91Ó°ÊÓ

The following is the solution to this problem:

4 amounts of bromate $$\begin{gathered} (\mathrm{in}\mathrm{μ²\mu }/\mathrm{L}):0.2,0.5,1.0,20 \\ \mathrm{Relative}\mathrm{standard}\mathrm{deviations}(\%):14.4,0.8,3.2,1.9 \\ {\mathrm{y}}_{\mathrm{blank}}=0 \end{gathered}$$

The concentration detection limits for four different concentrations, as well as the mean concentration detection limit, must be determined.

First solve the standard deviations for each concentration since we are given relative standard deviations, which are depending on the amount of concentration.

For the preliminary concentration,$(0.2\mathrm{μ²\mu }/\mathrm{L}):$

$$\begin{gathered} s-\mathrm{relative}\mathrm{standard}\mathrm{deviation}×\mathrm{concentration} \\ \mathrm{d}=0.144×0.2\mathrm{μ²\mu }/\mathrm{L} \\ \mathrm{s}=0.028\mathrm{μ²\mu }/\mathrm{L} \end{gathered}$$

The second concentration will be $(0.5\mathrm{μ²\mu }/\mathrm{L}):$

$$\begin{gathered} \mathrm{s}=\mathrm{relative}\mathrm{standard}\mathrm{deviation}×\mathrm{concentration} \\ =0.068×0.5\mathrm{μ²\mu }/\mathrm{L} \\ \mathrm{s}=0.034\mathrm{μ²\mu }/\mathrm{L} \\ \mathrm{The}\mathrm{third}\mathrm{concentration}\mathrm{consists}\mathrm{of}(1.0\mathrm{μ²\mu }/\mathrm{L}): \\ \mathrm{s}=\mathrm{relative}\mathrm{standard}\mathrm{deviation}×\mathrm{concentration} \\ =0.0.32×1.0\mathrm{μ²\mu }/\mathrm{L} \\ \mathrm{s}=0.032\mathrm{μ²\mu }/\mathrm{L} \\ \mathrm{The}\mathrm{fourth}\mathrm{concentration}\mathrm{consists}\mathrm{of}(2.0\mathrm{μ²\mu }/\mathrm{L}): \\ \mathrm{s}=\mathrm{relative}\mathrm{standard}\mathrm{deviation}×\mathrm{concentration} \\ =0.0.19×2.0\mathrm{μ²\mu }/\mathrm{L} \\ \mathrm{s}=0.0038\mathrm{μ²\mu }/\mathrm{L} \end{gathered}$$

Find the concentration detection limits for each concentration now that we have the s for each concentration:

For the preliminary concentration,$0.2\mathrm{μ²\mu }/\mathrm{L}$

role="math" localid="1655018837859" $$\begin{gathered} \mathrm{concentration}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s} \\ =-0.+\left(3\right)(0.0288\mathrm{μ²\mu }/\mathrm{L}) \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.086\mathrm{μ²\mu }/\mathrm{L} \\ \mathrm{The}\mathrm{second}\mathrm{concentration}\mathrm{will}\mathrm{be}(0.5\mathrm{μ²\mu }/\mathrm{L}): \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s} \\ =0+\left(3\right)(0.032\mathrm{μ²\mu }/\mathrm{L})\mathrm{Step}10 \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.142\mathrm{μ²\mu }/\mathrm{L} \\ \mathrm{The}\mathrm{third}\mathrm{concentration}\mathrm{consists}\mathrm{of}(1.0\mathrm{μ²\mu }/\mathrm{L}): \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s} \\ =0+\left(3\right)(0.032\mathrm{μ²\mu }/\mathrm{L}) \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.096\mathrm{μ²\mu }/\mathrm{L} \\ \mathrm{The}\mathrm{fourth}\mathrm{concentration}\mathrm{consists}\mathrm{of}(2.0\mathrm{μ²\mu }/\mathrm{L}): \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s} \\ =0+\left(3\right)(0.038\mathrm{μ²\mu }/\mathrm{L}) \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.114\mathrm{μ²\mu }/\mathrm{L} \end{gathered}$$

Find the mean of the four values of detection limits that we previously solved to get the mean concentration detection limit:

$$\begin{gathered} \mathrm{mean}\mathrm{concentration}\mathrm{detection}\mathrm{limit}=\frac{\mathrm{i}}{\mathrm{n}} \\ =\frac{\left(0.086+0.102+0.096+0.114\right)\mathrm{μ²\mu }/\mathrm{L}}{4} \\ =\frac{0.2954\mathrm{μ²\mu }/\mathrm{L}}{4} \\ =0.07385\mathrm{μ²\mu }/\mathrm{L} \\ \mathrm{mean}\mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.01\mathrm{μ²\mu }/\mathrm{L} \end{gathered}$$