Assume that a procedure yields a binomial distribution with

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 - 2 sigma and the maximum usual value mu + 2 sigma . n = 1584, p = 3/4

Solution

Here,

u = np = 1584*(3/4) = 1188 [ANSWER]

sigma = sqrt(np(1-p))= sqrt(1584*(3/4)*(1-3/4)) = 17.23368794 [ANSWER]

Thus,

u - 2*sigma = 1153.532624 [ANSWER]
u + 2*sigma = 1222.467376 [ANSWER]