2. An blood clinic tests patients from three different populations with 1/2 of i
ID: 3361255 • Letter: 2
Question
2. An blood clinic tests patients from three different populations with 1/2 of its patients from population X, 1/3 of its patients from population Y and 1/6 of its patients from population Z. The clinic tests hemoglobin levels of patients measured in grams per decimeter. I and only if a patient has a hemoglobin level of at least 16.5 then they are referred for special treatment. In population X the level of hemoglobin present in the blood is normally dis- tribution with a mean of 15 and a variance of 4. In population Y the level of hemoglobin present in the blood is normally dis- tribution with a mean of 16 and a variance of 6.25. In population Z the level of hemoglobin present in the blood is normally distri- bution with a mean of 15.35 and a variance of 5.29 If a patient at the clinic was referred for special treatment find the probability that the patient came from population Y. Give answer to the fourth decimal.Explanation / Answer
probability that patient from hospital X referred =P(X>16.5)=P(Z>(16.5-15)/41/2)=P(Z>0.75)=0.2266
probability that patient from hospital Y referred =P(X>16.5)=P(Z>(16.5-16)/6.251/2)=P(Z>0.2)=0.4207
probability that patient from hospital Z referred =P(X>16.5)=P(Z>(16.5-15.35)/5.291/2)=P(Z>0.5)=0.3085
here probability that patient referred for special treatment =P(from population X and referred+from population Y and referred+from population Z and referred)
=(1/2)*0.2266+(!/3)*(0.4207)+(1/6)*0.3085 =0.3050
hence probability that patient is from hospital Y given referred for special treatment
=P(from population Y and referred)/P(patient referred) =(1/3)*0.4207/0.3050 =0.4599
please revert for any clariification required
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