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Central Limit Theorem Question

  1. May 4, 2014 #1
    1. The problem statement, all variables and given/known data
    The Rockwell hardness of certain metal pins is known to have a mean of 50 and a standard deviation of 1.5.
    a)if the distribution of all such pin hardness measurements is known to be normal, what is the probability that the average hardness for a random sample of 9 pins is at least 52.
    b)what if sample is 40 pins?


    2. Relevant equations
    So I know that μ = 50 and σ(original) = 1.5, and n = 9

    3. The attempt at a solution
    σ(sample) = σ(original)/[itex]\sqrt{9}[/itex] = 1.5/3 = .5

    now find P([itex]\bar{x}[/itex] > 52) by relating it to the standard normal dist

    find z = (52 - 50)/.5 = 4.

    This seems like a huge z score to have and it pretty much much makes the probability 0. Am I doing something wrong here? It just seems like a pointless question if in both cases the probability comes to zero.
     
    Last edited: May 4, 2014
  2. jcsd
  3. May 4, 2014 #2

    ehild

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    The probabilities are really very small, but still differ from zero. There are online calculators http://www.solvemymath.com/online_math_calculator/statistics/cdf_calculator.php [Broken], for example.

    ehild
     
    Last edited by a moderator: May 6, 2017
  4. May 4, 2014 #3

    Zondrina

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    Your sample size is quite small. Shouldn't you be using a t-estimator instead?
     
  5. May 4, 2014 #4

    mfb

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    Staff: Mentor

    We know (or assume) that all pins follow a normal distribution. Then the mean of 9 pins follows a normal distribution as well.
    The calculation is fine, the result is just a small number.
     
  6. May 4, 2014 #5

    Ray Vickson

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    No. When the variance is known we use the normal distribution. When the variance is unknown (and is estimated using the data itself) we use the t-distribution.
     
  7. May 4, 2014 #6

    Ray Vickson

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    The probability is small, but nowhere near zero. (In fact, the ability to estimate very small probabilities is crucial in reliability and safety studies, etc.) For large ##x > 0## you can find simple, fairly accurate approximations to ##G(x) \equiv P(X > x)## for standard normal ##X##.
    Let [tex]\phi(t) = \frac{1}{\sqrt{2 \pi}} e^{-t^2/2}. [/tex]
    We have
    [tex] G(x) = \int_x^{\infty} \phi(t) \, dt \\
    A_1(x) = \frac{\phi(x)}{x} \;\; \text{approximation 1}\\
    A_2(x) = \frac{\phi(x)}{x} - \frac{\phi(x)}{x^3} \;\; \text{approximation 2} [/tex]
    The functions ##A_1(x)## and ##A_2(x)## are approximations to ##G(x)##, with ##A_2## being a bit better than ##A_1## for large ##x > 0## (but may be worse for small ##x > 0##). You can see how they perform from the following table:
    Code (Text):

      x       G(x)      A1(x)      A2(x)
     1.0  1.58655e-01 2.41971e-01 0.00000e+00
     1.5  6.68072e-02 8.63451e-02 4.79695e-02
     2.0  2.27501e-02 2.69955e-02 2.02466e-02
     2.5  6.20967e-03 7.01132e-03 5.88951e-03
     3.0  1.34990e-03 1.47728e-03 1.31314e-03
     3.5  2.32629e-04 2.49338e-04 2.28984e-04
     4.0  3.16712e-05 3.34576e-05 3.13665e-05
     
    These approximations are based in integration by parts, using the fact that ##d \phi(t)/dt = - t \phi(t)##:
    [tex] G(x) = \int_x^{\infty} \phi(t) \, dt = \int_x^{\infty} \frac{1}{t} \left(- \frac{d \phi(t)}{dt} \right) \, dt
    = \frac{\phi(x)}{x} - \int_x^{\infty} \frac{\phi(t)}{t^2} \, dt[/tex]
    The approximation ##A_1(x)## is just the first term above, and it should be a pretty good approximation for large ##x > 0## because the magnitude of the neglected term is less than ##G(x)/x^2##. The approximation ##A_2(x)## estimates the second term above through another integration by parts.
     
    Last edited: May 4, 2014
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