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Homework Help: I think this is about the Central Limit Theorem

  1. Nov 23, 2015 #1
    1. The problem statement, all variables and given/known data
    An engineer is measuring a quantity q. It is assumed that there is a random error in each measurement, so the engineer will take n measurements and reports the average of the measurements as the estimated value of q. Specifically, if Yi is the value that is obtained in the i'th measurement, we assume that

    where Xi is the error in the ith measurement. We assume that Xi's are i.i.d. with EXi=0 and Var(Xi)=4 units. The engineer reports the average of measurements

    Mn = (Y1 + .... Yn) / n

    many measurements does the engineer need to make until he is 95% sure that the final error is less than 0.1 units? In other words, what should the value of n be such that

    2. Relevant equations
    Central Limit Theorem states:
    Zn = (Mx - μ) / (σ / √n)

    3. The attempt at a solution
    So this is the formula I chose to use. It seems like a simple variable swap, but my problem is pulling n out.

    P(y1 ≤ Y ≤ y2)
    = P( (y1 - nμ) / (σ√n) ≤ ((X1 +... Xn) - nμ) / (σ√n)) ≤ (y2 - nμ) / (σ√n)

    Then this would give ((q + 0.1) / (2√n)) - ((q - 0.1) / (2√n)) = Φ-1 (0.95) = 1.64

    combining this would give:(q/2√n) + (0.1 / 2√n) - (q / 2√n)- (0.1 / 2√n) = 0.2 / 2√n = 1/√n = 1.64 ⇒ n = (1/1.64)2 = 0.37

    Now... this is not an integer, and makes absolutely no sense. I am semi-confident in my process, but I think I may have made too many assumptions about the central limit theorem.
  2. jcsd
  3. Nov 24, 2015 #2


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    Why do you change from ##\ \sigma\over\sqrt n\ ## to ##\ \sigma\sqrt n\ ## ?

    It seems your n is used for two different purposes as well

  4. Nov 24, 2015 #3
    When you multiply by n/n you get σ√n
  5. Nov 24, 2015 #4


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    I would call that multiplying by n, not multiplying by n/n

    Check it anyway !
  6. Nov 24, 2015 #5
    It has to be n/n because Mx is (X1 +...+Xn)/n.

    In my extrapolation of the formula you quoted, the y1 = q - 0.1, nμ = 0 and σ = √4
    Which portion did I misinterpret/use twice?
  7. Nov 24, 2015 #6


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    The "Then" part, where you divide by n

    Another comment: Your 1.64 looks one-sided to me. And your notation asks more from the reader than is reasonable; teachers shouldn't allow that ! (there is no using n twice; I got confused -- sorry)

    I end up with an awful lot of measurements required, which is rather to be expected if you start out with such a large variance !

    By the way, even "Var(Xi)=4 units" is confusing. Shouldn't it be Var(Xi)=4 units2 ?

    More (admittedly nitpicking)
    Central Limit Theorem states: Zn = (Mx - μ) / (σ / √n) is normally distributed with standard deviation 1

    Matter of completeness: for some readers Zn is something else: an impedance or whatever.

    Last edited: Nov 24, 2015
  8. Nov 24, 2015 #7
    Sorry, it ought to be. My book isn't the best despite my professor's claim. I guess they're assuming "units" is general enough.
    Maybe I'm not looking at it the same way you are, I don't divide by n. I multiply by n to get y1 and y2 = q-0.1, q+0.1, respectively. This is consistent with examples in my book, I think. Are you saying I mistook these values as y1,y2 when they were in fact y1 - nμ, y2 + nμ?

    I apologize for my not showing as many steps, or in the best way. It's much neater on my paper. I'll upload it when i get home if it remains unclear.
  9. Nov 24, 2015 #8


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    No need to apologize, I'm trying to help.
    In the top line you have y1 for nq, in the bottom line you have q only. You only divide the numerator by n.

    The 1.64 is for 5% in one tail, you want 2.5% in each tail.


    [edit] The more I read it, the more nonsensical it becomes:

    What does it say here ?
    Last edited: Nov 24, 2015
  10. Nov 24, 2015 #9

    Ray Vickson

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    I suggest you go back to square one. You need to figure out how to make a 95% probability interval (centered at the mean) come out with a length of 0.2 (i.e., from -0.1 to + 0.1). In terms of the unit normal distribution, how many standard deviations wide would that be? Then carefully figure out what that means in terms of n. When I do it I get a value of n between 1000 and 2000, but I will not state the precise value.
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