In an examination, the score, $X$, for paper 1 of a student is found to follow a normal distribution with mean 62 and standard deviation $\sigma$, and the score, $Y$, for paper 2 of a student is found to follow a normal distribution with mean 71 and standard deviation 8. The final score of a student for the examination is the average score of the 2 papers and it is assumed that $X$ and $Y$ are independent random variables.

1. Determine the distribution of the difference in scores between two randomly selected students.

Let $Y_A$ and $Y_B$ be the Paper 2 scores of students $A$ and $B$ respectively.
Since $Y_A, Y_B \sim \text{N}(71, 8^2)$ and they are independent randomly selected students,
$Y_A - Y_B \sim \text{N}(71 - 71, 8^2 + 8^2)$
$Y_A - Y_B \sim \text{N}(0, 128)$

2. Formulate the required inequality and calculate the probability.

The condition '$A$ has at most $2$ marks less than $B$' translates to $Y_A \ge Y_B - 2$, which rearranges to $Y_A - Y_B \ge -2$.
$P(Y_A - Y_B \ge -2) \approx 0.57016$
$P(Y_A - Y_B \ge -2) \approx 0.570 \text{ (to 3 s.f.)}$

1. Define the final score random variable and determine its mean and variance.

Let $F$ be the final score of a student. We are given $F = \frac{X+Y}{2}$.
$\mathrm{E}(F) = \frac{\mathrm{E}(X)+\mathrm{E}(Y)}{2} = \frac{62+71}{2} = 66.5$
$\mathrm{Var}(F) = \left(\frac{1}{2}\right)^2 (\mathrm{Var}(X)+\mathrm{Var}(Y)) = \frac{\sigma^2+8^2}{4} = \frac{\sigma^2+64}{4}$
Thus, $F \sim \mathrm{N}\left(66.5, \frac{\sigma^2+64}{4}\right)$.

2. Set up the probability equation for the final score.

We are given that $15\%$ of the students have at least a final score of $75$, so:
$\mathrm{P}(F \ge 75) = 0.15$

3. Standardise the normal variable and use the inverse normal function to find the corresponding $Z$-value.

Standardising the variable $F$ to $Z \sim \mathrm{N}(0, 1)$:
$\mathrm{P}\left(Z \ge \frac{75 - 66.5}{\sqrt{\frac{\sigma^2+64}{4}}}\right) = 0.15$
Using the inverse normal distribution (area to the left is $0.85$):
$\frac{8.5}{\frac{\sqrt{\sigma^2+64}}{2}} \approx 1.0364$
$\frac{17}{\sqrt{\sigma^2+64}} = 1.0364$

4. Solve for the standard deviation $\sigma$.

$\sqrt{\sigma^2+64} = \frac{17}{1.036433} \approx 16.4024$
$\sigma^2+64 = 16.4024^2 \approx 269.04$
$\sigma^2 = 205.04$
$\sigma = \sqrt{205.04} \approx 14.319$
$\sigma \approx 14.3 \text{ (to 3 s.f.)}$

1. Calculate $\sigma$ and the variance of the combined scores from the context in part (ii).

Since the value of $\sigma$ is needed, we must first determine it from the given conditions in part (ii).
Let $S = X + Y$ be the sum of scores. The final score is $\frac{S}{2}$.
We are given $\mathrm{P}\left(\frac{S}{2} \ge 75\right) = 0.15$, which means $\mathrm{P}(S \ge 150) = 0.15$.
$S \sim \mathrm{N}(62 + 71, \sigma^2 + 8^2) \implies S \sim \mathrm{N}(133, \sigma^2 + 64)$
Standardising the distribution:
$\mathrm{P}\left(Z \ge \frac{150 - 133}{\sqrt{\sigma^2 + 64}}\right) = 0.15$
From the graphing calculator, the $z$-value for the upper $15\%$ is $z \approx 1.0364$.
$\frac{17}{\sqrt{\sigma^2 + 64}} \approx 1.0364 \implies \sqrt{\sigma^2 + 64} \approx 16.402$
$\sigma^2 + 64 \approx 269.04 \implies \sigma \approx 14.3$

2. Define the distribution for the difference in scores between Paper 1 and Paper 2.

Let $W = X - Y$ represent the difference in scores between Paper 1 and Paper 2.
$\mathrm{E}(W) = \mathrm{E}(X) - \mathrm{E}(Y) = 62 - 71 = -9$
Since $X$ and $Y$ are independent random variables:
$\operatorname{Var}(W) = \operatorname{Var}(X) + \operatorname{Var}(Y) = \sigma^2 + 8^2 = \sigma^2 + 64$
From our previous calculation, we already found that $\sigma^2 + 64 \approx 269.04$.
Therefore, $W \sim \mathrm{N}(-9, 269.04)$.

3. Calculate the probability that the student performs better in Paper 1.

A student performs better in Paper 1 if $X > Y$, which translates to $W > 0$.
$\mathrm{P}(X > Y) = \mathrm{P}(W > 0)$
Using the graphing calculator:
$\mathrm{P}(W > 0) \approx 0.29160$
Rounding to 3 significant figures, the probability is $0.292$.

1. State whether the answers are valid and explain the realistic relationship between the two variables.

The answers obtained are likely invalid.

This is because the calculation relies on the assumption that $X$ and $Y$ are independent. In reality, this assumption is unrealistic as a single student's performance in Paper 1 and Paper 2 of the same examination are likely to be dependent (e.g., positively correlated).