In this question you should state the parameters of any distributions you use. An orchard sells kiwis of two varieties, Type $A$ and Type $B$. The masses, in grams, of Type $A$ and Type $B$ kiwis follow the distributions $\mathrm{N}(110, 6^2)$ and $\mathrm{N}(85, 8^2)$ respectively. It is assumed that these two distributions are independent. Kiwis are selected randomly and packed into bags. Type $A$ kiwis are packed in bags of 30, while Type $B$ kiwis are packed in bags of 60.

1. State the distribution of the sample mean mass of Type $A$ kiwis.

Let $\bar{X}$ be the mean mass of a random sample of $30$ Type $A$ kiwis.
Since the mass of a Type $A$ kiwi follows $\mathrm{N}(110, 6^2)$, the sample mean is exactly normally distributed:
$\bar{X} \sim \mathrm{N}\left(110, \frac{6^2}{30}\right)$
$\bar{X} \sim \mathrm{N}(110, 1.2)$

2. Calculate the probability that the mean mass is between 109 grams and 115 grams.

Using a graphing calculator to find the normal probability:
$\mathrm{P}(109 < \bar{X} < 115) = 0.8193498\dots$
$\mathrm{P}(109 < \bar{X} < 115) \approx 0.819 \text{ (to 3 s.f.)}$

1. Define the random variables for the sample means of the masses of Type A and Type B kiwis.

Let $\bar{X}$ and $\bar{Y}$ be the sample means of the masses in a bag of Type $A$ (30 kiwis) and Type $B$ (60 kiwis) respectively.
$\bar{X} \sim \mathrm{N}\left(110, \frac{6^2}{30}\right) \Rightarrow \bar{X} \sim \mathrm{N}(110, 1.2)$
$\bar{Y} \sim \mathrm{N}\left(85, \frac{8^2}{60}\right) \Rightarrow \bar{Y} \sim \mathrm{N}\left(85, \frac{16}{15}\right)$

2. Formulate the random variable for the difference in the average selling prices per kiwi.

The selling prices per gram are $\frac{\char36{}25}{1000} = \char36{}0.025$ for Type $A$ and $\frac{\char36{}15}{1000} = \char36{}0.015$ for Type $B$.
Let $W$ be the difference between the average selling price of a kiwi from a bag of Type $A$ and a bag of Type $B$.
$W = 0.025\bar{X} - 0.015\bar{Y}$
$\mathrm{E}(W) = 0.025(110) - 0.015(85) = 2.75 - 1.275 = 1.475$
$\mathrm{Var}(W) = 0.025^2(1.2) + 0.015^2\left(\frac{16}{15}\right) = 0.00075 + 0.00024 = 0.00099$
So $W \sim \mathrm{N}(1.475, 0.00099)$

3. Calculate the required probability.

The probability that the average price of a Type $A$ kiwi is more than a Type $B$ kiwi by at most $\char36{}1.50$ is $\mathrm{P}(0 < W \leqslant 1.50)$.
$\mathrm{P}(0 < W \leqslant 1.50) \approx 0.78656 \approx 0.787$ (to 3 s.f.)

1. Find the distributions of the total masses of a bag of Type A kiwis and a bag of Type G kiwis.

Let $M_A$ be the total mass of a bag of 30 Type $A$ kiwis, and $M_G$ be the total mass of a bag of 50 Type $G$ kiwis.

Since $A \sim \mathrm{N}(110, 6^2)$:
$M_A = A_1 + A_2 + \dots + A_{30} \sim \mathrm{N}(30 \times 110, 30 \times 6^2) \Rightarrow M_A \sim \mathrm{N}(3300, 1080)$

For Type $G$ kiwis, the distribution is unknown. However, since the sample size $n=50 \ge 30$ is large, by the Central Limit Theorem:
$M_G = G_1 + G_2 + \dots + G_{50} \sim \mathrm{N}(50 \times 90, 50 \times 5^2) \Rightarrow M_G \sim \mathrm{N}(4500, 1250)$ approximately.

2. Define the distribution for the total mass of the Assortment Box and find the probability of a box being fit for sale.

Let the total mass of an Assortment Box be $T = M_A + M_G$.
Since $M_A$ is normally distributed and $M_G$ is approximately normally distributed (and they are independent):
$\mathrm{E}(T) = 3300 + 4500 = 7800$
$\mathrm{Var}(T) = 1080 + 1250 = 2330$
Thus, $T \sim \mathrm{N}(7800, 2330)$ approximately.

A box is fit for sale if its total mass is at least $7.85 \text{ kg} = 7850 \text{ g}$.
$\mathrm{P}(T \ge 7850) \approx 0.150146$

3. Model the number of fit boxes with a Binomial distribution and calculate the final probability.

Let $Y$ be the number of boxes out of 10 that are fit for sale.
Assuming the masses of the 10 boxes are independent, $Y$ follows a Binomial distribution:
$Y \sim \mathrm{B}(10, 0.150146)$

The probability that none of the boxes are fit for sale is:
$\mathrm{P}(Y = 0) = (1 - 0.150146)^{10} = (0.849854)^{10} \approx 0.19652$

Correct to 3 significant figures, the probability is $0.197$.

1. State the approximation used for the distribution of Type G kiwis.

Approximation: Since the sample size of 50 Type $G$ kiwis is sufficiently large ($n \ge 30$), by the Central Limit Theorem, the total mass of the 50 Type $G$ kiwis in a bag is approximately normally distributed.

2. State the assumption(s) required to combine the distributions.

Assumption(s): The mass of any Type $G$ kiwi is independent of the mass of any other Type $G$ kiwi. Furthermore, the masses of Type $A$ kiwis and the masses of Type $G$ kiwis are independent of each other.