在本题中，你应该清晰地写出你所定义的任何随机变量的分布。 随机抽取的小瓶香水容量（以 ml 为单位）$S$ 均值为 20，方差为 $\sigma^2$。

1. Calculate the probability of a negative volume if $S$ follows $\text{N}(20, 15^2)$.

If $S \sim \text{N}(20, 15^2)$, the probability of a negative volume is:
$\text{P}(S < 0) = \text{P}\left(Z < \frac{0 - 20}{15}\right) \approx 0.0912$

2. Explain why this probability implies the normal model is inappropriate.

Since the volume of perfume must be strictly non-negative ($S \ge 0$), a normal distribution that predicts a significant probability (about $9.12\%$) of a negative volume is not an appropriate model.

1. Set up the probability equation in terms of the standard normal variable $Z$.

Given $S \sim \text{N}(20, \sigma^2)$
$P(S > 23) = 0.0668$
$P\left(Z > \frac{23 - 20}{\sigma}\right) = 0.0668$

2. Use the inverse normal distribution to find the $z$-value and solve for $\sigma$.

$P\left(Z < \frac{3}{\sigma}\right) = 1 - 0.0668 = 0.9332$
Using a graphing calculator (Inverse Normal),
$\frac{3}{\sigma} \approx 1.500$
$\sigma = 2$

1. Define a random variable for the total volume and calculate its expected value.

Let $T = S_1 + S_2 + \dots + S_6 + L_1 + L_2 + \dots + L_9$.
$E(T) = 6E(S) + 9E(L)$
$E(T) = 6(20) + 9(100) = 120 + 900 = 1020$

2. Calculate the variance of the total volume, using the property of independence for randomly chosen bottles.

$Var(T) = 6Var(S) + 9Var(L)$
$Var(T) = 6(4) + 9(25) = 24 + 225 = 249$

3. State the combined normal distribution, convert the target volume to matching units, and calculate the probability.

$T \sim \text{N}(1020, 249)$
Since $1 \text{ litre} = 1000 \text{ ml}$, we want to find $P(T \ge 1000)$.
Using a graphing calculator:
$P(T \ge 1000) \approx 0.89749$
$\approx 0.897 \text{ (to 3 s.f.)}$

1. Define the linear combination for the difference and calculate its expectation and variance.

Let $W = L - 6S$ be the difference between the volume of a large bottle and 6 times the volume of a small bottle.
$\text{E}(W) = \text{E}(L) - 6\text{E}(S) = 100 - 6(20) = -20$
$\text{Var}(W) = \text{Var}(L) + 6^2\text{Var}(S) = 25 + 36(4) = 169$

2. State the distribution of the linear combination.

Since $L$ and $S$ are independent normally distributed random variables, any linear combination of them is also normally distributed.
$W \sim \text{N}(-20, 169)$

3. Formulate and calculate the required probability for the absolute difference.

We require the probability that the magnitude of the difference is more than 25 ml, which is $\text{P}(|W| > 25)$.
$\text{P}(|W| > 25) = \text{P}(W < -25) + \text{P}(W > 25)$
Using a graphing calculator:
$\text{P}(|W| > 25) \approx 0.350265 + 0.000269 = 0.35053$
$\approx 0.351 \text{ (to 3 s.f.)}$

1. State the necessary assumption in the context of the problem.

To use the property $\operatorname{Var}(aX + bY) = a^2\operatorname{Var}(X) + b^2\operatorname{Var}(Y)$ or to add variances of multiple random variables, the random variables must be independent. In this context, this means the volume of perfume in any chosen bottle does not affect the volume of perfume in any other bottle.