Here is the question:
A random sample of size $12$ from a normal population $\mathcal{N}(\mu,\sigma^2)$ has mean $\bar{x}=27.8$ and standard deviation $s^2=3.24$. If we use $\bar{X}$ to estimate $\mu$, what is the probability that the absolute error is less than 1?
I don't believe there is enough information to answer this question. We're trying to compute $\mathbb{P}(|\bar{X}-\mu|<1)$. We know $\bar{X}-\mu\sim\mathcal{N}(0,\frac{\sigma^2}{12})$ so the probability we seek is
$$\mathbb{P}\left(|Z|<\frac{\sqrt{12}}{\sigma}\right)=2\int_{0}^{\sqrt{12}/{\sigma}}\frac{1}{\sqrt{2\pi} }e^{-z^2/2}\mathrm{d}z$$
Here $Z\sim \mathcal{N}(0,1)$. This question doesn't give any information about $\sigma$ and I hesitate to use $s$ in the place of $\sigma $ since the problem doesn't explicitly mention to make this estimation. Thoughts?
How about this? Define $f:(0,\infty)\rightarrow(0,1)$ by $f(\sigma)=2\int_0^{\sqrt{12}/\sigma}\frac{1}{\sqrt{2\pi}}e^{-z^2/2}\mathrm{d}z$. Note $f$ is a decreasing bijection, so $p^{-1}$ exists.
Let $x_1,...,x_{12}$ be our observed sample of size $12$.
The "best" value for $p=f(\sigma)$ is the one that maximizes the likelihood function $$\begin{eqnarray*}l(\mu,p)&=&f_{X_1}(x_1)\times \dots \times f_{X_{12}}(x_{12}) \\&=&\frac{1}{(f^{-1}(p))^{12}(\sqrt{2\pi})^{12}}\exp\Big\{-\frac{1}{2(f^{-1}(p))^2}\sum_{i=1}^{12}(x_i-\mu)^2\Big\}\end{eqnarray*}$$ It's not hard to show that $l(\mu,p)$ is optimized when $\mu=\bar{x}=27.8$ and $p=f\left(s\sqrt{\frac{11}{12}}\right)\approx 95.56$%. Thoughts?