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Normal distribution problem

  1. Apr 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A student sits a test and is told that the marks follow a normal distribution with mean 100. The student receives a mark of 124 and is told that he is at the 68th percentile. Calculate the variance of the distribution.

    2. Relevant equations
    https://lh3.googleusercontent.com/proxy/XW4iFBVEU-T7v_I-U5zV6mo7CvCYxVfS6kbEiAuBBsiEcJWb-P7AHpF6ciRBv4skZW5pqDilEHh9F5zdeCfu0N0ztdfDS_lm-A=w685-h215-nc

    3. The attempt at a solution
    I replaced 0.68 for f (x) and 100 for μ and 124 for x. I tried to rearange to solve for σ but was unsuccessful only getting as far as,
    0.553 = -288/σ^2 - lnσ
    I think my approach might be completely wrong, or is there a way of doing this with the aid of a graphic calculator?
     
  2. jcsd
  3. Apr 4, 2016 #2

    Ray Vickson

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    Your ##f##-formula is incorrect: is should be
    [tex] f(x) = \frac{1}{\sqrt{2 \pi} \delta} e^{- \displaystyle \frac{(x-\mu)^2}{2 \delta^2}} [/tex]
    Anyway, that is not relevant for this problem: a percentile is a point on the cumulative distribution, not on the probability density function.

    There is no closed-form formula for the cumulative distribution, so you need to use tables of the cumulative, or use appropriate software. Some scientific calculators even have a "normal cumulative" button.
     
    Last edited: Apr 4, 2016
  4. Apr 4, 2016 #3

    SteamKing

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    Your formula shows the PDF for the normal distribution with mean μ and SD δ. The percentile of the student's score represents the area under the PDF distribution curve from -∞ to x. For example, the 50th percentile means that x = μ. Like Ray Vickson posted, you'll have to use standard normal tables to find out the SD.
     
  5. Apr 4, 2016 #4
    ∫ f (x)dx=0.68

    Where f (x) is the PDF

    Will this work as a definite integral from 0 to 124 and I try to solve again for σ ?
    Well this seemed like it would work until I actually tried to integrate it:)
     
    Last edited: Apr 4, 2016
  6. Apr 4, 2016 #5

    SteamKing

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    No, that's not how the standard normal tables are constructed.

    The standard normal distribution has μ = 0 and δ = 1.0. In order to accommodate normal distributions with other values of μ and δ, a transformation formula is used:

    ##Z = \frac{x-μ}{δ}##

    The total area under the standard normal distribution curve is

    ##\int_{-∞}^{+∞} f(Z)dZ = 1.0##,

    but since the curve is symmetrical about μ = 0

    ##\int_{-∞}^0 f(Z)dZ = 0.5##

    Knowing the value of the cumulative distribution, you want to find Z which corresponds to that amount from a standard normal table, like this one:

    https://www.mathsisfun.com/data/standard-normal-distribution-table.html

    Once you find the correct Z value, then using the student's score and the mean of the distribution, you should be able to work back and solve for the S.D., δ.
     
  7. Apr 4, 2016 #6
    So I set it up so that
    P (X≤124) = 0.68
    and
    P (Z≤(124-100)/σ)
    Using invNorm on my calculator for the standard (z) distribution I found
    (124-100)/σ≈0.468
    σ≈51.3
    Variance = 10051
    Nice :)

    Thank you
     
  8. Apr 4, 2016 #7

    SteamKing

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    Everything looks OK, except the variance should be equal to σ2. IDK how you got 10051.
     
  9. Apr 4, 2016 #8
    Your right, I must have put the wrong number in the calculator.
     
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