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

Answer:395, 17.212 (360.576,429.424)Step-by-step explanation:Given that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p.Here [tex]n=1580and p =1/4 =0.25\\q = 0.75\\np = 395 and\\nq = 1185[/tex]Since np and nq are greater than 5 by rule of thumb we can approximate binomial to normal.Mean = np = 395Variance = npq = [tex]395*0.75=296.25[/tex]Std dev = 17.212Thus X no of successes is N(395, 17.212)THe the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma would be[tex]395-2(17.212), 395+2(17.212)\\= 360.576,429.424[/tex]