SOLUTION: A recent report stated that when eating a sandwich, 32% of U.S. adults cut it in half before eating it. If 9 U.S. adults are selected at random, what is the probability th

Algebra ->  Probability-and-statistics -> SOLUTION: A recent report stated that when eating a sandwich, 32% of U.S. adults cut it in half before eating it. If 9 U.S. adults are selected at random, what is the probability th      Log On


   

Question 106761: A recent report stated that when eating a sandwich, 32% of U.S. adults cut it in half before eating it. If 9 U.S. adults are selected at random, what is the probability that exactly 4 cut their sandwich in half before eating it?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
It is a binominal problem with n=9, p=0.32, x = 4
P(x=4) = 9C4(0.32)^4(0.68)^5 = 0.192094...
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I used binompdf(9,0.32,4) on a TI-83 calculator.
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Cheers,
Stan H.