In this question you should state clearly the values of the parameters of any normal distributions you use. A fruit seller sells two kinds of fruits, namely durians and soursops. The masses, in kg, of a durian and a soursop are modelled as having normal distributions with means and standard deviations as shown in the following table.

	Mean mass (kg)	Standard deviation (kg)
Durian	2.2	0.4
Soursop	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:
$\mathrm{P}(C \ge 225) \approx 0.38992$
$\mathrm{P}(C \ge 225) = 0.390 \text{ (to 3 s.f.)}$