Sampling with replacement - standard deviation

• Dec 28th 2009, 09:45 PM
southliguang
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?
• Jan 1st 2010, 10:31 PM
CaptainBlack
Quote:

Originally Posted by southliguang
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 $p=N^{-n}$. So the distribution of the number of unmarked balls is $B(N,p)$. The standard deviation of the number of marked balls is equal to that of the number of unmarked balls.

CB