e.) 110.6 mg/dL Having a bit confusion with these 5 problems. Could you help me?
ID: 529766 • Letter: E
Question
e.) 110.6 mg/dL
Having a bit confusion with these 5 problems. Could you help me?
The known concentration of glucose in a blood sample is 102.0 mg/dL. During testing of a new at home glucose monitor, the device gives the following determinations of glucose in the blood sample: 92.2, 106.8, 102.9, 102.3, and 110.6 mg/dL. Calculate the absolute error and relative error for each measurement made by the glucose monitor. a) 92.2 mg/dL absolute error = relative error = b) 06.8 mg/dL absolute error = relative error = c) 102.9 mg/dL absolute error = relative error = d) 102.3 mg/dL absolute error = relative error =Explanation / Answer
mean of the samples = sum of samples/5= (92.2+106.8+102.9+102.3+110.6)/5 =102.96
absoulte error (mg/dL)
absolute error= (92.2-102.96)=-10.76 (sample-1), relative error= 100*(10.76/92.2)= 11.67
absoute error= (106.8-102.96)=3.84 ( sample-2), relative error= 100*3.84/106.8 = 3.55
absolute error= 102.9-102.96= -0.06 ( sample-3), relative error= 100*0.06/102.9= 0.058
absoulte error= 102.3-102.96= -0.66 ( sample -4), relative error= 100*0.66/102.3= 0.645
absoulte error= 110.6-102.96= 7.64 ( sample-5), relative errot= 100*7.64/110.6= 6.91
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