(New page: the variance of a binomial random variable: *var(X) = E[X^2] - (E[X])^2 * = E[X(X-1)] + E[X] - (E[X])^2 * = n*(n-1)*P^2 + n*p - (n*p)^2 * = n*p - n*p^2 *now, set n = 100...)
 
 
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the variance of a binomial random variable:
 
the variance of a binomial random variable:
*var(X) = E[X^2] - (E[X])^2
+
      = E[X^2] - (E[X])^2
*      = E[X(X-1)] + E[X] - (E[X])^2
+
      = E[X(X-1)] + E[X] - (E[X])^2
*      = n*(n-1)*P^2 + n*p - (n*p)^2
+
      = n*(n-1)*P^2 + n*p - (n*p)^2
*      = n*p - n*p^2
+
      = n*p - n*p^2
  
*now, set n = 1000, take derivative with respect to p, set equal to zero, solve for p, and plug into var(x) equation to solve for the maximum value.
+
now, set n = 1000, take derivative with respect to p, set equal to zero, solve for p, and plug p value into var(x) equation to solve for the maximum value.
  
*i end up getting p = .5 so max var(X) = 250 (when X is a binomial random variable with parameter 1000)
+
i end up getting p = .5 so max var(X) = 250 (when X is a binomial random variable with parameter 1000)
 +
 
 +
i think this is correct but let me know if there's a mistake..

Latest revision as of 06:00, 3 November 2008

the variance of a binomial random variable:

      = E[X^2] - (E[X])^2
      = E[X(X-1)] + E[X] - (E[X])^2
      = n*(n-1)*P^2 + n*p - (n*p)^2
      = n*p - n*p^2

now, set n = 1000, take derivative with respect to p, set equal to zero, solve for p, and plug p value into var(x) equation to solve for the maximum value.

i end up getting p = .5 so max var(X) = 250 (when X is a binomial random variable with parameter 1000)

i think this is correct but let me know if there's a mistake..

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