Q:

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).

Accepted Solution

A:
Part 1:

The probability of a binary distribution x ~ bin(n, p) is given by

[tex]P(X=x)=(^n_x)p^x(1-p)^{n-x}[/tex]

[tex]P(x\leq3)=P(0)+P(1)+P(2)+P(3) \\ \\ =(^{25}_0)(0.05)^0(1-0.05)^{25-0}+(^{25}_1)(0.05)^1(1-0.05)^{25-1} \\ +(^{25}_2)(0.05)^2(1-0.05)^{25-2}+(^{25}_3)(0.05)^3(1-0.05)^{25-3} \\ \\ =1\cdot1\cdot(0.95)^{25}+25\cdot(0.05)\cdot(0.95)^{24}+300\cdot(0.0025)\cdot(0.95)^{23} \\ +2300\cdot(0.000125)\cdot(0.95)^{22} \\ \\ =0.2774+0.3650+0.2305+0.0930 \\ \\ =\bold{0.9659}[/tex]



Part 2:

[tex]P(X\ \textless \ 3)=P(0)+P(1)+P(2) \\ \\ =P(X\leq3)-P(X=3)=0.9659-0.0930 \\ \\ =\bold{0.8729}[/tex]



Part 3:

[tex]P(X\geq4)=P(4)+P(5)+ . . . +P(25) \\ \\ =1-P(x\ \textless \ 4)=1-[P(0)+P(1)+P(2)+P(3)] \\ \\ =1-0.9659=\bold{0.0341}[/tex]