Question #305749

1.     Suppose we know that 70% of all Canadians have or had a pet in their household. A researcher takes a simple random sample of 20 Canadians. What is the probability that at least 9 people in the sample have never had a pet in their household? Explain how you got the number. (2)



1
Expert's answer
2022-03-05T04:59:04-0500

Let X=X= the number of people who have never had a pet in their household: XBin(n,p).X\sim Bin(n, p).

Given n=20,q=0.7,p=0.3n=20, q=0.7, p=0.3


P(X9)=1P(X<9)=1P(X=0)P(X=1)P(X\ge9)=1-P(X<9)=1-P(X=0)-P(X=1)

P(X=2)P(X=3)P(X=4)-P(X=2)-P(X=3)-P(X=4)

P(X=5)P(X=6)P(X=7)-P(X=5)-P(X=6)-P(X=7)

P(X=8)=1(200)(0.3)0(0.7)20-P(X=8)=1-\dbinom{20}{0}(0.3)^0(0.7)^{20}

(201)(0.3)1(0.7)19(202)(0.3)2(0.7)18-\dbinom{20}{1}(0.3)^1(0.7)^{19}-\dbinom{20}{2}(0.3)^2(0.7)^{18}

(203)(0.3)3(0.7)17(204)(0.3)4(0.7)16-\dbinom{20}{3}(0.3)^3(0.7)^{17}-\dbinom{20}{4}(0.3)^4(0.7)^{16}

(205)(0.3)5(0.7)15(206)(0.3)6(0.7)14-\dbinom{20}{5}(0.3)^5(0.7)^{15}-\dbinom{20}{6}(0.3)^6(0.7)^{14}

(207)(0.3)7(0.7)13(208)(0.3)8(0.7)12-\dbinom{20}{7}(0.3)^7(0.7)^{13}-\dbinom{20}{8}(0.3)^8(0.7)^{12}

0.11333146288\approx0.11333146288




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