91Ó°ÊÓ

• The standard deviation to mean ratio is known as the coefficient of variation (CV).
• The larger the dispersion around the mean, the higher the coefficient of variation. In most cases, it's given as a percentage.

(a)

Solve for the least predicted cofficients of variance, cv percent, in this issue (both parts a and b).

It has provided the following for the first component (a):

concentration of analyte (i) 1 wt\% and (ii) 1 part per trillion

It can easily solve for the CV percent using the analyte concentration. It can enter the value into the equation as follows:

a-i.)

$$\begin{gathered} \mathrm{CV}\%=2^{\left(1-0.5\mathrm{logC}\right)} \\ =2^{\left(1-0.5\log 0.01\right)} \\ =2^2 \\ \mathrm{CV}\%=4\% \end{gathered}$$

a-ii.)

$$\begin{gathered} \mathrm{CV}\%=2^{\left(1-0.5\mathrm{logC}\right)} \\ =2\left[1-0.5\log \left(1×{10}^{-12}\right)\right] \\ =2^2 \\ \mathrm{CV}\%=128\% \end{gathered}$$

(b)

It provided the following for component b:

between-laboratory variance of about 0.5 to 0.7 an unknown containing$10\mathrm{wt}\%{\mathrm{NH}}_3$

When calculating the CV percent, we just enter the specified values into the algorithm. Then there's:

$$\begin{gathered} \mathrm{CV}\%=0.5×2^{\left(1-0.5\mathrm{logC}\right)} \\ =0.{52}^{\left[1-0.5\log 0.5\right]} \\ =0.5×2^{1.5} \\ \mathrm{CV}\%=1.4\% \end{gathered}$$

To figure out what the coefficient of variance is. However, we are requested to find the minimum in this problem, and we are given the values ( 0.5 to 0.7 ). As a result, we consider the stated minimum variation (which is 0.5 ).