Answer:
The Public Citizen’s Health Research Group studied the serious disciplinary actions that were taken during a recent year on nonfederal medical doctors in the United States. The national average was 3.05 serious actions per 1000 doctors. The state with the lowest number was Minnesota with only 1.07 serious actions per 1000 doctors. Assume that the numbers of serious actions per 1000 doctors in both the United States and in Minnesota are Poisson distributed. (Round the intermediate values to 4 decimal places. Round your answers to 4 decimal places.) a. What is the probability of randomly selecting 1000 U.S. doctors and finding one serious action taken? b. What is the probability of randomly selecting 2000 U.S. doctors and finding four serious actions taken? c. What is the probability of randomly selecting 3000 Minnesota doctors and finding fewer than seven serious actions taken?
Explanation:
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The probability will be:
(a) 0.455
(b) 5.5 × 10⁻⁴
(c) 0.9548
According to the question,
Since X being the no. of actions and is Poisson distributed.
Hence,
→ [tex]P(X) = \frac{e^{-\mu} \mu^x}{x!}[/tex]
(a)
→ [tex]P(X-1) = \frac{e^{-3.05} 3.05}{3!}[/tex]
[tex]= 0.1445[/tex]
(b)
→ [tex]P(X =4) = \frac{2^{-6.1}(6.1)^4}{4!}[/tex]
[tex]= 5.7\times 10^{-4}[/tex]
(c)
→ [tex]P(X<7) = P(0)+P(1)+P(2)+...+P(6)[/tex]
[tex]= e^{-3.21}\times 3.21^0 +e^{-3.21} 3.21^1+ \frac{e^{-3.21}3.21^2}{2} +...[/tex]
[tex]= 0.9548[/tex]
Thus the above answers are correct.
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