某公司生产的科学计算器平均每 40 台中有一台是有故障的。一家教育用品店的经理决定向该公司订购 $n$ 台科学计算器。随机选择一台科学计算器是有故障的这一事件与随机选择另一台计算器是否有故障是相互独立的。
这些有故障的科学计算器的数量用随机变量 $X$ 表示。

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 $33$.