1. DonationAcolyte
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    20 Jun '04 23:16
    (Apologies for the nature of this puzzle. It's generally seen as a good thing to show 17-year-old maths students who want something a little harder than standard A-level questions.)

    The standard normal distribution is defined by the probability density function
    f(x) = K*exp(-x^2), where K is constant. Find K.
  2. Standard memberroyalchicken
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    21 Jun '04 01:02
    Originally posted by Acolyte
    (Apologies for the nature of this puzzle. It's generally seen as a good thing to show 17-year-old maths students who want something a little harder than standard A-level questions.)

    The [b]standard normal
    distribution is defined by the probability density function
    f(x) = K*exp(-x^2), where K is constant. Find K.[/b]
    (Disclaimer: I don't know any stats beyond the AP curriculum, which consists of a lot of messing about with calculators, so I didn't even know the form of the normal distribution until Acolyte just said it. For shame!)

    Well, since it's a continuous probability distribution, integrating it from zero to infinity wrt x should give 1, so I guess:

    K = [INTEGRAL(0 to infinity) exp(-x^2) dx]^-1

    Let's call that integral I, 'cause it's typographically mean. We have two options here, namely to integrate the Maclaurin series of the integrand term-by-term, which would give an answer in terms of gamma(1/2), which won't do a lot of good. Past experience is making a double integral in polar coordinates look like the sexy thing to do, so, noting that the integrand is always positive, we will find I^2 and take the positive square root:

    I^2 = INT(0 to inf)INT(0 to inf) exp(-(x^2 + y^2)) dydx

    = INT(0 to pi/2)INT(0 to inf)r*exp(-r^2) drdq

    = pi/2*INT(0 to inf)r*exp(-r^2)dq

    = pi/4

    So that I = pi^(1/2)/2 and thus K = 2*pi^(-1/2). That right?



  3. DonationAcolyte
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    21 Jun '04 15:15
    Originally posted by royalchicken
    Past experience is making a double integral in polar coordinates look like the sexy thing to do
    Bah! You've seen the trick before. But have you solved my other as-yet-unsolved problems involving a spherical path and patterns on a grid?
  4. Standard memberroyalchicken
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    21 Jun '04 20:06
    Originally posted by Acolyte
    Bah! You've seen the trick before. But have you solved my other as-yet-unsolved problems involving a spherical path and patterns on a grid?
    Same trick, very slightly different context. I haven't solved the spherical path one yet, and haven't seen the grid one yet. Will do.

    I'm taking STEP next week -- is it a bad idea to refer to one's strategies as 'the sexy thing to do'?

    *goes and looks*
  5. Standard memberroyalchicken
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    22 Jun '04 05:33
    Which is the grid one?
  6. DonationAcolyte
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    22 Jun '04 08:00
    Originally posted by royalchicken
    Which is the grid one?
    It's in the thread "Odd and Even?"
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