Question

sleep disorders a survey finds that 19% of americans suffer from a sleep disorder. for a randomly selected sample of 15 people, find each probability. round intermediate calculations and final answers to at least three decimal places.
your answer
part: 0 / 3
part 1 of 3
(a) at least 3 people have a sleep disorder
p(at least 3 people have a sleep disorder) =

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n = 15\), \(p=0.19\), \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Calculate \(P(X\geq3)=1-(P(X = 0)+P(X = 1)+P(X = 2))\)

• For \(k = 0\):

\(C(15,0)=\frac{15!}{0!(15-0)!}=1\)
\(P(X = 0)=1\times(0.19)^{0}\times(1 - 0.19)^{15-0}=(0.81)^{15}\approx0.055\)

• For \(k = 1\):

\(C(15,1)=\frac{15!}{1!(15 - 1)!}=15\)
\(P(X = 1)=15\times(0.19)^{1}\times(0.81)^{14}=15\times0.19\times(0.81)^{14}\approx0.157\)

• For \(k = 2\):

\(C(15,2)=\frac{15!}{2!(15-2)!}=\frac{15\times14}{2\times1}=105\)
\(P(X = 2)=105\times(0.19)^{2}\times(0.81)^{13}=105\times0.0361\times(0.81)^{13}\approx0.237\)

Step3: Compute \(P(X\geq3)\)

\(P(X\geq3)=1-(0.055 + 0.157+0.237)=1 - 0.449 = 0.551\)

Answer:

\(0.551\)