A statistical risk for infectious disease. Please show work and explain
1. The rate of diarrhoea who ate pizza and ice cream is
39/52
The rate of diarrhoea who ate pizza only is 14/40
The rate of diarrhoea who ate ice cream only is 11/15
The rate who didn’t eat neither pizza nor ice cream is 9/30
The question is the attack rate for person who ate pizza is the
addition of both the attack rate who ate pizza with ice cream and
pizza only.
The attack rate is= 39/52 +14/40 = 53/92
The answer is option C.
2. The percentage of rate of diarrhoea who ate pizza and ice cream
is 39/52 = 75%
The percentage of rate of diarrhoea who ate pizza only is 14/40 =
35%
The percentage of rate of diarrhoea who ate ice cream only is 11/15
= 73.3%
The percentage of rate of diarrhoea who ate neither pizza nor ice
cream is 9/30 = 30%
The answer is option D. Ice cream only because the rate percentage
is highest.
A statistical risk for infectious disease. Please show work and explain Calculate the attack rate for...
1. Using the table from the last question, calculate the overall attack rate in persons who did not eat ice cream. Ate ice cream Diarrhea 39 11 Total 52 15 Did not eat ice cream Diarrhea 14 Total 40 30 Ate pizza Did not eat pizza
I could use some help with the five following questions regarding epidemiology. Moving to another question will save this response uestion1 Iculate the attack rate for person who did eat pizza Ate Ice Cream Did Not Eat Ice Cream DIARRHEATOTAL DIARRHEA 39 TOTA ATE PIZZA 52 15 14 DID NOT EAT PIZZA 40 30 50187 92 45 Moving othe estion will save this response
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