Question

sleep disorders a survey finds that 21% 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.
part: 0 / 3
part 1 of 3
(a) at least 2 people have a sleep disorder
p (at least 2 people have a sleep disorder) =
try again

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 \(n = 15\), \(p=0.21\), \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Calculate \(P(X\lt2)=P(X = 0)+P(X = 1)\)

• For \(k = 0\):

\(C(15,0)=\frac{15!}{0!(15-0)!}=1\)
\(P(X = 0)=1\times(0.21)^{0}\times(1 - 0.21)^{15-0}=(0.79)^{15}\approx0.031\)

• For \(k = 1\):

\(C(15,1)=\frac{15!}{1!(15 - 1)!}=\frac{15!}{1!14!}=15\)
\(P(X = 1)=15\times(0.21)^{1}\times(0.79)^{14}\)
\(=15\times0.21\times(0.79)^{14}\approx15\times0.21\times0.039\approx0.123\)

Step3: Calculate \(P(X\geq2)\)

\(P(X\geq2)=1-(P(X = 0)+P(X = 1))\)
\(=1-(0.031 + 0.123)=1 - 0.154=0.846\)

Answer:

\(0.846\)