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Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean mu and standard deviation sigma. Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma. n=1580,=1/4
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Use expected value to calculate for the mean. For binomial distributions \(B(n,p,k)\), mean or expected value is \(np\) The standard deviation or Sx is \(√(np(1-p)\)
For a moment I forgot range rule of thumb, I think it was that the range is always 4 times the standard deviation of the population/sample
But once you get your expected value and Sqx, you can find the minimum usual and maximum usual values.
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