在本题中，你应该清晰地写出你所使用的任何正态分布的参数值。 一个水果商贩销售两种水果，即榴莲和红毛榴莲。以 kg 为单位，单个榴莲和单个红毛榴莲的质量被建模为服从正态分布，其均值和标准差如下表所示。

	平均质量 (kg)	标准差 (kg)
榴莲	2.2	0.4
红毛榴莲	1.3	0.2

1. Define the normal distribution for the mass of a soursop and set up the probability equation.

Let $S$ be the mass of a soursop in kg.
We are given $S \sim \text{N}(1.3, 0.2^2)$.
We need to find the mass $s$ such that $\text{P}(S > s) = 0.95$.
This is equivalent to $\text{P}(S \le s) = 1 - 0.95 = 0.05$.

2. Use the inverse normal function on a graphing calculator to find the value of $s$.

Using the inverse normal function on the graphing calculator with area $= 0.05$, $\mu = 1.3$, and $\sigma = 0.2$:
$s \approx 0.97103$
$s = 0.971\text{ kg}$ (to 3 s.f.)

1. Define the distribution for the mass of a durian and find the probability of mass being less than 2.3 kg.

Let $X$ be the mass of a randomly chosen durian in kg.
$X \sim \mathrm{N}(2.2, 0.4^2)$
$P(X < 2.3) = P\left(Z < \frac{2.3 - 2.2}{0.4}\right)$
$= P(Z < 0.25) \approx 0.59871$

2. Find the probability of a durian having a mass more than 2.5 kg.

$P(X > 2.5) = P\left(Z > \frac{2.5 - 2.2}{0.4}\right)$
$= P(Z > 0.75) \approx 0.22663$

3. Calculate the probability of the combined event for 3 durians, accounting for all possible arrangements.

There are $\frac{3!}{2!1!} = 3$ ways to choose which 2 durians out of the 3 have a mass less than 2.3 kg.

$\text{Required probability} = 3 \times [P(X < 2.3)]^2 \times P(X > 2.5)$
$= 3 \times (0.59871)^2 \times 0.22663$
$\approx 0.24370$
$\approx 0.244 \text{ (to 3 s.f.)}$

1. Determine the expectation and variance of the linear combination $D - 2S$ and state its distribution.

Let $D$ and $S$ be the masses (in kg) of a randomly chosen durian and soursop respectively. Let $W = D - 2S$.
$\text{E}(W) = \text{E}(D) - 2\text{E}(S) = 2.2 - 2(1.3) = -0.4$
$\text{Var}(W) = \text{Var}(D) + 2^2\text{Var}(S) = 0.4^2 + 4(0.2^2) = 0.16 + 0.16 = 0.32$
As required, the distribution is $D - 2S \sim \text{N}(-0.4, 0.32)$.

2. Set up the required probability taking into account the absolute difference and consistent units.

We need the probability that the mass differs by more than $300\text{ g}$, which is $0.3\text{ kg}$.
$\text{P}(|D - 2S| > 0.3) = \text{P}(W < -0.3) + \text{P}(W > 0.3)$

3. Evaluate the probability using the normal distribution.

Using a graphing calculator:
$\text{P}(W < -0.3) \approx 0.57016$
$\text{P}(W > 0.3) \approx 0.10798$
$\text{Total Probability} = 0.57016 + 0.10798 = 0.67814 \approx 0.678 \text{ (to 3 s.f.)}$

1. State the assumption required to calculate the variance of the linear combination of the two random variables

In part (c), calculating $\mathrm{Var}(D - 2S) = \mathrm{Var}(D) + 2^2 \mathrm{Var}(S)$ requires the two variables to be independent.

Assumption: The mass of a randomly chosen durian and the mass of a randomly chosen soursop are independent.

1. Define the random variable for the total mass of 5 durians and find its distribution.

Let $T_D = D_1 + D_2 + D_3 + D_4 + D_5$ be the total mass of 5 randomly chosen durians.
$\mathrm{E}(T_D) = 5 \times 2.2 = 11$
$\mathrm{Var}(T_D) = 5 \times 0.4^2 = 0.8$
$T_D \sim \mathrm{N}(11, 0.8)$

2. Determine the distribution of the total cost.

Let $C$ be the total cost of the 5 durians. Since durians are sold at $\char36{}20$ per kg, $C = 20T_D$.
$\mathrm{E}(C) = 20 \times 11 = 220$
$\mathrm{Var}(C) = 20^2 \times 0.8 = 320$
$C \sim \mathrm{N}(220, 320)$

3. Calculate the probability that the total cost is at least $\$225$.

We want to find $\mathrm{P}(C \ge 225)$.
Using a graphing calculator:
$\mathrm{P}(C \ge 225) \approx 0.38992$
$\mathrm{P}(C \ge 225) = 0.390 \text{ (to 3 s.f.)}$