Solve the first part of the question only Solve for the stand deviation at start mixing Solve for standard deviation during mixing a d mixing index

The total mass of the batch is the sum of the dough and yeast masses.
Total mass $= 95 \text{ kg (dough)} + 5 \text{ kg (yeast)} = 100 \text{ kg}$.
The concentration of yeast ($C$) is the mass of yeast divided by the total mass.

$$C = \frac{\text{Mass of yeast}}{\text{Total mass}} = \frac{5 \text{ kg}}{100 \text{ kg}} = 0.05$$

For a binary mixture, the standard deviation at the start of mixing, representing the unmixed state, is given by the formula:

$$\sigma_0 = \sqrt{C(1-C)}$$

Substitute the value of $C$:

$$\sigma_0 = \sqrt{0.05(1-0.05)}$$ $$\sigma_0 = \sqrt{0.05 \times 0.95}$$ $$\sigma_0 = \sqrt{0.0475}$$ $$\sigma_0 \approx 0.2179$$

The standard deviation at the start of mixing is 0.2179.