = 0, you make the transformation f(X) to f(X-mu), but shouldn't you make the transformation f(X) to f(X+mu) ?
    No because you take values from right to left, thus a shift to the right. A shift to the right is represented through a "-" sign.
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    (Original post by BananaPie)
    No because you take values from right to left, thus a shift to the right. A shift to the right is represented through a "-" sign.
    So the mean, if positive, is to the right of the origin. Then why isn't the transformation f(X+mu)? Or is it but in disguise?
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    (Original post by metaltron)
    So the mean, if positive, is to the right of the origin. Then why isn't the transformation f(X+mu)? Or is it but in disguise?
    No you've gotten it mixed up. A shift to the right by "a" units is represented by f(x-a). This is because the (x) of f(x) must equate to 0. So if you've shifted to the right by "a" units, you subtract "a" units inside the bracket to get back to 0.
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    (Original post by BananaPie)
    No you've gotten it mixed up. A shift to the right by "a" units is represented by f(x-a). This is because the (x) of f(x) must equate to 0. So if you've shifted to the right by "a" units, you subtract "a" units inside the bracket to get back to 0.
    I'm not quite sure I have it mixed up, at least I wouldn't get them mixed up in a C1 question. You're shifting it back to the y-axis which is a shift to the left which is f(x+ mu).
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    (Original post by metaltron)
    Hi,

    I'm a bit confused about how the normal standardisation formula is derived. We know if:

     X \sim N(\mu,\sigma^2)

    The pdf f(X) is:

     f(X) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(X-\mu)^2}{2\sigma^2}}

    and:

     if \ Z = \frac{X-\mu}{\sigma}


     f(Z) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{Z^2}{2}}

    My question is why is there still a sigma on the bottom of the fraction? Have I done something wrong with the algebra? My S2 book insists that it is a simple C1 transformation which makes me feel a bit stupid not understanding where this standardisation formula comes from! Can somebody please explain it for me? Thanks
    The standard normal distribution id defined as being Z \sim N(0,1) i.e. \mu=0 and \sigma=1.

    This means the probability density function is:

    f(Z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}
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    (Original post by Asklepios)
    The standard normal distribution id defined as being Z \sim N(0,1) i.e. \mu=0 and \sigma=0.

    This means the probability density function is:

    f(Z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}
    Thanks for replying This thread has suddenly burst into life. I know that formula for f(Z) but what I'm saying is I somehow have a sigma at the bottom. Help me get rid of it please!!! Its driving me crazy.
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    (Original post by metaltron)
    Hi,

    I'm a bit confused about how the normal standardisation formula is derived. We know if:

     X \sim N(\mu,\sigma^2)

    The pdf f(X) is:

     f(X) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(X-\mu)^2}{2\sigma^2}}

    and:

     if \ Z = \frac{X-\mu}{\sigma}


     f(Z) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{Z^2}{2}}

    My question is why is there still a sigma on the bottom of the fraction? Have I done something wrong with the algebra? My S2 book insists that it is a simple C1 transformation which makes me feel a bit stupid not understanding where this standardisation formula comes from! Can somebody please explain it for me? Thanks
    where did you get that expression for f(Z) from - is it part of your working or out of the book?

    The starting point for the transformation is that whatever you have as your pdf, total probability must equal 1, so the integral from -infinity to +infinity of your pdf must equal 1.

    If you're familiar with the legendary e^{-x^2} integral, then that's what you should be aiming at, and there shouldn't be any sigma-dependence.
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    (Original post by davros)
    where did you get that expression for f(Z) from - is it part of your working or out of the book?

    The starting point for the transformation is that whatever you have as your pdf, total probability must equal 1, so the integral from -infinity to +infinity of your pdf must equal 1.

    If you're familiar with the legendary e^{-x^2} integral, then that's what you should be aiming at, and there shouldn't be any sigma-dependence.
    Yeah it is part of my working and is clearly wrong. I tried substituting in:

     Z = \frac{X-\mu}{\sigma}

    to try and get a standard normal distribution in terms of Z from the general normal distribution formula in terms of X. However I am stuck with a sigma at the bottom somehow.
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    (Original post by metaltron)
    Thanks for replying This thread has suddenly burst into life. I know that formula for f(Z) but what I'm saying is I somehow have a sigma at the bottom. Help me get rid of it please!!! Its driving me crazy.
    For f(Z), sub in \mu=0 and \sigma=1 into the formula for f(X).
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    (Original post by Asklepios)
    The standard normal distribution id defined as being Z \sim N(0,1) i.e. \mu=0 and \sigma=1.

    This means the probability density function is:

    f(Z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}
    Fixed your sigma from a 0 to a 1 there.




    @OP: Ror the standard curve sigma=1. So yes, you're dividing by a sigma that is the same as dividing by 1. Also, as Asklepios said, you can just sub in mu and sigma individually instead of subbing in z.
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    (Original post by davros)
    If you're familiar with the legendary e^{-x^2} integral, then that's what you should be aiming at, and there shouldn't be any sigma-dependence.
    Like how you phrased things. Legendary. Nearly drove me nuts trying to crack it previously. Peace.
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    (Original post by Asklepios)
    For f(Z), sub in \mu=0 and \sigma=1 into the formula for f(X).
    Maybe I haven't explained myself properly. If you have:

     X \sim N(6,25)

    you can use the standardisation formula to get another variable Z which has normal distribution:

     Z \sim N(0,1)

    However I'm struggling to see how the standardisation formula is derived.
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    (Original post by metaltron)
    Maybe I haven't explained myself properly. If you have:

     X \sim N(6,25)

    you can use the standardisation formula to get another variable Z which has normal distribution:

     Z \sim N(0,1)

    However I'm struggling to see how the standardisation formula is derived.
    Is the "standardization formula" you are asking for the formula z=(x-m)/s or rather the probability density function f(z)=blah?

    If it's f(z)=blah as your first post implied then you should reread the thread because we explain its reason pretty clearly.

    If you're wondering about the formula, then it's simple: the probabilities with respect to each curve's standard deviation doesn't change after translations and stretches. So first, translate each point left equal to the mean so that the mean is at 0, and divide each point by the stdev so that the stdev is 1.
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    (Original post by aznkid66)
    Fixed your sigma from a 0 to a 1 there.




    @OP: Ror the standard curve sigma=1. So yes, you're dividing by a sigma that is the same as dividing by 1. Also, as Asklepios said, you can just sub in mu and sigma individually instead of subbing in z.
    Hi, thanks to everybody who has replied to this thread and I hope I am not now becoming annoying!

    I understand subbing in mu=0 and sigma= 1 gives you the standard normal distribution pdf, but I am trying to derive the standardisation formula we used. I explained above as well.
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    (Original post by aznkid66)
    Is the "standardization formula" you are asking for the formula z=(x-m)/s or rather the probability density function f(z)=blah?
    (x-m)/s ... how do you derive it or show it works?
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    (Original post by metaltron)
    Yeah it is part of my working and is clearly wrong. I tried substituting in:

     Z = \frac{X-\mu}{\sigma}

    to try and get a standard normal distribution in terms of Z from the general normal distribution formula in terms of X. However I am stuck with a sigma at the bottom somehow.
    Sorry if I'm telling you something you already know, but it's not just a case of relabelling the variable in the pdf. You start with the condition that the pdf must integrate to 1 over the entire range, so you're actually changing a variable inside the integral. That means when you convert from dX to dZ you get an extra \sigma factor which cancels the one you have.

    Basically, you always need the appropriate normalisation constant for your pdf.
 
 
 
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