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Thread: FInding mean and standard deviation in a normal distribution.

  1. #1
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    FInding mean and standard deviation in a normal distribution.

    Hi all. I'm stuck on this problem that one of my students brought along to tutoring.

    Of a large group of men, 3% are under 1.6125m tall and 50% are between 1.7m and 1.8m tall. If the heights are normally distributed, find the mean and the standard deviation.

    Using the 3% bit , I can get 1.6125 = \mu - 1.881\sigma.

    I just can't seem to get the second simultaneous equation from the 50% bit.

    I know that  P(X<1.8) - P(X<1.7) =0.5, but then what?

    (Edited: This implies that that 1.7<\mu<1.8 ... so I could solve it using excel and trial&error, but there must be a better way)

    Any help appreciated.
    Last edited by Debsta; Aug 24th 2017 at 02:23 AM.
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  2. #2
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    Re: FInding mean and standard deviation in a normal distribution.

    Well, using calculus and numerical analysis, this question is answerable. I am not sure how one would answer it without those tools.

    So, you have one formula:

    $1.6125 = \mu - 1.881 \sigma$

    Another two:

    $1.7 = \mu + z_1 \sigma$

    $1.8 = \mu + z_2 \sigma$

    And then one from calculus:

    $\displaystyle \sqrt{ \dfrac{2}{\pi} } \int_{z_1}^{z_2} e^{-\tfrac{x^2}{2}} dx = 0.5$

    This final equation can also be written:

    $z_2 = \sqrt{2} \text{erf}^{-1}\left[ 0.5 + \text{erf}\left( \dfrac{z_1}{\sqrt{2}} \right) \right]$

    where

    $\text{erf}^{-1}(z)$ is given here: https://en.wikipedia.org/wiki/Error_...erse_functions
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    Re: FInding mean and standard deviation in a normal distribution.

    Using some numerical analysis, I found $1.751833 < \mu < 1.752873$ and $0.074074 < \sigma < 0.074627$.

    Tool: Excel spreadsheet
    Column A: Z-score (increments of 0.01)
    Column B: Percentage between mean and z-score
    Column C: Column B - 0.5 (to get the negative percentage needed for 0.5 percent to be between them)
    Column E: Negative Column B (to be used as a lookup match column)
    Column F: Negative Column A (to be used as a lookup column)
    Column D: =INDEX(F:F,MATCH(C2,E:E,-1))

    Column H: the calculated sigma: =0.1/(A2-D2)
    Column I: the calculated mean: =1.6125+1.881*H2
    Column J: the error: =I2+H2*A2-1.8

    Column J will tell you how close you are to the correct answer. The closer to zero, the better.
    Last edited by SlipEternal; Aug 24th 2017 at 07:25 AM.
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    Re: FInding mean and standard deviation in a normal distribution.

    The feeling when you can't answer something as a tutor :/

    It is one thing to peer tutor for National Honor Society and another when you are making 60 smacks an hour.

    At least this question required numerical techniques, so it is understandable to not have an immediate answer at your disposal.
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    Re: FInding mean and standard deviation in a normal distribution.

    Thank you SlipEternal for your time in finding a solution. The question was from a Year 12 student, so the calculus approach would be beyond them at this stage. I think it may have been a case of the textbook author writing a "nice" question before thinking about how the answer would be found. A good learning experience nonetheless. Thanks again.

    Thanks for your comment, abender. With due respect, this happens to me very rarely. I do get paid for tutoring but I don't waste time (and student's money) pondering over a problem when I don't see an immediate solution. We discuss the difficulties (and there is learning taking place while we do) and move on. I then spend my own time away from the student trying to solve the problem (I won't let a problem beat me) at a level which they will understand, and get back to them with a solution, if there is one. I can assure you my students get their money's worth. But you raise a valid point. Cheers.
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    Re: FInding mean and standard deviation in a normal distribution.

    Quote Originally Posted by Debsta View Post
    Thank you SlipEternal for your time in finding a solution. The question was from a Year 12 student, so the calculus approach would be beyond them at this stage. I think it may have been a case of the textbook author writing a "nice" question before thinking about how the answer would be found. A good learning experience nonetheless. Thanks again.

    Thanks for your comment, abender. With due respect, this happens to me very rarely. I do get paid for tutoring but I don't waste time (and student's money) pondering over a problem when I don't see an immediate solution. We discuss the difficulties (and there is learning taking place while we do) and move on. I then spend my own time away from the student trying to solve the problem (I won't let a problem beat me) at a level which they will understand, and get back to them with a solution, if there is one. I can assure you my students get their money's worth. But you raise a valid point. Cheers.
    Oh, I didn't mean to insinuate that it happens to you often. But it happens to everyone. I like your approach. I frequently email follow-ups in LaTeX if there were any questions we either didn't get to, were confusing or I didn't feel I answered adequately.
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    Re: FInding mean and standard deviation in a normal distribution.

    Thinking about it, I bet they wanted you to assume the mean was 1.75 (it comes really close). If you do that, the problem is really easy to solve.
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    Re: FInding mean and standard deviation in a normal distribution.

    Quote Originally Posted by SlipEternal View Post
    Thinking about it, I bet they wanted you to assume the mean was 1.75 (it comes really close). If you do that, the problem is really easy to solve.
    Yes that did cross my mind, but then the info about the 3% becomes irrelevant (and incorrect). It was a poorly worded question if that was the author's intention.
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    Re: FInding mean and standard deviation in a normal distribution.

    Quote Originally Posted by Debsta View Post
    Yes that did cross my mind, but then the info about the 3% becomes irrelevant (and incorrect). It was a poorly worded question if that was the author's intention.
    Suppose we assume $\mu = 1.75$. Then by the equation you came up with: $1.6125 = 1.75 - 1.881\sigma$, so $\sigma = 0.073$.

    Now, we have that $1.7 = 1.75-0.685\sigma$ and $1.8 = 1.75+0.685\sigma$.

    We find that 49% of people are between 1.7 and 1.8 (pretty close to 50%). So, that may have been the approach the author was going for. Not a great question, certainly.
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    Re: FInding mean and standard deviation in a normal distribution.

    Quote Originally Posted by SlipEternal View Post
    Suppose we assume $\mu = 1.75$. Then by the equation you came up with: $1.6125 = 1.75 - 1.881\sigma$, so $\sigma = 0.073$.

    Now, we have that $1.7 = 1.75-0.685\sigma$ and $1.8 = 1.75+0.685\sigma$.

    We find that 49% of people are between 1.7 and 1.8 (pretty close to 50%). So, that may have been the approach the author was going for. Not a great question, certainly.
    Or, ignoring the 3% bit for now, we get  1.8 = 1.75 + 0.67449\sigma which gives \sigma=0.0741.

    Then, testing this result, we find that P(X<1.6125)=0.0318 not 0.03 or 3% as stated in the question.

    Don't like the question at all, especially in a high school textbook.

    Interesting nonetheless! Thanks for your input.
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