Module 2.1 Practice (2)

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Yes (A) 180 107 No (A') 43 14 1. Give an example of a simple event. 2. Give an example of a joint event. 7. Given a client in this sample who ha 8. Test whether “client has health insu Insurance factor : Does the individual have h Insurance factor Normal (B) Obese (B') 3.What is the probability that a rando 4. What is the probability that a rando 5. What is the probability that a rando 6.What is the probability that a rando
Count of weight status Underweight (B'') 116 40 as a weight status “obese”, what is the probability that he/she has health insurance urance” and “is obese” are independent events. health insurance and weight status of the client. omly selected client has health insurance? omly selected client has normal weight? omly selected client has health insurance and normal weight? omly selected client has health insurance or is obese?
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Survey of cross-sectional data from NSW Australia revealed that 81% clients have health insurance. If none would have health insurance. P (X = 0) 6.131066E-08 all would have health insurance. P (X= 10) 0.121576655 no more than two people would have health insurance. P(x<=2) 5.28182E-05 less than five would have health insurance. P X<5) 0.004875725 at least three would have health insurance P(x>=3) 0.999947182
f 10 people are randomly selected in NSW, determine the probability that P(X<5) P(X <=2) 6.131066E-08 6.131066E-08 2.61377E-06 2.61377E-06 5.014312E-05 5.014312E-05 0.000570048 5.28182E-05 0.004252859 0.004875725
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Survey of cross-sectional data from NSW Australia revealed that 45% clients have normal weight. If 12 peop What is the probability that zero clients have normal weight? What is the probability that no more than 2 clients have normal weight? What is the probability that three or more clients have normal weight? What are the mean and standard deviation?
ple are randomly selected in NSW, determine the following.
Traffic accidents on any given day in Sydney is an example of Poisson distribution. The mean number of accid
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dents reported is 6 per day. The probability that an accident involves a driver with no comprehensive car ins
surance is 0.15.
c. What is the mean and standard deviation?
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