在此题中，你应该写明你所使用的任何分布的参数。 一家零食公司生产两种薯片：Rays 和 Luffles。一包普通装 Rays 薯片的质量（单位：克）服从分布 $\text{N}(100, \sigma^2)$，而一包普通装 Luffles 薯片的质量（单位：克）服从分布 $\text{N}(120, 16)$。所有普通装薯片的质量相互独立。

1. Set up the probability inequality and standardise to $Z$

Let $R$ be the mass of a packet of Rays, $R \sim \text{N}(100, \sigma^2)$.
Given the condition, we have:
$\text{P}(R \le 100 - \sigma^2) < 0.2$
Standardising the variable to the standard normal $Z$:
$\text{P}\left(Z \le \frac{(100 - \sigma^2) - 100}{\sigma}\right) < 0.2$
$\text{P}(Z \le -\sigma) < 0.2$

2. Solve for $\sigma$ using the inverse normal function

From the standard normal distribution, the $z$-value corresponding to a cumulative probability of $0.2$ is found using a graphing calculator:
$-\sigma < \Phi^{-1}(0.2)$
$-\sigma < -0.84162$
Multiplying by $-1$ and flipping the inequality sign:
$\sigma > 0.84162$
$\sigma > 0.842 \text{ (to 3 s.f.)}$

1. Define the random variable for the total mass and compute its expected value.

Let $T = R_1 + R_2 + L_1 + L_2 + L_3$ be the total mass of the 5 packets.
$\text{E}(T) = 2\text{E}(R) + 3\text{E}(L)$
$\text{E}(T) = 2(100) + 3(120) = 560$

2. Compute the variance of the total mass, using the fact that the packets are independent.

Since the masses of all regular packets are independent,
$\text{Var}(T) = 2\text{Var}(R) + 3\text{Var}(L)$
$\text{Var}(T) = 2(3^2) + 3(16)$
$\text{Var}(T) = 18 + 48 = 66$

3. State the distribution of the total mass, convert the units to match, and calculate the required probability.

Therefore, the distribution of $T$ is $T \sim \text{N}(560, 66)$.
Convert $0.55\text{ kg} = 550\text{ g}$.
$\text{P}(T > 550) \approx 0.89082 \approx 0.891$ (to 3 s.f.)

1. Determine the expected value and variance of the total mass of the Mega Jumbo Pack.

Let $M$ be the mass of a Mega Jumbo Pack.
$M = R_1 + R_2 + \dots + R_{24-n} + L_1 + L_2 + \dots + L_n$
$\text{E}(M) = (24 - n)\text{E}(R) + n\text{E}(L)$
$= (24 - n)(100) + n(120) = 2400 + 20n$
$\text{Var}(M) = (24 - n)\text{Var}(R) + n\text{Var}(L)$
$= (24 - n)(9) + n(16) = 216 + 7n$

2. Find the expected value and variance for 20 times the mass of a regular packet of Luffles.

Let $L_0$ be the mass of a randomly chosen regular packet of Luffles.
$\text{E}(20L_0) = 20(120) = 2400$
$\text{Var}(20L_0) = 20^2(16) = 6400$

3. Formulate the normal distribution for the difference in mass.

Let $Y = M - 20L_0$.
$\text{E}(Y) = 2400 + 20n - 2400 = 20n$
$\text{Var}(Y) = 216 + 7n + 6400 = 6616 + 7n$
Hence, $Y \sim \text{N}(20n, 6616 + 7n)$.

4. Set up the probability inequality and solve for the minimum integer value of $n$.

We are given that $\text{P}(Y > 500) \ge 0.1$.
Standardising, we have $\text{P}\left(Z > \frac{500 - 20n}{\sqrt{6616 + 7n}}\right) \ge 0.1$.
Using a graphing calculator to find $\text{P}(Y > 500)$ for integer values of $n$:
When $n = 19$, $\text{P}(Y > 500) \approx 0.0720 < 0.1$
When $n = 20$, $\text{P}(Y > 500) \approx 0.1119 \ge 0.1$

Therefore, the minimum value of $n$ is 20.