ssume that a procedure yields a binomial distribution with n

ssume 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 muand standard deviation sigma. Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigmaand the maximum usual value mu plus 2 sigma

n=1455, p=2/5

Solution

Note that

mu = n p = 1455*(2/5) = 582

sigma = sqrt[n p (1-p)] = sqrt(1455*(2/5)*(1-2/5)) = 18.6868938

Thus,

mu - 2*sigma = 544.6262124
mu + 2*sigma = 619.3737876

Thus, the usual values are between these two values. [ANSWER]