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(75887)   (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,
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Binomial Problem with n= 10 and p = 0.04
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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
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b) at most 3 people cut their sandwhich in half before eating it
Ans: binomcdf(10,0.04,3) = 0.9996
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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
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Cheers,
Stan H.
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