Average = (1.10+ 1.90+ 1.20+ 1.30+ 1.60+ 1.70+ 1.70)/7
=1.50 abs units
Let x1= 1.10
Standard deviation, S1= |x1-1.50|= 0.40 abs units
S2 = |1.90- 1.50|= 0.40abs units
S3= |1.20-1.50| = 0.30abs units
S4 = |1.30-1.50|= 0.20abs units
S5= |1.60-1.50|= 0.10abs units
S6=S7= |1.70-1.50|= 0.20abs units
A student measured the concentration of a metal in soil samples with FAAS, Reagent blanks give si...
A student measured the concentration of a metal in soil samples with FAAS. Reagent blanks give signals of 1.90, 0.60, 1.60, 0.50, 1.70, 0.90, and 0.60 abs units. The slope of the calibration curve is m = 0.325 abs units/ppm. Calculate the average of blank signals with appropriate significant figures. Calculate the standard deviation of blank signals with appropriate significant figures. Calculate the minimum detectable concentration (detection limit in ppm) with appropriate significant figures.