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Math Help - Fill In the Blanks

  1. #1
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    Fill In the Blanks

    In one toss of a fair coin, we expect ___0.5_____ heads, give or take _________ heads or so.

    In n=100 tosses, we expect ___50______ heads, give or take _________ heads or so.

    In n=400 tosses, we expect ___200_____ heads, give or take _________ heads or so.

    Note that the standard deviation is proportional to the square root of the number of tosses.

    I think I got the first fill-in-the-blanks right, but not sure what the 2nd blanks are about. I'm guessing standard deviation. However, I am confused when they define SD as proportional to the sq rt of the number of tosses. Does that mean, sqrt(1) = 1, and sqrt(100) = 10, and sqrt(400) = 20.
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  2. #2
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    Discrete Probability Distribution

    I tried the following solution, not sure about the accuracy of the results though

    Coin Toss

    Number----------- Probability
    of Heads--------- of Outcome
    x --------------- P(x)
    0 --------------- 1/2 = 0.5
    1 --------------- 1/2 = 0.5
    Total------------ 2/2 = 1


    Mean of a probability distribution

    \mu = \sum [xP(x)]
    \mu = 0(0.5)+1(0.5)
    \mu = 0.5

    Variance
    \sigma^2 = \sum [(x-\mu)^2P(x)]
    \sigma^2 = [(0-0.5)^2 (0.5) + (1-0.5)^2 (0.5)]
    \sigma^2 = [(-0.5)^2 (0.5) + (0.5)^2 (0.5)]
    \sigma^2 = [(0.25)(0.5) + (0.25)(0.5)]
    \sigma^2 = [(0.125) + (0.125)]
    \sigma^2 = 0.25

    Standard Deviation
    \sigma = 0.5

    In one toss of a fair coin, we expect ___0.5_____ heads, give or take ___0.5____ heads or so.

    In n=100 tosses, we expect ___50______ heads, give or take ___5___ heads or so

    In n=400 tosses, we expect ___200_____ heads, give or take ___10____ heads or so.
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  3. #3
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    For SD, you can also use sqrt[(n)(pi)(1-pi)]. I agree with your answers.

    Quote Originally Posted by dexteronline View Post
    I tried the following solution, not sure about the accuracy of the results though

    Coin Toss

    Number----------- Probability
    of Heads--------- of Outcome
    x --------------- P(x)
    0 --------------- 1/2 = 0.5
    1 --------------- 1/2 = 0.5
    Total------------ 2/2 = 1


    Mean of a probability distribution

    \mu = \sum [xP(x)]
    \mu = 0(0.5)+1(0.5)
    \mu = 0.5

    Variance
    \sigma^2 = \sum [(x-\mu)^2P(x)]
    \sigma^2 = [(0-0.5)^2 (0.5) + (1-0.5)^2 (0.5)]
    \sigma^2 = [(-0.5)^2 (0.5) + (0.5)^2 (0.5)]
    \sigma^2 = [(0.25)(0.5) + (0.25)(0.5)]
    \sigma^2 = [(0.125) + (0.125)]
    \sigma^2 = 0.25

    Standard Deviation
    \sigma = 0.5

    In one toss of a fair coin, we expect ___0.5_____ heads, give or take ___0.5____ heads or so.

    In n=100 tosses, we expect ___50______ heads, give or take ___5___ heads or so

    In n=400 tosses, we expect ___200_____ heads, give or take ___10____ heads or so.
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