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 1 / 3
part 2 of 3
(b) 5 or 6 people have a sleep disorder
p(5 or 6 people have a sleep disorder) =

Question

Accepted Answer

# Explanation:
## Step1: Recall binomial probability formula
The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 20$, $p=0.2$, $C(n,k)=\frac{n!}{k!(n - k)!}$

## Step2: Calculate $P(X = 5)$
$$
\begin{align*}
C(20,5)&=\frac{20!}{5!(20 - 5)!}=\frac{20\times19\times18\times17\times16}{5\times4\times3\times2\times1}=15504\
P(X = 5)&=C(20,5)\times(0.2)^{5}\times(0.8)^{15}\
&=15504\times0.00032\times0.035184372088832\
&\approx0.1745
\end{align*}
$$

## Step3: Calculate $P(X = 6)$
$$
\begin{align*}
C(20,6)&=\frac{20!}{6!(20 - 6)!}=\frac{20\times19\times18\times17\times16\times15}{6\times5\times4\times3\times2\times1}=38760\
P(X = 6)&=C(20,6)\times(0.2)^{6}\times(0.8)^{14}\
&=38760\times0.000064\times0.04398046511104\
&\approx0.1091
\end{align*}
$$

## Step4: Calculate $P(X = 5\ or\ X = 6)$
Since $P(A\ or\ B)=P(A)+P(B)$ (for mutually - exclusive events $A$ and $B$), $P(X = 5\ or\ X = 6)=P(X = 5)+P(X = 6)$
$P(X = 5\ or\ X = 6)=0.1745 + 0.1091=0.2836\approx0.284$

# Answer:
$0.284$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall binomial probability formula

Step2: Calculate \(P(X = 5)\)

Step3: Calculate \(P(X = 6)\)

Step4: Calculate \(P(X = 5\ or\ X = 6)\)

Answer: