Quick questions on Estimation and confidence intervals explained: H2 Further Mathematics

Dividing the sum of squared deviations by $n$ underestimates the population variance, because the deviations are taken about the sample mean. The unbiased estimate uses $n - 1$:

For a sample of size $n$ from a population with mean $\mu$ and variance $\sigma^2$, the sample mean has

A confidence interval gives a range of plausible values for $\mu$. With the population standard deviation $\sigma$ known (or a large sample),

A $95\%$ confidence interval does not mean there is a $95\%$ probability the true mean lies in this particular interval. It means the procedure produces an interval that captures the true mean in $95\%$ of repeated samples. Wider confidence (say $99\%$) gives a wider interval; a larger sample narrows it.

Why does the unbiased estimate of variance divide by $n - 1$? [2 marks]

State the $95\%$ confidence interval formula for a mean with known $\sigma$. [1 mark]

Does a $95\%$ confidence interval mean a $95\%$ chance the true mean is inside it? [1 mark]