Question

sleep disorders a survey finds that 20% of americans suffer from a sleep disorder. for a randomly selected sample of 20 people, find each probability. round intermediate calculations and final answers to at least three decimal places.
part 1 of 3
(a) at least 2 people have a sleep disorder
p(at least 2 people have a sleep disorder) = 0.931
part 2 of 3
(b) 5 or 6 people have a sleep disorder
p(5 or 6 people have a sleep disorder) = 0.284
part 3 of 3
(c) exactly 2 people have a sleep disorder
p(exactly 2 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 \(n = 20\) (sample size), \(p=0.2\) (probability of success), \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Calculate \(P(X = 2)\)

• Calculate \(C(20,2)=\frac{20!}{2!(20 - 2)!}=\frac{20\times19}{2\times1}=190\)
• \(p = 0.2\), \(1-p=0.8\), \(k = 2\)
• \(P(X = 2)=C(20,2)\times(0.2)^{2}\times(0.8)^{18}\)
• \((0.2)^{2}=0.04\), \((0.8)^{18}\approx0.018\)
• \(P(X = 2)=190\times0.04\times0.018\)
• \(190\times0.04 = 7.6\)
• \(7.6\times0.018=0.137\)

Answer:

\(0.137\)