从均值为 $\mu$、方差为 10 的总体中抽取了一个包含 $n$ 个观测值的样本。求所需的最小样本容量，使得样本均值介于 $\mu - 0.5$ 和 $\mu + 0.5$ 之间的概率大于 0.89。写出为了进行此计算而必须做出的一个假设。

1. Identify the distribution of the sample mean and formulate the probability inequality.

Let $\bar{X}$ be the sample mean.
Assuming the population is normally distributed, $\bar{X} \sim \text{N}\left(\mu, \frac{10}{n}\right)$.

We require:
$\text{P}(\mu - 0.5 < \bar{X} < \mu + 0.5) > 0.89$

2. Standardise the probability statement.

Standardising to the standard normal variable $Z \sim \text{N}(0, 1)$:
$\text{P}\left( \frac{\mu - 0.5 - \mu}{\sqrt{10/n}} < Z < \frac{\mu + 0.5 - \mu}{\sqrt{10/n}} \right) > 0.89$
$\text{P}\left( -\frac{0.5}{\sqrt{10/n}} < Z < \frac{0.5}{\sqrt{10/n}} \right) > 0.89$

3. Relate the symmetric interval to a single tail probability.

By the symmetry of the standard normal distribution:
$2\,\text{P}\left( Z < \frac{0.5}{\sqrt{10/n}} \right) - 1 > 0.89$
$2\,\text{P}\left( Z < \frac{0.5}{\sqrt{10/n}} \right) > 1.89$
$\text{P}\left( Z < \frac{0.5}{\sqrt{10/n}} \right) > 0.945$

4. Use the inverse normal function to find the critical value and set up an inequality for $n$.

From the GC, the critical $z$-value for an area of $0.945$ is approximately $1.59819$.
$\frac{0.5}{\sqrt{10/n}} > 1.59819$
$\frac{0.5\sqrt{n}}{\sqrt{10}} > 1.59819$

5. Solve for $n$ and deduce the least integer sample size.

$\sqrt{n} > \frac{1.59819 \times \sqrt{10}}{0.5}$
$\sqrt{n} > 10.1079$
$n > 102.169$
Since the sample size $n$ must be an integer, the least sample size required is $103$.

6. State the necessary assumption for the calculations to be valid.

Assumption: We assume that the population follows a normal distribution.
(Note: Since $n$ is initially unknown, we cannot guarantee $n \ge 30$ to invoke the Central Limit Theorem upfront, making this assumption necessary to proceed.)