在本题中，你应该清晰地写出你使用的所有分布，以及相应参数的值。 当地的一家超市销售两种土豆：褐土豆（Russet potatoes）和荷兰土豆（Holland potatoes）。褐土豆的质量（以克为单位）服从分布 $\mathrm{N}(200, 30^2)$，荷兰土豆的质量（以克为单位）服从分布 $\mathrm{N}(150, 24^2)$。

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

Let $H$ and $R$ be the masses of a randomly chosen Holland and Russet potato respectively.
Let $W = H_1 + H_2 + H_3 - 2R$, where $H_1, H_2, H_3$ are independent observations of $H$.
$\mathrm{E}(W) = 3\mathrm{E}(H) - 2\mathrm{E}(R)$
$= 3(150) - 2(200) = 50$

2. Calculate the variance of the linear combination and state its overall normal distribution as required by the question context.

$\mathrm{Var}(W) = 3\mathrm{Var}(H) + (-2)^2\mathrm{Var}(R)$
$= 3(24^2) + 4(30^2)$
$= 1728 + 3600 = 5328$
Since any linear combination of independent normal variables is also normally distributed, we state the distribution as:
$W \sim \mathrm{N}(50, 5328)$

3. Find the probability that the combination is less than zero.

We require $\mathrm{P}(H_1 + H_2 + H_3 < 2R)$, which is equivalent to $\mathrm{P}(W < 0)$.
Using a graphing calculator:
$\mathrm{P}(W < 0) \approx 0.24667$
$\approx 0.247 \text{ (to 3 s.f.)}$

1. Define the random variable for the variety pack and calculate its expected value and variance.

Let $V$ be the total mass of the variety pack.
$V = R_1 + R_2 + R_3 + R_4 + R_5 + H_1 + H_2 + H_3 + H_4$
$\text{E}(V) = 5(200) + 4(150) = 1600$
$\text{Var}(V) = 5(30^2) + 4(24^2) = 6804$
$V \sim \text{N}(1600, 6804)$

2. Set up the probability equation based on the given information.

$\text{P}(1600 - k < V < 1600 + k) = 0.775$

3. Solve for $k$ using the symmetry of the normal distribution and the inverse normal function.

By the symmetry of the normal distribution:
$\text{P}(V < 1600 + k) = 0.5 + \frac{0.775}{2} = 0.8875$
Using the inverse normal function on a graphing calculator:
$1600 + k \approx 1700.082$
$k \approx 100.082$
$k = 100 \text{ (to 3 s.f.)}$

1. Identify the binomial model and the probability of a pack not being within $k$ grams.

Let $Y$ be the number of variety packs out of $40$ that are not within $k$ grams of $1600$ grams.
From part (ii), the probability that a pack is within $k$ grams is $0.775$.
Thus, the probability that a pack is not within $k$ grams is $1 - 0.775 = 0.225$.
$Y \sim \mathrm{B}(40, 0.225)$

2. Calculate the probability that at most 5 packs are not within $k$ grams.

We require $\mathrm{P}(Y \le 5)$.
Using a graphing calculator to compute the cumulative binomial probability:
$\mathrm{P}(Y \le 5) = 0.086522\dots$
$\mathrm{P}(Y \le 5) \approx 0.0865 \text{ (to 3 s.f.)}$

1. State the necessary assumption for combining random variables and using the binomial distribution

The calculations for the sum and difference of normal variables in parts (i) and (ii), as well as the binomial trials of variety packs in part (iii), all mathematically require the individual random events to be independent.

1. Define the sample mean random variable and state its approximate distribution using the Central Limit Theorem.

Let $\bar{W}$ be the sample mean mass of $100$ New potatoes.
Since the sample size $n = 100$ is large ($n \ge 30$), by the Central Limit Theorem, $\bar{W}$ is approximately normally distributed.
$\bar{W} \sim \mathrm{N}\left(55, \frac{13^2}{100}\right) \text{ approximately}$
$\bar{W} \sim \mathrm{N}(55, 1.69)$

2. Calculate the probability that the sample mean is not more than 58 grams.

We need to find $\mathrm{P}(\bar{W} \le 58)$.
Using a graphing calculator,
$\mathrm{P}(\bar{W} \le 58) \approx 0.989495$
$\mathrm{P}(\bar{W} \le 58) = 0.989 \text{ (to 3 s.f.)}$