水果按重量出售。苹果的售价为每千克 $2，橙子的售价为每千克 $3。苹果和橙子的质量（以千克为单位）被建模为具有独立的正态分布，其均值和标准差如表中所示。

	平均质量	标准差
苹果	0.2	0.05
橙子	0.3	0.08

1. Define the random variable for the difference and calculate its mean and variance.

Let $A$ be the mass of an apple and $R$ be the mass of an orange, such that $A \sim \mathrm{N}(0.2, 0.05^2)$ and $R \sim \mathrm{N}(0.3, 0.08^2)$.
Let $W$ be the difference in mass between 3 randomly chosen apples and 1 orange. Thus, $W = A_1 + A_2 + A_3 - R$.

$\mathrm{E}(W) = 3\mathrm{E}(A) - \mathrm{E}(R) = 3(0.2) - 0.3 = 0.3$
$\mathrm{Var}(W) = 3\mathrm{Var}(A) + \mathrm{Var}(R) = 3(0.05^2) + 0.08^2 = 0.0139$

So, $W \sim \mathrm{N}(0.3, 0.0139)$.

2. Evaluate the probability that the absolute difference is less than 0.2 kg.

We require $\mathrm{P}(|W| < 0.2)$, which is $\mathrm{P}(-0.2 < W < 0.2)$.
Using a graphing calculator:
$\mathrm{P}(-0.2 < W < 0.2) \approx 0.19816$
$\mathrm{P} \approx 0.198 \text{ (to 3 s.f.)}$

1. Define the random variable for the total selling price and calculate its expected value.

Let $C$ be the total selling price of 5 apples and 5 oranges.
$C = 2(A_1 + A_2 + \dots + A_5) + 3(O_1 + O_2 + \dots + O_5)$
$\mathrm{E}(C) = 2(5 \times 0.2) + 3(5 \times 0.3) = 2(1) + 3(1.5) = 6.5$

2. Calculate the variance of the total selling price.

$\mathrm{Var}(C) = 2^2(5 \times 0.05^2) + 3^2(5 \times 0.08^2)$
$= 4(0.0125) + 9(0.032) = 0.05 + 0.288 = 0.338$

3. State the distribution of $C$ and calculate the required probability.

Since the individual masses are normally distributed, $C \sim \mathrm{N}(6.5, 0.338)$.
$\mathrm{P}(C > 5) \approx 0.995059$
$\mathrm{P}(C > 5) \approx 0.995 \text{ (to 3 s.f.)}$

1. State the distribution of the sample mean weight of $n$ apples.

Let $\bar{X}$ be the mean weight of $n$ randomly chosen apples.
$\bar{X} \sim \mathrm{N}\left(0.2, \frac{0.05^2}{n}\right)$

2. Set up the probability inequality and standardise.

We require $\mathrm{P}(\bar{X} > 0.21) < 0.3$.
Standardising the normal variable gives:
$\mathrm{P}\left(Z > \frac{0.21 - 0.2}{0.05 / \sqrt{n}}\right) < 0.3$
$\mathrm{P}(Z > 0.2\sqrt{n}) < 0.3$

3. Solve for $n$ using the inverse normal distribution.

For the standard normal distribution, the value $z$ such that $\mathrm{P}(Z > z) = 0.3$ is approximately $0.5244$.
Therefore, we need the test statistic to be strictly greater than this value:
$0.2\sqrt{n} > 0.5244$
$\sqrt{n} > 2.622$
$n > 6.875$
Since $n$ must be an integer, the least value of $n$ is $7$.