The random variable $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$.

1. Standardise the variable $X$ to the standard normal variable $Z$.

$\mathrm{P}(X > \mu + 0.05\sigma) = \mathrm{P}\left(\frac{X - \mu}{\sigma} > 0.05\right)$
$= \mathrm{P}(Z > 0.05)$

2. Evaluate the probability.

$\mathrm{P}(Z > 0.05) \approx 0.48006$
$\approx 0.480 \text{ (to 3 s.f.)}$

1. State the distribution of the sample mean $\overline{X}$.

Since $X \sim \mathrm{N}(\mu, \sigma^2)$, the sample mean of $n$ independent observations is distributed as:
$\overline{X} \sim \mathrm{N}\left(\mu, \frac{\sigma^2}{n}\right)$

2. Formulate the condition given in the question and use symmetry.

The probability that $\overline{X}$ differs from $\mu$ by at least $0.05\sigma$ is less than 0.4:
$\mathrm{P}(|\overline{X} - \mu| \ge 0.05\sigma) < 0.4$
By the symmetry of the normal distribution,
$\mathrm{P}(\overline{X} - \mu \ge 0.05\sigma) < 0.2$

3. Standardise the variable and form an inequality for $n$.

Standardising using $Z = \frac{\overline{X} - \mu}{\sigma/\sqrt{n}}$:
$\mathrm{P}\left( Z \ge \frac{0.05\sigma}{\sigma / \sqrt{n}} \right) < 0.2$
$\mathrm{P}(Z \ge 0.05\sqrt{n}) < 0.2$

4. Solve the inequality to find the least integer $n$.

Using the graphing calculator, the value of $z$ such that $\mathrm{P}(Z \ge z) = 0.2$ is approximately $0.84162$.
Because $\mathrm{P}(Z \ge z)$ is a strictly decreasing function of $z$, we require:
$0.05\sqrt{n} > 0.84162$
$\sqrt{n} > 16.8324$
$n > 283.33$
Since $n$ must be an integer, the least value of $n$ is $284$.

1. Identify the success probability from part (a) and define the binomial distribution model.

From part (a), the probability that a single observation exceeds $\mu + 0.05\sigma$ is:
$p = \mathrm{P}\left(Z > \frac{\mu + 0.05\sigma - \mu}{\sigma}\right) = \mathrm{P}(Z > 0.05) \approx 0.48006$
Let $Y$ be the number of observations out of $m$ that exceed $\mu + 0.05\sigma$.
$Y \sim \mathrm{B}(m, 0.48006)$

2. Formulate the probability inequality for the required condition.

We are given that the probability of at least 4 such observations is at most $0.95$:
$\mathrm{P}(Y \ge 4) \le 0.95$
Rewriting this using the complement:
$1 - \mathrm{P}(Y \le 3) \le 0.95$
$\mathrm{P}(Y \le 3) \ge 0.05$

3. Solve for the greatest value of $m$ using a graphing calculator.

Using the table function on a graphing calculator to test integer values of $m$:
When $m = 13$, $\mathrm{P}(Y \le 3) \approx 0.0526 \ge 0.05$
When $m = 14$, $\mathrm{P}(Y \le 3) \approx 0.0337 < 0.05$
Hence, the greatest integer value of $m$ satisfying the condition is 13.