Probability Assignment Help

Probability Assignment Help

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A restaurant in your town had 163 diners on Friday night. By Monday evening, 24 of those diners had reported to the Emergency Department, outpatient clinics, or saw their physicians with symptoms of gastrointestinal distress, suggesting food poisoning.

Stool cultures for Salmonella were positive in all 24 diners.

Problem 1 What is the estimated* probability of a diner having contracted food poisoning at that restaurant on that Friday night? [*Note: This is an estimated probability because it doesn’t include persons whose symptoms may not have been severe enough for them to have sought medical attention (the actual probability could be higher); statisticsassignmenthelp.com

nor does it exclude the possibility that some of those symptomatic people may have contracted food poisoning from some other source (the actual probability could be lower); but the estimated probability would give epidemiologists a base line figure to work from.

Solution We are given that; A restaurant in your town had 163 diners on Friday night. By Monday evening, 24 of those diners had reported to the Emergency Department, outpatient clinics, or saw their physicians with symptoms of gastrointestinal distress, suggesting food poisoning. n = total number of dinners = 163 x = number of dinners had reported to the Emergency Department, outpatient clinics, or saw their physicians with symptoms of gastrointestinal distress, suggesting food poisoning = 24 x = 24 then n - x = 163 - 24 = 139 = those we do not have reported to the Emergency Department.

What is the estimated* probability of a diner having contracted food poisoning at that restaurant on that Friday night? statisticsassignmenthelp.com

P( a diner having contracted food poisoning at that restaurant on that Friday night ) = x / n P( a diner having contracted food poisoning at that restaurant on that Friday night ) = 24 / 163 P( a diner having contracted food poisoning at that restaurant on that Friday night ) = 0.1472 Problem 2: In 2017, Texas led the nation in the percentage of people who lacked health insurance (17% of the population). It is known that, nationally, 5% of patients account for 50% of the costs of healthcare. These are the “high cost” patients.

Assume* that: Being a high cost patient and being uninsured are independent characteristics Insured and uninsured people become “patients” at the same rate The uninsured and high cost patients in Texas are evenly distributed across the state, and that high cost patients are evenly distributed across insured and uninsured patient populations.

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a. What is the probability that a patient in a Texas healthcare facility will be a high cost patient who is uninsured?

Solution Assume I and IC be the set of all insured and uninsured respectively Then, we have P(I) = 1-0.216 = 0.784 Assume H and HC to be the highest and lowest patients respectively Then, we have P(H) = 0.05 P(Patient is High cost and uninsured) = P(HՈIC)=P(H).P(IC) =0.05X0.216=0.108

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