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| Question 805466:  Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial p.  Use the given values of n & p to find the mean and standard deviation.  Also, use the range rule of thumb to find the minimum usual value mean - 2standard deviation and the maximum usual value mean + 2standard deviation. n = 150, p = 0.8
 Answer by stanbon(75887)
      (Show Source): 
You can put this solution on YOUR website! Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial p. Use the given values of n & p to find the mean and standard deviation. mean = np
 std = sqrt(npq)
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 Also, use the range rule of thumb to find the minimum usual value mean - 2standard deviation and the maximum usual value mean + 2standard deviation.
 n = 150, p = 0.8
 mean = 150*0.8 = 120
 std = sqrt[120*0.2] = sqrt(24) = 2sqrt(6)
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 Cheers,
 Stan H.
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