The masses, in grams, of the packets of semolina flour follow the distribution $N(225,\;25^2)$ and the masses, in grams, of the packets of millet flour follow the distribution $N(\mu,\;\sigma^2)$.

1. Define the random variable and formulate the required probability, ensuring units are consistent.

Let $S$ be the mass, in grams, of a packet of semolina flour.
$S \sim \text{N}(225, 25^2)$

Convert kilograms to grams:
$0.85 \text{ kg} = 850 \text{ g}$
$1.05 \text{ kg} = 1050 \text{ g}$

We require the probability $P(850 < 4S < 1050)$.

2. Simplify the inequality and evaluate the probability using a graphing calculator.

$P(850 < 4S < 1050) = P\left(\frac{850}{4} < S < \frac{1050}{4}\right)$
$= P(212.5 < S < 262.5)$

Using the graphing calculator:
$P(212.5 < S < 262.5) \approx 0.62465$
$\approx 0.625 \text{ (to 3 s.f.)}$

1. Express the expected value of $M$ in terms of $\mu$.

Let $S$ and $Y$ represent the mass of a packet of semolina flour and millet flour respectively. The mean mass of the 5 packets is:
$M = \frac{S_1 + S_2 + S_3 + Y_1 + Y_2}{5}$
$\mathrm{E}(M) = \frac{3\mathrm{E}(S) + 2\mathrm{E}(Y)}{5} = \frac{3(225) + 2\mu}{5} = \frac{675 + 2\mu}{5}$

2. Use the symmetry of the normal distribution to deduce the mean of $M$ and show $\mu = 150$.

Since $\mathrm{P}(M < 125) = \mathrm{P}(M > 265) = 0.02$, the distribution of $M$ is symmetric about its mean. Therefore, the mean of $M$ lies exactly halfway between 125 and 265:
$\mathrm{E}(M) = \frac{125 + 265}{2} = 195$
Equating the expressions for the mean:
$\frac{675 + 2\mu}{5} = 195$
$675 + 2\mu = 975$
$2\mu = 300 \implies \mu = 150 \text{ (shown)}$

3. Express the variance of $M$ in terms of $\sigma^2$.

Since the masses of individual packets are independent:
$\mathrm{Var}(M) = \frac{1}{5^2} [3\mathrm{Var}(S) + 2\mathrm{Var}(Y)]$
$\mathrm{Var}(M) = \frac{1}{25} [3(25^2) + 2\sigma^2] = \frac{1875 + 2\sigma^2}{25} = 75 + 0.08\sigma^2$

4. Formulate an equation for $\sigma$ using the standard normal distribution.

Given $\mathrm{P}(M < 125) = 0.02$, we standardise the variable $M \sim N(195, 75 + 0.08\sigma^2)$:
$\mathrm{P}\left(Z < \frac{125 - 195}{\sqrt{75 + 0.08\sigma^2}}\right) = 0.02$
Using the inverse normal function, the z-value corresponding to an area of 0.02 is approximately $-2.0537$.
$\frac{-70}{\sqrt{75 + 0.08\sigma^2}} = -2.0537$

5. Solve the equation to find the value of $\sigma$.

$\sqrt{75 + 0.08\sigma^2} = \frac{-70}{-2.0537}$
$\sqrt{75 + 0.08\sigma^2} \approx 34.084$
Squaring both sides:
$75 + 0.08\sigma^2 \approx 1161.72$
$0.08\sigma^2 \approx 1086.72$
$\sigma^2 \approx 13584$
$\sigma \approx 116.55 \approx 117 \text{ (to 3 s.f.)}$