SOLUTION: NBC news reported on May 2, 2013, that 1 in 20 children in the United States have a food allergy of some sort. Consider selecting a random sample of 25 children and let X be the nu
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-> SOLUTION: NBC news reported on May 2, 2013, that 1 in 20 children in the United States have a food allergy of some sort. Consider selecting a random sample of 25 children and let X be the nu
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Question 1054952: NBC news reported on May 2, 2013, that 1 in 20 children in the United States have a food allergy of some sort. Consider selecting a random sample of 25 children and let X be the number in the sample who have a food allergy. Then X~Bin(25, 0.05)
a) Determine both P(X<=3) and P(X<3)
b) Determine P(X>=4) Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! For x=3
25C3(0.05)^3(0.95)^22; 25C3=2300
=0.0930
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x=2
25C2(0.05)^2(0.95)^23=0.2305
x=1 25*0.05*0.95^24=0.3650
x=0 0.95^25=0.2774
The sum of those is 0.8729. That is p x<3.
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Probability x >=4 is 1-(0.8729+0.0930)=0.0341
