由机器 $A$ 装入罐中的汤的质量（$X$ 克）服从均值为 $\mu$ 克、标准差为 $\sigma$ 克的正态分布。

1. Deduce the mean $\mu$ using the symmetry of the normal distribution.

$P(X < 112) = P(X > 128)$
By the symmetry of the normal curve, the mean lies exactly halfway between $112$ and $128$:
$\mu = \frac{112 + 128}{2} = 120$

2. Standardise the value $X = 112$ and set up an equation using the inverse normal function.

$P(X < 112) = 0.09121$
Standardising the variable:
$P\left(Z < \frac{112 - 120}{\sigma}\right) = 0.09121$
Using the inverse normal function on a graphing calculator, the $z$-score corresponding to a cumulative area of $0.09121$ is approximately $-1.3333$:
$\frac{-8}{\sigma} = -1.3333$

3. Solve for the standard deviation $\sigma$.

$\sigma = \frac{-8}{-1.3333} = 6.00 \text{ (to 3 s.f.)}$

1. Find the probabilities of a single can satisfying each condition

Let $X$ be the mass of soup in a can. $X \sim \mathrm{N}(125, 8^2)$.
Using a graphing calculator:
$P(X > 124) \approx 0.54974$
$P(120 < X < 123) \approx 0.13531$

2. Calculate the combined probability for 3 randomly chosen cans

The probability that exactly 2 cans are $> 124\text{ g}$ and 1 can is between $120\text{ g}$ and $123\text{ g}$ is given by:
$\text{Required probability} = \binom{3}{2} \times [P(X > 124)]^2 \times P(120 < X < 123)$
$= 3 \times (0.54974)^2 \times (0.13531)$
$= 0.12268 \dots$
$\approx 0.123 \text{ (to 3 s.f.)}$

1. State the distribution of the sample mean.

Let $\bar{X}$ be the sample mean mass of soup in $n$ cans.
Since the mass of a single can is normally distributed as $\mathrm{N}(125, 8^2)$ based on the adjusted settings from part (ii),
$\bar{X} \sim \mathrm{N}\left(125, \frac{8^2}{n}\right)$

2. Set up the probability inequality and standardise.

We are given that $\mathrm{P}(\bar{X} > 127) \le 0.1$.
Standardising the variable:
$\mathrm{P}\left(Z > \frac{127 - 125}{8/\sqrt{n}}\right) \le 0.1$
$\mathrm{P}\left(Z > \frac{\sqrt{n}}{4}\right) \le 0.1$

3. Use the inverse standard normal distribution to find a relationship for $n$.

From the standard normal distribution, the critical $z$-value for the upper $10\%$ tail (i.e. $\mathrm{P}(Z > z) = 0.1$) is approximately $1.28155$.
Thus, we require:
$\frac{\sqrt{n}}{4} \ge 1.28155$
$\sqrt{n} \ge 5.1262$
$n \ge 26.278$

4. State the least integer value for $n$.

Since the sample size $n$ must be a whole number, the least value of $n$ is $27$.

1. Define the linear combination for the difference and calculate its expected value.

Let $T_Y = Y_1 + Y_2 + Y_3 + Y_4$ be the total mass of 4 randomly chosen packets and $T_X = X_1 + X_2 + X_3$ be the total mass of 3 randomly chosen cans.
Let $W = T_Y - 1.5 T_X$ be the required difference.
Using the adjusted mean of $X$ from part (ii), $\text{E}(X) = 125$ and $\text{E}(Y) = 150$.
$\text{E}(W) = 4\text{E}(Y) - 1.5(3\text{E}(X))$
$\text{E}(W) = 4(150) - 4.5(125) = 600 - 562.5 = 37.5$

2. Calculate the variance of the linear combination.

Using the adjusted standard deviation of $X$ from part (ii), $\text{Var}(X) = 8^2 = 64$ and $\text{Var}(Y) = 7.5^2 = 56.25$.
Because the samples are independent:
$\text{Var}(W) = 4\text{Var}(Y) + 1.5^2(3\text{Var}(X))$
$\text{Var}(W) = 4(56.25) + 2.25(3 \times 64)$
$\text{Var}(W) = 225 + 2.25(192) = 225 + 432 = 657$

3. State the distribution of the difference and evaluate the probability.

Since linear combinations of normal variables are also normally distributed:
$W \sim \text{N}(37.5, 657)$
We need the probability that this difference is at most 50 grams, which means $-50 \le W \le 50$.
$\text{P}(|W| \le 50) = \text{P}(-50 \le W \le 50)$
Using a graphing calculator:
$\text{P}(-50 \le W \le 50) \approx 0.68679$
$\text{P}(-50 \le W \le 50) = 0.687 \text{ (to 3 s.f.)}$