渔业公司 A 拥有两种类型的渔船：“标准”型渔船和“大型”型渔船。在一次常规的捕鱼行程中，各自定义渔船所捕获的鱼量均服从正态分布，其均值和标准差如表所示。

1. Formulate the probability equation for the Standard vessel's catch.

Let $S$ be the mass of fish caught by a Standard vessel.
$S \sim \mathrm{N}(300, \sigma^2)$
It is given that $\mathrm{P}(S > 400) = 0.08$.

2. Standardise the variable and use the inverse normal function.

$\mathrm{P}\left(Z > \frac{400 - 300}{\sigma}\right) = 0.08$
$\mathrm{P}\left(Z < \frac{100}{\sigma}\right) = 1 - 0.08 = 0.92$
Using the graphing calculator to find the inverse normal of $0.92$:
$\frac{100}{\sigma} \approx 1.40507$

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

$\sigma \approx \frac{100}{1.40507}$
$\sigma \approx 71.171$
$\sigma \approx 71.2 \text{ (shown)}$

1. Define the random variable for the total catch and calculate its mean and variance.

Let $T = S_1 + S_2 + S_3 + L_1 + L_2$ be the total amount of fish caught by the 5 vessels.
Using $\sigma = 71.171$ from part (a):
$\text{E}(T) = 3\text{E}(S) + 2\text{E}(L)$
$\text{E}(T) = 3(300) + 2(540) = 1980$

$\text{Var}(T) = 3\text{Var}(S) + 2\text{Var}(L)$
$\text{Var}(T) = 3(71.171)^2 + 2(110)^2$
$\text{Var}(T) = 15195.9 + 24200 = 39395.9$

Thus, $T \sim \text{N}(1980, 39395.9)$.

2. Calculate the probability that the total catch meets the target.

We need to find the probability that the total catch is at least 2100 kg, which is $\text{P}(T \ge 2100)$.

Using a graphing calculator:
$\text{P}(T \ge 2100) = 0.27272\dots$
$\text{P}(T \ge 2100) \approx 0.273 \text{ (to 3 s.f.)}$

1. State the statistical assumption required to sum the variances of the individual random variables.

In part (b), the variance of the total catch is calculated by summing the variances of the individual catches. This requires the assumption that the amount of fish caught by each vessel is independent of the amount caught by the other vessels.

1. Define the random variable for the total catch of Company B and calculate its expectation.

Let $T_B$ be the total catch for Company B.
Since the vessels catch scalar multiples of the original amounts, $T_B = 3S + 2L$.
$\text{E}(T_B) = 3\text{E}(S) + 2\text{E}(L)$
$\text{E}(T_B) = 3(300) + 2(540) = 1980$

2. Calculate the variance of the total catch, taking care to square the scaling constants.

Using the more precise value $\sigma \approx 71.171$ from part (a):
$\text{Var}(T_B) = 3^2\text{Var}(S) + 2^2\text{Var}(L)$
$\text{Var}(T_B) = 9(71.171)^2 + 4(110^2)$
$\text{Var}(T_B) \approx 45587.5 + 48400 = 93987.5$

3. Find the probability that the target of at least 2100kg is met.

$T_B \sim \text{N}(1980, 93987.5)$
$\text{P}(T_B \ge 2100) \approx 0.34773 \approx 0.348 \text{ (to 3 s.f.)}$

1. State the distribution of the sample mean.

Let $L$ be the amount of fish caught by a Large vessel. From the given information, $L \sim \text{N}(540, 110^2)$.
For a sample of $n$ trips, the sample mean $\bar{L}$ follows a normal distribution:
$\bar{L} \sim \text{N}\left(540, \frac{110^2}{n}\right)$

2. Set up an inequality by standardising the normal variable.

We are given that $\text{P}(\bar{L} < 552) \ge 0.7$.
Standardising the variable gives:
$\text{P}\left(Z < \frac{552 - 540}{110/\sqrt{n}}\right) \ge 0.7$
$\text{P}\left(Z < \frac{12\sqrt{n}}{110}\right) \ge 0.7$

3. Solve the inequality to find the least possible integer value for $n$.

Using the inverse normal distribution, for $\text{P}(Z < z) = 0.7$, the critical $z$-value is $z \approx 0.52440$.
$\frac{12\sqrt{n}}{110} \ge 0.52440$
$\sqrt{n} \ge \frac{110 \times 0.52440}{12}$
$\sqrt{n} \ge 4.8070$
$n \ge 23.107$
Since the sample size $n$ must be an integer, the least possible value of $n$ is $24$.