In this question, you should state the parameters of any distribution that you use. A local fish market is famous for two types of fish it sells. The masses in kg of each type of fish may be assumed to follow independent normal distributions, with means and standard deviations as shown in the following table.

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

Let $G$ and $S$ be the masses of a Garoupa and a Snapper respectively.
$G \sim \mathrm{N}(0.50, 0.025^2)$
$S \sim \mathrm{N}(0.45, 0.020^2)$

Let $W = G_1 + G_2 + S_1 + S_2 + S_3$ be the total mass of 2 Garoupas and 3 Snappers.
$\mathrm{E}(W) = 2\mathrm{E}(G) + 3\mathrm{E}(S)$
$\mathrm{E}(W) = 2(0.50) + 3(0.45) = 2.35 \text{ kg}$

2. Calculate the variance of the total mass.

Since the masses of the randomly chosen fishes are independent,
$\mathrm{Var}(W) = 2\mathrm{Var}(G) + 3\mathrm{Var}(S)$
$\mathrm{Var}(W) = 2(0.025^2) + 3(0.020^2)$
$\mathrm{Var}(W) = 0.00125 + 0.0012 = 0.00245$

3. State the distribution of the total mass and compute the required probability.

As $W$ is a linear combination of independent normal variables, it is also normally distributed.
$W \sim \mathrm{N}(2.35, 0.00245)$

We require $\mathrm{P}(W > 2.25)$. Using a graphing calculator:
$\mathrm{P}(W > 2.25) \approx 0.97832 \approx 0.978 \text{ (to 3 s.f.)}$

1. Determine the normal distribution parameters for the total cost of $k$ Snappers.

Let $S_i$ be the mass of the $i$-th Snapper. We are given $S_i \sim \mathrm{N}(0.45, 0.020^2)$.
The total cost for $k$ Snappers is $C = 18.50(S_1 + S_2 + \dots + S_k)$.
Since the masses are independent,
$\mathrm{E}(C) = 18.50 \times k \times 0.45 = 8.325k$
$\mathrm{Var}(C) = 18.50^2 \times k \times 0.020^2 = 342.25 \times 0.0004k = 0.1369k$
$\therefore C \sim \mathrm{N}(8.325k, 0.1369k)$

2. Set up the probability inequality based on Mrs Tan's budget.

Mrs Tan brings $\char36{}50$ and wants to be at least $95\%$ certain that she has enough money to pay for the purchase. This means we require:
$\mathrm{P}(C \le 50) \ge 0.95$

3. Standardise the normal variable to form an inequality in terms of $k$.

Standardising the distribution of $C$:
$\mathrm{P}\left(Z \le \frac{50 - 8.325k}{\sqrt{0.1369k}}\right) \ge 0.95$
Using the standard normal distribution, $\mathrm{P}(Z \le 1.64485) \approx 0.95$, so we require the upper limit to be at least $1.64485$:
$\frac{50 - 8.325k}{0.37\sqrt{k}} \ge 1.64485$
$50 - 8.325k \ge 0.60859\sqrt{k}$
$8.325k + 0.60859\sqrt{k} - 50 \le 0$

4. Solve the quadratic inequality to find the maximum integer value of $k$.

Let $y = \sqrt{k}$. The inequality is $8.325y^2 + 0.60859y - 50 \le 0$.
Solving the corresponding quadratic equation $8.325y^2 + 0.60859y - 50 = 0$:
$y = \frac{-0.60859 \pm \sqrt{0.60859^2 - 4(8.325)(-50)}}{2(8.325)}$
$y = \frac{-0.60859 \pm \sqrt{0.37038 + 1665}}{16.65}$
$y \approx 2.4144 \quad \text{or} \quad y \approx -2.4875$
Since $y = \sqrt{k} \ge 0$, we have $0 \le \sqrt{k} \le 2.4144$.
$k \le 2.4144^2 \implies k \le 5.829$
Since $k$ must be a whole number, the maximum number of Snappers Mrs Tan could buy is 5.

(Alternatively, checking values using a graphing calculator gives $\mathrm{P}(C \le 50) \approx 1.000$ for $k=5$ and $\approx 0.522$ for $k=6$, confirming max $k=5$.)

1. Find the probability that a single Garoupa weighs more than 0.48 kg.

Let $X$ be the mass of a Garoupa in kg.
$X \sim \mathrm{N}(0.50, 0.025^2)$

$P(X > 0.48) \approx 0.78814$

2. Set up a binomial distribution for the number of Garoupas weighing more than 0.48 kg.

Let $W$ be the number of Garoupas out of 6 that weigh more than 0.48 kg.
Assuming their masses are independent, $W \sim \mathrm{B}(6, 0.78814)$.

3. Calculate the probability that more than 3 Garoupas weigh more than 0.48 kg.

We require $P(W > 3)$.
$P(W > 3) = P(W \ge 4) = 1 - P(W \le 3)$

Using a graphing calculator:
$P(W \ge 4) \approx 0.88579$
$P(W > 3) = 0.886 \text{ (to 3 s.f.)}$

1. Find the probability that a single Garoupa has a mass greater than 0.48 kg.

Let $G$ be the mass of a Garoupa.
$G \sim \mathrm{N}(0.50, 0.025^2)$
$\mathrm{P}(G > 0.48) \approx 0.78814$

2. Deduce the required condition and calculate the final probability for all 6 Garoupas.

The lightest Garoupa has a mass greater than $0.48\text{ kg}$ if and only if all $6$ Garoupas have a mass greater than $0.48\text{ kg}$.
Assuming the masses of the Garoupas are independent,
$\mathrm{P}(\text{all } 6 \text{ Garoupas} > 0.48) = [\mathrm{P}(G > 0.48)]^6$
$\approx (0.78814)^6$
$\approx 0.23964 \approx 0.240$ (to 3 s.f.)