Assume that a procedure yields a binomial distribution with n trials and the pro

Question

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 = 170, p = 0.25 mu = sigma = (Round to one decimal place as needed.) mu - 2 sigma = (Round to one decimal place as needed.) mu + 2 sigma = (Round to one decimal place as needed.)

Explanation / Answer

mu=n*p=170*0.25=42.5

sigma is sqrt(np(1-p))=sqrt(170*0.25*0.75)=5.6

mu-2sigma =42.5-2*5.6=31.3

mu+2sigma=42.5+2*5.6=53.7

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