一家杂货店销售两种苹果：品牌 $A$ 和品牌 $S$。这两种品牌苹果的质量（以克为单位）可以建模为相互独立的正态分布，其均值和标准差如下表所示。

	均值	标准差
品牌 $A$	82.2	2.2
品牌 $S$	78.8	3.1

你可以假设以下苹果样本是随机抽取的。

1. State the distribution for the mass of a Brand $A$ apple.

Let $X$ be the mass of a Brand $A$ apple in grams.
$X \sim \mathrm{N}(82.2, 2.2^2)$

2. Calculate the required probability using a graphing calculator.

$\mathrm{P}(80 < X < 84) = 0.634718\dots \approx 0.635$ (to 3 s.f.)

1. Find the expected value and variance of the total mass of 5 independent Brand $A$ apples.

Let $T = A_1 + A_2 + A_3 + A_4 + A_5$ where $A_i$ represents the mass of a Brand $A$ apple.
$\mathrm{E}(T) = 5 \times 82.2 = 411$
$\mathrm{Var}(T) = 5 \times 2.2^2 = 24.2$

2. State the distribution of the total mass.

Since the mass of each apple is normally distributed and independent, their sum is also normally distributed.
$T \sim \mathrm{N}(411, 24.2)$

3. Calculate the probability that the total mass is more than 408 grams.

$\mathrm{P}(T > 408) = 0.72901 \approx 0.729 \text{ (to 3 s.f.)}$

1. Define a new random variable for the difference in mass and find its expected value.

Let $D = S - 0.9A$, where $S$ is the mass of a Brand $S$ apple and $0.9A$ is the mass of a peeled Brand $A$ apple.
$\mathrm{E}(D) = \mathrm{E}(S) - 0.9\mathrm{E}(A)$
$\mathrm{E}(D) = 78.8 - 0.9(82.2)$
$\mathrm{E}(D) = 78.8 - 73.98 = 4.82$

2. Calculate the variance of the difference in mass.

Since $S$ and $A$ are independent,
$\mathrm{Var}(D) = \mathrm{Var}(S) + 0.9^2\mathrm{Var}(A)$
$\mathrm{Var}(D) = 3.1^2 + 0.81(2.2^2)$
$\mathrm{Var}(D) = 9.61 + 0.81(4.84) = 13.5304$
Thus, $D \sim \mathrm{N}(4.82, 13.5304)$.

3. Formulate the probability that the absolute difference exceeds one gram.

The phrase "difference in mass" implies the absolute difference. We need to find $\mathrm{P}(|D| > 1)$.
$\mathrm{P}(|D| > 1) = \mathrm{P}(D > 1) + \mathrm{P}(D < -1)$

4. Evaluate the probability using a graphing calculator.

Using the normal cumulative distribution function on a graphing calculator:
$\mathrm{P}(D > 1) \approx 0.85048$
$\mathrm{P}(D < -1) \approx 0.05680$
$\mathrm{P}(|D| > 1) = 0.85048 + 0.05680 = 0.90728$
$\mathrm{P}(|D| > 1) \approx 0.907 \text{ (to 3 s.f.)}$