# SOLUTION: a recent report stated that 4% of adults cut their sandwhich in half before eating it. if 10 u.s. adults are selected randomly, what is the probability that a) at least 6 people c

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: a recent report stated that 4% of adults cut their sandwhich in half before eating it. if 10 u.s. adults are selected randomly, what is the probability that a) at least 6 people c      Log On

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 Question 427985: a recent report stated that 4% of adults cut their sandwhich in half before eating it. if 10 u.s. adults are selected randomly, what is the probability that a) at least 6 people cut their sandwhich in half before eating b) at most 3 people cut their sandwhich in half before eating it c) (use binomial formula) exactly 4 people cut their sandwhich in half before eating it Thanks!Answer by stanbon(57239)   (Show Source): You can put this solution on YOUR website!a recent report stated that 4% of adults cut their sandwhich in half before eating it. if 10 u.s. adults are selected randomly, --- Binomial Problem with n= 10 and p = 0.04 ------- what is the probability that a) at least 6 people cut their sandwhich in half before eating Ans: P(6<= x <=10) = 1-binomcdf(10,0.04,5) = 0.00000074824 -------- b) at most 3 people cut their sandwhich in half before eating it Ans: binomcdf(10,0.04,3) = 0.9996 ========= c) (use binomial formula) exactly 4 people cut their sandwhich in half before eating it P(x= 4) = 10C4*(0.04)^4*(0.96)^6 = binompdf(10,0.04,4) =0.0004208 ============== Cheers, Stan H. ================