Checking an exercise on confidence intervals

Question

I have the following exercise on confidence intervals:

Given a population $X \sim \mathcal N( \mu, 4)$.

1. Consider a random sample of size $n = 10$. Determine a $95$% confidence interval for the mean $\mu$ choosing it in a way such that it has minimum length.

2. Now assume you have extracted a random sample of size $n = 5$, obtaining the following values $-3.5 , -1 , 0.5, 1.5, 3$. Determine an interval that has a probability of $95$% of containing $\mu$.

My answers:

1. So I know that \begin{align} 1-\alpha =\mathbb{P}\left[-z_{1-\alpha/2} < Z < z_{1-\alpha/2}\right] & \sim \mathbb{P}\left[-z_{1-\alpha/2} <\frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}<z_{1-\alpha/2}\right]\\ &= \mathbb{P}\left[\bar{X}-z_{1-\alpha/2} \frac{\sigma}{\sqrt{n}} <\mu <\bar{X}+z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}}\right] \end{align} So the confidence interval will be $$ \left[\bar{X} - 1.96 \frac{4}{10}, \bar{X} + 1.96 \frac{4}{10}\right]$$ this also has minimal length because the Normal distribution is symmetric correct? I don't find it explicitly?

2. $\bar{X}$ is equal to $0.5 / 5$ so the interval just becomes $$\left[0.5 / 5 - 1.96 \frac{4}{10}, 0.5 / 5 + 1.96 \frac{4}{10}\right]$$

correct? I feel like I am missing something on the first answer.

Answer 1

You are just making an error in your maths, since $\sqrt{n}$ equals $\sqrt{10}$ in the first answer and $\sqrt{5}$ in the second one, not $10$.

About the fact that the shortest interval is that centred on the mean, it's quite obvious from the shape of the normal distribution, not just because it's symmetrical but because density probability lowers as you move away from the mean. By geometric reasoning or using the average value theorem for integrals, it should be easy to prove that any interval with the same length would have a lower probability than the symmetrical one.