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Math Help - Standard Deviation of a Coin Toss

  1. #1
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    Standard Deviation of a Coin Toss

    I am trying to calculate the standard deviation of a coin toss where the probability of one coin toss being either a 'head' or a 'tail' is 50%.

    I have made the calculations using the following formulas:

    μ (Mean)= Nπ (N = number of trials, π = probability of success)

    σ2 (Variance) = Nπ(1-π)

    σ (Standard Deviation) = (Nπ (1-π))Square Rooted

    (I appologise for the way I have expressed the last formula, but I don't know how to show Square Root signs on the forum)

    So far I have made calculations for a sample of 12 and a sample of 10,000

    Sample of 12:
    μ = Nπ= (12)(0.5) = 6
    σ2 = Nπ(1-π)= (12)(0.5)(1.0 - 0.5) = 3.0
    σ = 1.73

    Sample of 10,000:
    μ = Nπ= (10,000)(0.5) = 5,000
    σ2 = Nπ(1-π)= (10,000)(0.5)(1.0 - 0.5) = 2,500
    σ = 50

    These calculations don't seem right to me. I would expect the standard deviation to be roughly between 1.05 to 1.10. I would also expect the
    standard deviation to become reasonbaly constant once over 10,000 trials have been calculated, but if I calculate the stanard deviation for 1,000,000
    trials I get, σ = 500

    Where am I going wrong?

    Dan
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  2. #2
    Super Member TheChaz's Avatar
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    Quote Originally Posted by danw76 View Post
    I am trying to calculate the standard deviation of a coin toss where the probability of one coin toss being either a 'head' or a 'tail' is 50%.

    I have made the calculations using the following formulas:

    μ (Mean)= Nπ (N = number of trials, π = probability of success)

    σ2 (Variance) = Nπ(1-π)

    σ (Standard Deviation) = (Nπ (1-π))Square Rooted

    (I appologise for the way I have expressed the last formula, but I don't know how to show Square Root signs on the forum)

    So far I have made calculations for a sample of 12 and a sample of 10,000

    Sample of 12:
    μ = Nπ= (12)(0.5) = 6
    σ2 = Nπ(1-π)= (12)(0.5)(1.0 - 0.5) = 3.0
    σ = 1.73

    Sample of 10,000:
    μ = Nπ= (10,000)(0.5) = 5,000
    σ2 = Nπ(1-π)= (10,000)(0.5)(1.0 - 0.5) = 2,500
    σ = 50

    These calculations don't seem right to me. I would expect the standard deviation to be roughly between 1.05 to 1.10. I would also expect the
    standard deviation to become reasonbaly constant once over 10,000 trials have been calculated, but if I calculate the stanard deviation for 1,000,000
    trials I get, σ = 500

    Where am I going wrong?

    Dan
    See the crosspost:
    http://www.mymathforum.com/viewtopic.php?f=24&t=20870
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