MATH SOLVE

8 months ago

Q:
# A recent study reported that the prevalence of hyperlipidemia (defined as total cholesterol over 200) is 30% in children 2-6 year of age. If 12 children are analyzed:a.What is the probability that at least 3 are hyperlipidemic?

Accepted Solution

A:

Answer:The probability is 0.74719Step-by-step explanation:Let's start defining the random variable X.X : ''Number of children with hyperlipidemia out of 12 children''X can be modeled as a binomial random variable.X ~ Bi (n,p)Where n is the sample size and p is the ''success probability''.We defining as a success to find a child that has hyperlipidemia.The probability function for X is : [tex]P(X=x)=(nCx).p^{x}.(1-p)^{n-x}[/tex]Where nCx is the combinatorial number define as :[tex]nCx=\frac{n!}{x!(n-x)!}[/tex]We are looking for [tex]P(X\geq 3)[/tex][tex]P(X\geq 3)=1-P(X\leq 2)[/tex][tex]P(X\geq 3)=1-[P(X=0)+P(X=1)+P(X=2)][/tex][tex]P(X\geq 3)=1-[(12C0)0.3^{0}0.7^{12}+(12C1)0.3^{1}0.7^{11}+(12C2)0.3^{2}0.7^{10}][/tex][tex]P(X\geq 3)=1-(0.7^{12}+0.07118+0.16779)=1-0.25281=0.74719[/tex]There is a probability of 0.74719 that at least 3 children are hyperlipidemic.