A sample of $n$ observations was taken from a population of mean $\mu$ and variance 10. Find the least sample size required such that the probability of the sample mean lying between $\mu - 0.5$ and $\mu + 0.5$ is more than 0.89. State an assumption that you have to make in order to proceed with this calculation.

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.)