捕鱼公司 A 有两种类型的渔船：“标准型”渔船和“大型”渔船。在一次典型的捕鱼航行中，各渔船捕获的鱼量服从正态分布，其均值和标准差如下表所示。

	均值 (kg)	标准差 (kg)
标准型	300	$\sigma$
大型	540	110

1. Set up the probability equation and standardise the variable.

Let $S$ be the mass of fish caught by a Standard vessel in kg.
$S \sim \mathrm{N}(300, \sigma^2)$
Given $\mathrm{P}(S > 400) = 0.08$, we standardise to the standard normal distribution $Z$:
$\mathrm{P}\left(Z > \frac{400 - 300}{\sigma}\right) = 0.08$
$\mathrm{P}\left(Z < \frac{100}{\sigma}\right) = 1 - 0.08 = 0.92$

2. Use the inverse normal function to solve for the standard deviation.

Using the graphing calculator (inverse normal):
$\frac{100}{\sigma} \approx 1.40507$
$\sigma = \frac{100}{1.40507}$
$\sigma \approx 71.171$
$\sigma \approx 71.2 \text{ (shown)}$

1. Define the total catch random variable and calculate its expected value.

Let the total catch be $T = S_1 + S_2 + S_3 + L_1 + L_2$.

$\text{E}(T) = 3\text{E}(S) + 2\text{E}(L)$
$\text{E}(T) = 3(300) + 2(540) = 1980$

2. Calculate the variance of the total catch.

$\text{Var}(T) = 3\text{Var}(S) + 2\text{Var}(L)$

Using the unrounded standard deviation $\sigma \approx 71.171$ (derived from part a):
$\text{Var}(T) = 3(71.171)^2 + 2(110)^2$
$\text{Var}(T) \approx 15195.9 + 24200 = 39395.9$

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

Since $T$ is a linear combination of independent normal variables, $T \sim \text{N}(1980, 39395.9)$.

We require $\text{P}(T \ge 2100)$:
$\text{P}(T \ge 2100) \approx 0.27273$
$\text{P}(T \ge 2100) \approx 0.273 \text{ (to 3 s.f.)}$

1. Identify the statistical assumption necessary for the linear combination of random variables

In part (b), we calculate the variance of the total catch by summing the variances of the individual catches: $\mathrm{Var}(S_1 + S_2 + S_3 + L_1 + L_2) = 3\mathrm{Var}(S) + 2\mathrm{Var}(L)$. This requires the assumption that the random variables are independent.

1. Define the distribution for the total catch by Company B's vessels, noting that their catches are multiples of the standard and large vessels.

Let $W$ be the total catch by $1$ Premium-Standard and $1$ Premium-Large vessel.
Since the Premium-Standard catches $3$ times as much as a Standard vessel, its catch is $3S$. The Premium-Large catches $2$ times as much as a Large vessel, so its catch is $2L$.
$W = 3S + 2L$

$\text{E}(W) = 3\text{E}(S) + 2\text{E}(L)$
$\text{E}(W) = 3(300) + 2(540) = 1980 \text{ kg}$

$\text{Var}(W) = 3^2\text{Var}(S) + 2^2\text{Var}(L)$
Using $\sigma \approx 71.171$ from part (a):
$\text{Var}(W) = 9(71.171)^2 + 4(110)^2$
$\text{Var}(W) \approx 45587.8 + 48400 = 93987.8$

Thus, $W \sim \text{N}(1980, 93987.8)$.

2. Calculate the probability that the total catch is at least 2100 kg.

We require $\text{P}(W \ge 2100)$.
Using the normal cumulative distribution function on a graphing calculator:
$\text{P}(W \ge 2100) \approx 0.34774$
$\text{P}(W \ge 2100) = 0.348 \text{ (to 3 s.f.)}$

1. State the distribution of the sample mean.

Let $\bar{L}$ be the sample mean catch of a Large vessel for $n$ trips.
Since the catch of a Large vessel is normally distributed with mean $540$ kg and standard deviation $110$ kg, the sample mean follows:
$\bar{L} \sim \mathrm{N}\left(540, \frac{110^2}{n}\right)$

2. Set up the probability inequality and standardise.

We are given that $\mathrm{P}(\bar{L} < 552) \ge 0.7$.
Standardising the variable to the standard normal distribution $Z \sim \mathrm{N}(0, 1)$:
$\mathrm{P}\left(Z < \frac{552 - 540}{110/\sqrt{n}}\right) \ge 0.7$
$\mathrm{P}\left(Z < \frac{12\sqrt{n}}{110}\right) \ge 0.7$

3. Solve the inequality for $n$ using the inverse normal function.

Using the inverse normal function for an area of $0.7$, we find the critical $z$-value is approximately $0.52440$.
$\frac{12\sqrt{n}}{110} \ge 0.52440$
$\sqrt{n} \ge \frac{110 \times 0.52440}{12}$
$\sqrt{n} \ge 4.8070$
$n \ge 23.107$
Since $n$ must be an integer (representing the number of trips), the least possible value of $n$ is $24$.