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In a study, 44% of adults questioned reported that their health was excellent. A researcher wishes to study the health of peo
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Answer #1

p=0.44

n=14

x=3

n P(X = 1) = p (1 – p)-2

P(X=x)=\binom{14}{3}0.44^x(1-0.44)^{14-x}

To find

P(X>3)

P(X>3)=1-P(X \leq 3)=1-\sum_0^3\binom{14}{x}0.44^x(1-0.44)^{14-x}=1-0.073=0.927

So probability that when 14 adults are randomly selected, more than 3 are in excellent health is 0.927

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