An blood clinic tests patients from three different populations with 1/2 of its

Question

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. If 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 referred for special treatment =P(special treatment)

=P(patient from population X and referred for special treatment +patient from population Y and referred for special treatment +patient from population Z and referred for special treatment )

=(1/2)*P(Z>(16.5-15)/41/2)+(1/3)*(P(Z>(16.5-16)/6.251/2)+(1/6)*P(Z>(16.5-15.35)/5.291/2)

=(1/2)*P(Z>0.75)+(1/3)*P(Z>0.2)+(1/6)*P(Z>0.5)

=(1/2)*P(Z>0.75)+(1/3)*P(Z>0.2)+(1/6)*P(Z>0.5) =0.5*0.2266+(1/3)*0.4207+(1/6)*0.1667=0.3050

therefore probability that patient came from population Y given  referred for special treatment

=P(patient from population Y and referred for special treatment |special treatment)

=(1/3)*0.4207/0.3050 =0.4599

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