Confidence Interval, why is this unusual?

Question

A Normal population has a known mean of $\mu = 50$ and a known variance of $\sigma = \sqrt 2$. A random sample of size $n = 16$ is selected from this population and find that the sample mean is $x = 52$. This is an out-of-the-box sample result. common Why? Answer using a $95\%$ confidence interval.

I got the interval $(51.30, 52.69)$... But what should I do after?

Answer 1

Let X denote the distribution of the existing population.

$X \sim N(50, 2)$

$X_1 + X_2 + ... + X_{16} \sim N(16*50, 16 * 2)$

$X_1 + X_2 + ... + X_{16} \sim N(800, 32)$

Let S denote the distribution of the sample.

$S =\frac{X_1 + X_2 + ... + X_{16}}{16} \sim N(800/16, (1/16)^2 * 32) $

$S = \frac{X_1 + X_2 + ... + X_{16}}{16} \sim N(50, 1/8) $

So, our 95% interval (2 standard deviations) is $(50-2*\sqrt{1/8},50+2*\sqrt{1/8}) = (49.3, 50.7)$

So if we did this experiment, 95% of the time, our sample mean would lie between 49.3 and 50.7. So it could be said that finding the sample mean to be 52 is a bit unusual.

Answer 2

If $X\sim N(50;2)$ then $\overline{X}_{16}\sim N(50;1/8)$

This is the graph of the normal density...

as you can see... 52 is a very unusual value for $\overline{X}_{16}$