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 $N(110,6^2)$ and $N(85,8^2)$ 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. The selling prices of Type A and Type B kiwis are $25 per kilogram and $15 per kilogram respectively. The orchard also produces genetically modified Type G kiwis, which are packed in bags of 50. The masses of Type G kiwis are identically distributed with a mean of 90 grams and a standard deviation of 5 grams. Each Assortment Box contains one bag of Type A kiwis and one bag of Type G kiwis. A box is fit for sale if the total mass of the kiwis in the box is at least 7.85 kilograms.

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

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

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

Using a graphing calculator,
$P(109 < \bar{X}_A < 115) \approx 0.81934$
$\approx 0.819 \text{ (to 3 s.f.)}$

1. Define the random variables for the average selling price of a kiwi from a bag of Type A and Type B kiwis.

Let $\bar{A}$ and $\bar{B}$ be the sample mean masses of kiwis in a bag of Type A ($n=30$) and Type B ($n=60$) respectively.
Let $S_A$ and $S_B$ be the average selling price of a kiwi from a bag of Type A and Type B kiwis.

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.
Therefore, $S_A = 0.025\bar{A}$ and $S_B = 0.015\bar{B}$.

2. Find the expectation and variance for the average selling price of a kiwi from a bag of Type A.

$\mathrm{E}(S_A) = 0.025 \times 110 = 2.75$
$\mathrm{Var}(S_A) = 0.025^2 \mathrm{Var}(\bar{A}) = 0.000625 \left(\frac{6^2}{30}\right) = 0.00075$

3. Find the expectation and variance for the average selling price of a kiwi from a bag of Type B.

$\mathrm{E}(S_B) = 0.015 \times 85 = 1.275$
$\mathrm{Var}(S_B) = 0.015^2 \mathrm{Var}(\bar{B}) = 0.000225 \left(\frac{8^2}{60}\right) = 0.00024$

4. Define the distribution for the difference in average selling prices and calculate the required probability.

Let $D = S_A - S_B$.
$\mathrm{E}(D) = 2.75 - 1.275 = 1.475$
$\mathrm{Var}(D) = \mathrm{Var}(S_A) + \mathrm{Var}(S_B) = 0.00075 + 0.00024 = 0.00099$

$D \sim \mathrm{N}(1.475, 0.00099)$

We require the probability that $S_A$ is more than $S_B$ by at most $\char36{}1.50$, which means $0 < D \leqslant 1.50$.
$\mathrm{P}(0 < D \leqslant 1.50) \approx 0.78656 \approx 0.787$ (to 3 s.f.)

1. Find the distribution, mean, and variance of the total mass of an Assortment Box

Let $M$ be the total mass of the kiwis in an Assortment Box, consisting of 30 Type A kiwis and 50 Type G kiwis.

$\text{E}(M) = 30(110) + 50(90) = 7800$
$\text{Var}(M) = 30(6^2) + 50(5^2) = 1080 + 1250 = 2330$
Since the sample size of Type G kiwis ($n=50$) is large, by the Central Limit Theorem, their total mass follows an approximate normal distribution. Thus, the total mass $M$ is approximately normally distributed:
$M \sim \text{N}(7800, 2330)$

2. Calculate the probability that a single Assortment Box is fit for sale

A box is fit for sale if its total mass is at least $7.85 \text{ kg} = 7850 \text{ g}$.
Using the normal distribution $M \sim \text{N}(7800, 2330)$:
$P(M \ge 7850) = 0.149954$

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

Let $Y$ be the number of Assortment Boxes fit for sale out of 10 randomly selected boxes.
$Y \sim \text{B}(10, 0.149954)$
We require the probability that none of the boxes are fit for sale ($Y = 0$):
$P(Y = 0) = (1 - 0.149954)^{10} = 0.19698 \approx 0.197 \text{ (to 3 s.f.)}$

1. State the approximation needed for the distribution of the mass of Type G kiwis.

Approximation: The distribution of the masses of Type G kiwis is not given. However, since the sample size of $50$ Type G kiwis in a bag is sufficiently large ($n \ge 30$), we use the Central Limit Theorem to approximate the total mass of the $50$ Type G kiwis as being normally distributed.

2. State the assumption needed regarding the relationship between the masses of the kiwis.

Assumption: The question states that the masses of Type G kiwis are 'identically distributed' but does not explicitly state they are independent. We must assume that the mass of any Type G kiwi is independent of the mass of other Type G kiwis so that we can sum their variances and apply the Central Limit Theorem.
(Another valid assumption: The total mass of the bag of Type A kiwis is independent of the total mass of the bag of Type G kiwis in the same Assortment Box).