在这道题中，你应该说明你所使用的任何分布的参数。 当地的一个鱼市因其销售的两种鱼而闻名。可以假设每种鱼的质量（单位：kg）服从独立的正态分布，其均值和标准差如下表所示。

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.)