Fruits are sold by weight. Apples are sold at $2 per kg and oranges at $3 per kg. The masses, in kg, of apples and oranges are modelled as having independent normal distributions with means and standard deviations as shown in the table.

	Mean mass	Standard deviation
Apples	0.2	0.05
Oranges	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 $7$.