在本题中，你应该写明你所使用的任何分布的参数。 一个果园销售两种品种的奇异果：A型和B型。A型和B型奇异果的质量（以克为单位）分别服从分布 $N(110,6^2)$ 和 $N(85,8^2)$。假设这两个分布是相互独立的。 奇异果被随机选择并装入袋中。A型奇异果每袋装 30 个，而B型奇异果每袋装 60 个。 A型和B型奇异果的售价分别为每公斤 $25 和每公斤 $15。 该果园还生产转基因G型奇异果，每袋装 50 个。G型奇异果的质量呈同分布，均值为 90 克，标准差为 5 克。 每个礼盒包含一袋A型奇异果 and 一袋G型奇异果。如果礼盒中奇异果的总质量至少为 7.85 公斤，则该礼盒符合销售标准。

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).