Question

A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is used to measure the glucose concentration in the blood sample five times. The measured glucose concentrations are 109.7, 92.4, 101.2, 104.5, and 104.2 mg/dL. Calculate the absolute error and relative error for each measurement made by the glucose monitor. A. 109.7 mg/dL absolute error = mg/dL relative error = B. 92.4 mg/dL absolute error = mg/dL relative error = C. 101.2 mg/dL absolute error = mg/dL relative error = D. 104.5 mg/dL absolute error = mg/dL relative error = c E. 104.2 mg/dL absolute error = C m mg/dL ydl relative error = relative error = C

Answer 1

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Absolute errorlea) = I measured value - true valuel or la = |xm-x+1 Relative error (sier) = absolute error (ea] measured value (m) - xm = 109.7 mg/dL + = 102.0 mglde a. la = 110907 - 10200 mg/de la = 707 mg/dll → ex = la = 7.9 mg/dL =10.07 Togo 7 malde - 100 xm = 92.4 mg/dl x+= 102° 0 mg/de a la = 192.4 - 102.00 malde = 9.6 mg/dl) >ex = ea/xm = 9.6/204 =10.103.1 6 7 xm= 101.2 mgld. xt = 1020 mg/dl. i la = 1 xm- xz1 = 110102-102.01 molde - 0.8 mg/dll er = Calam ²0 og milde 10.0079. 101.2 mg/de xm = 10405 mglde xt = 102.0 mg lde i la = 1104.5 - 102-01 mg/dl. = (2-5 mg/dl. 10.024 ] @ lalam 11045 @ → la = 1104.2 - 102.01 mg/dL (1-2 mg/dL) → er = lasem - 2020.0215 104.2