Suppose the number of children in a household has a binomial distribution with p

Question

Suppose the number of children in a household has a binomial distribution with parameters n =15 and p = 80%. Find the probability of a household having:

a) 15C2(.8)2(.2)11 + 15C11(.8)11(.2)? Getting stuck on the last exponent...not sure what it is

b) 15C0(.8)0(.2)15 + 15C1(.8)1(.2)15 + 15C2(.8)2(.2)15 + 15C3(.8)3(.2)15 + 15C4(.8)4(.2)15 + 15C5(.8)5(.2)15 + 15C6(.8)6(.2)15 + 15C7(.8)7(.2)15 + 15C8(.8)8(.2)15 + 15C9(.8)9(.2)15

c) ?

d)

15C0(.8)0(.2)15 + 15C1(.8)1(.2)15 + 15C2(.8)2(.2)15 + 15C3(.8)3(.2)15 + 15C4(.8)4(.2)15 + 15C5(.8)5(.2)15 + 15C6(.8)6(.2)15 + 15C7(.8)7(.2)15 + 15C8(.8)8(.2)15 + 15C9(.8)9(.2)15 +  15C10(.8)10(.2)15

e) ?

(1 point) Suppose the number of children in a household has a binomial distribution with parameters n = 15 and p-80% Find the probability of a household having (a) 2 or 11 children (b) 9 or fewer children (c) 12 or more children (d) fewer than 11 children (e) more than 9 children

Explanation / Answer

For binomial disrtibution,

p(x) = nCx px (1-p)n-x

a)

p(2 or 11 children) = p(2) + p(11)

= 15C2 0.82 0.213 +  15C11 0.811 0.24

= 0.00000006+0.187604

= 0.187604

b)

p(x<=9) = p(0) + p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) + p(8) + p(9)

= 0.06105

c)

p( x>=12) = p(12) + p(13) + p(14) + p(15)

= 15C12 0.812 0.23 + 15C13 0.813 0.22 + 15C14 0.814 0.2 + 15C15 0.815 0.20

= 0.64816

d)

p(x < 11) = 1 - p( x >=11)

= 1 - [ p(11) + p(12) + p(13) + p(14) + p(15)]

= 0.16423

e)

p(x > 9) = p (x >=10)

= p(10) + p(11) + p(12) + p(13) + p(14) + p(15)

= 15C10 0.810 0.25 + 15C11 0.811 0.24 + 15C12 0.812 0.23 + 15C13 0.813 0.22 + 15C14 0.814 0.2 + 15C15 0.815 0.20

= 0.93895

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