On average, one in 40 scientific calculators manufactured by a company is faulty. An educational store manager decides to order $n$ scientific calculators from the company. The event that a randomly chosen scientific calculator is faulty is independent of that of another randomly chosen calculator.
The number of these scientific calculators which are faulty is denoted by the random variable $X$.

1. State the assumption required for the binomial distribution that is not already explicitly guaranteed by the question stem.

The conditions for a binomial distribution are a fixed number of trials, two possible outcomes, independent trials, and a constant probability of success. The question states that there is a fixed number of calculators ($n$) and that their faultiness is independent. Therefore, the required assumption is that the probability of success (a calculator being faulty) remains constant for each calculator.

1. Calculate the mean and variance of $X$.

Given $p = \frac{1}{40} = 0.025$ and $n = 30$,
$X \sim \text{B}(30, 0.025)$

$\text{Mean, } \mu = np = 30(0.025) = 0.75$

$\text{Variance, } \sigma^2 = np(1-p) = 30(0.025)(0.975) = 0.73125$ \quad (or $\frac{117}{160}$)

2. Translate the given condition into an inequality involving $X$.

$\text{Standard deviation, } \sigma = \sqrt{0.73125} \approx 0.85513$

We need the probability that $X$ differs from its mean by less than $2\sigma$, which is $\text{P}(|X - \mu| < 2\sigma)$:
$\text{P}(0.75 - 2(0.85513) < X < 0.75 + 2(0.85513))$
$\text{P}(-0.960 < X < 2.460)$

3. Identify valid values of $X$ and compute the probability.

Since $X$ is the number of faulty calculators, it can only take discrete integer values. Therefore, $X = 0, 1, \text{ or } 2$.

$\text{P}(X \le 2) = \text{P}(X=0) + \text{P}(X=1) + \text{P}(X=2)$

Using a graphing calculator's binomial cumulative distribution function:
$\text{P}(X \le 2) \approx 0.96161$
$\text{P}(X \le 2) = 0.962 \text{ (to 3 s.f.)}$

1. Formulate the probability inequality for the given condition.

Let $X \sim \mathrm{B}(n, 0.025)$.
We are given that $\mathrm{P}(X > 2) \le 0.05$.
$1 - \mathrm{P}(X \le 2) \le 0.05$
$\mathrm{P}(X \le 2) \ge 0.95$

2. Evaluate the probability for consecutive integer values of $n$ using a graphing calculator.

Using the table or trial function on a graphing calculator for $\mathrm{P}(X \le 2)$:
When $n = 33$, $\mathrm{P}(X \le 2) \approx 0.95119 > 0.95$
When $n = 34$, $\mathrm{P}(X \le 2) \approx 0.94723 < 0.95$

3. Identify the largest integer that satisfies the inequality.

As $n$ increases, the probability $\mathrm{P}(X \le 2)$ strictly decreases. Therefore, the largest possible number of calculators the manager can order is $33$.