Using this slove for stand deviation during mixing

For a perfectly mixed binary system, the concentration of yeast in any sample is ideally uniform and equal to the overall average concentration $C$. Therefore, the standard deviation of concentrations in a perfectly mixed system is zero.

$$\sigma_p = 0$$

The standard deviation during mixing ($\sigma$) represents the actual measured standard deviation of samples taken at a particular point in time during the mixing process. It is related to the standard deviation at the start of mixing ($\sigma_0$) and the mixing index ($M$) by the formula:

$$M = \frac{\sigma_0 - \sigma}{\sigma_0 - \sigma_p}$$

From the previous calculation, $\sigma_0 = 0.2179$. Substituting this value and $\sigma_p = 0$ into the formula:

$$M = \frac{0.2179 - \sigma}{0.2179 - 0}$$ $$M = \frac{0.2179 - \sigma}{0.2179}$$

Rearranging the equation to solve for $\sigma$:

$$0.2179 M = 0.2179 - \sigma$$ $$\sigma = 0.2179 - 0.2179 M$$ $$\sigma = 0.2179 (1 - M)$$

Since no value for the mixing index ($M$) or experimental data from during mixing is provided, a specific numerical value for $\sigma$ cannot be determined from the information given.