$X$ is a normal random variable with mean 1 and variance 1, and $Y$ is also a normal random variable with mean $\mu$ and variance 2.

1. Find the distribution of the sum $X + Y$.

Assuming $X$ and $Y$ are independent (as will be stated in part (c)),
$E(X + Y) = E(X) + E(Y) = 1 + \mu$
$\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) = 1 + 2 = 3$
Let $W = X + Y$, then $W \sim N(1 + \mu, 3)$.

2. Set up the inequality for the given probability.

We are given that $P(0 \le X + Y < 3) > 0.44$, which means:
$P(0 \le W < 3) > 0.44$

3. Solve the inequality for $\mu$ using a graphing calculator.

Using a graphing calculator to plot the probability function $Y_1 = \text{normalcdf}(0, 3, 1+X, \sqrt{3})$ against $\mu$, and the line $Y_2 = 0.44$.
Finding the points of intersection:
$\mu \approx -1.1012 \quad \text{and} \quad \mu \approx 2.1012$
From the graph, the probability $Y_1$ is strictly greater than $0.44$ when $\mu$ is between these two values.
$-1.10 < \mu < 2.10 \text{ (to 3 s.f.)}$

1. Formulate the normal distribution for the sum of $n$ independent observations of $X$.

Let $W = X_1 + X_2 + \dots + X_n$.
Since each $X_i \sim \mathrm{N}(1, 1)$ independently:
$\mathrm{E}(W) = n \times 1 = n$
$\mathrm{Var}(W) = n \times 1 = n$
So $W \sim \mathrm{N}(n, n)$.

2. Find the distribution of the difference between $W$ and $2Y$.

Given $Y \sim \mathrm{N}(10, 2)$, let $V = W - 2Y$.
$\mathrm{E}(V) = \mathrm{E}(W) - 2\mathrm{E}(Y) = n - 2(10) = n - 20$
$\mathrm{Var}(V) = \mathrm{Var}(W) + 2^2\mathrm{Var}(Y) = n + 4(2) = n + 8$
So $V \sim \mathrm{N}(n - 20, n + 8)$.

3. Standardise the variable and set up an inequality using the given probability.

We are given $\mathrm{P}(W > 2Y + 10) < 0.03$, which means $\mathrm{P}(V \ge 10) < 0.03$.
Standardising the normal variable:
$\mathrm{P}\left(Z \ge \frac{10 - (n - 20)}{\sqrt{n + 8}}\right) < 0.03$
$\mathrm{P}\left(Z \ge \frac{30 - n}{\sqrt{n + 8}}\right) < 0.03$
From the standard normal distribution, if $\mathrm{P}(Z \ge z) = 0.03$, then $z \approx 1.8808$.
For the probability to be strictly less than $0.03$, the $z$-value must be strictly greater than this critical value:
$\frac{30 - n}{\sqrt{n + 8}} > 1.8808$

4. Solve the inequality to find the largest possible integer value of $n$.

We can test values of $n$ or solve the equation algebraically.
If $n = 20$:
$\frac{30 - 20}{\sqrt{20 + 8}} = \frac{10}{\sqrt{28}} \approx 1.8898 > 1.8808$
If $n = 21$:
$\frac{30 - 21}{\sqrt{21 + 8}} = \frac{9}{\sqrt{29}} \approx 1.6713 < 1.8808$
Alternatively, solving the quadratic equation $(30 - n)^2 = 1.8808^2(n + 8)$ yields $n \approx 20.04$.
Since we need $\frac{30-n}{\sqrt{n+8}} > 1.8808$, we must have $n \le 20.04$.
Thus, the largest possible integer $n$ is $20$.

1. State the assumption necessary for the calculation of the variance of linear combinations such as $X+Y$ and $\sum X_i - 2Y$.

$X$ and $Y$ are independent.