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Thread: Sampling with replacement - standard deviation

  1. #1
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    Sampling with replacement - standard deviation

    Can anyone help me with this problem?

    Suppose there are N balls in total. At each sampling, a ball is randomly taken out, marked with a mark, and put back to the samples. If a ball is taken out for a second time during a sampling, just put back without doing anything. Let X denote the number of balls being marked after n repeated samplings. The expectation of X can be calculated as: N*(1 - (1 - 1/N)^n) . But how to calculate the standard deviation of X?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by southliguang View Post
    Can anyone help me with this problem?

    Suppose there are N balls in total. At each sampling, a ball is randomly taken out, marked with a mark, and put back to the samples. If a ball is taken out for a second time during a sampling, just put back without doing anything. Let X denote the number of balls being marked after n repeated samplings. The expectation of X can be calculated as: N*(1 - (1 - 1/N)^n) . But how to calculate the standard deviation of X?
    The probability that a given ball is not marked is $\displaystyle p=N^{-n}$. So the distribution of the number of unmarked balls is $\displaystyle B(N,p)$. The standard deviation of the number of marked balls is equal to that of the number of unmarked balls.

    CB
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