A grocery store sells two types of apples, Brand $A$ and Brand $S$. The masses (in grams) of the apples for each brand can be modelled by independent normal distributions with means and standard deviations as shown in the table.

	Mean	Standard Deviation
Brand $A$	82.2	2.2
Brand $S$	78.8	3.1

You may assume that the following samples of apples are randomly chosen.

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:
$\mathrm{P}(D > 1) \approx 0.85048$
$\mathrm{P}(D < -1) \approx 0.05680$
$\mathrm{P}(|D| > 1) = 0.85048 + 0.05680 = 0.90728$
$\mathrm{P}(|D| > 1) \approx 0.907 \text{ (to 3 s.f.)}$