Fishing Company A has two types of fishing vessels, the "Standard" vessel and the "Large" vessel. The amount of fish caught by the respective vessels on a typical fishing trip are normally distributed with means and standard deviations as shown in the table. Fishing Company B operates in the same seas as Fishing Company A. However, Company B has the improved versions of the two types of fishing vessels that Company A has: the "Premium-Standard" vessel and the "Premium-Large" vessel. In a fishing trip, the Premium-Standard vessel is capable of catching 3 times as much fish as the Standard vessel and the Premium-Large vessel is capable of catching 2 times as much fish as Large vessel.

1. Define the random variable and set up the probability equation using the given information

Let $X$ be the amount of fish caught by a Standard vessel in kg.
$X \sim \mathrm{N}(300, \sigma^2)$
Given $\mathrm{P}(X > 400) = 0.08$

2. Standardise the variable and express the probability in terms of $Z$

$\mathrm{P}\left(Z > \frac{400 - 300}{\sigma}\right) = 0.08$
$\mathrm{P}\left(Z < \frac{100}{\sigma}\right) = 1 - 0.08 = 0.92$

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

From the inverse normal distribution (or GC):
$\frac{100}{\sigma} \approx 1.40507$
$\sigma = \frac{100}{1.40507} \approx 71.171$
$\sigma \approx 71.2 \text{ (shown)}$

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

Let $T$ be the total catch of 3 Standard vessels and 2 Large vessels. Since the vessels are distinct, $T = S_1 + S_2 + S_3 + L_1 + L_2$.

2. Determine the distribution of $T$ and calculate the probability of meeting the target.

Since $S$ and $L$ are normally distributed, $T$ is also normally distributed.
$T \sim \mathrm{N}(1980, 39408.32)$
We want to find $\mathrm{P}(T \ge 2100)$.
Using a graphing calculator,
$\mathrm{P}(T \ge 2100) \approx 0.27276$
$\mathrm{P}(T \ge 2100) = 0.273 \text{ (to 3 s.f.)}$

1. State the key statistical assumption made when combining the random variables for the different vessels.

The calculation of the variance of the total catch ($S_1 + S_2 + S_3 + L_1 + L_2$) relies on adding the individual variances, which requires the random variables to be independent.

1. Formulate the expectation of the new random variable $T$ for the total catch of the two Premium vessels.

Let $T$ be the total catch of the Premium-Standard and Premium-Large vessels.
Since the vessels catch 3 times and 2 times their standard counterparts respectively, $T = 3S + 2L$.
$\text{E}(T) = 3\text{E}(S) + 2\text{E}(L)$
$\text{E}(T) = 3(300) + 2(540) = 1980$

2. Calculate the variance of $T$.

$\text{Var}(T) = 3^2\text{Var}(S) + 2^2\text{Var}(L)$
Using the more exact value of $\sigma = 71.171$ from part (a):
$\text{Var}(T) = 9(71.171)^2 + 4(110)^2$
$\text{Var}(T) \approx 45588 + 48400 = 93988$
(Note: Using the shown value $\sigma = 71.2$ gives $\text{Var}(T) \approx 94025$, which is also acceptable.)

3. Find the probability that the total catch is at least 2100kg.

$T \sim \text{N}(1980, 93988)$
We want to find $\text{P}(T \ge 2100)$.
Using the normal cumulative distribution function on a graphing calculator:
$\text{P}(T \ge 2100) = 0.34773... \approx 0.348 \text{ (to 3 s.f.)}$

1. State the distribution of the sample mean $\bar{Y}$.

Let $Y$ be the amount of fish caught by a Large vessel on a typical trip.
We are given $Y \sim \mathrm{N}(540, 110^2)$.
Let $\bar{Y}$ be the sample mean catch over $n$ randomly chosen trips.
Since the parent population is normal, the sample mean follows a normal distribution exactly:
$\bar{Y} \sim \mathrm{N}\left(540, \frac{110^2}{n}\right)$

2. Set up the probability inequality and standardise the variable.

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

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

Using the inverse normal distribution, the $z$-value corresponding to a cumulative probability of $0.7$ is approximately $0.52440$.
$\frac{12\sqrt{n}}{110} \ge 0.52440$
$\sqrt{n} \ge \frac{0.52440 \times 110}{12}$
$\sqrt{n} \ge 4.8070$
$n \ge 23.107$
Since $n$ must be an integer (representing the number of fishing trips), the least possible value of $n$ is $24$.

(Alternatively, by using a graphing calculator table to evaluate $\mathrm{P}(\bar{Y} < 552)$ for integer values of $n$: when $n=23$, prob $\approx 0.6995 < 0.7$; when $n=24$, prob $\approx 0.7034 \ge 0.7$. Hence $n=24$.)