Question

(Statistics) [Let X1, X2,...,Xn be a collection of independent normal variables with mean "mu" and variance sigma^2]. If sigma^2 = 16 and n = 25 what is the probability that the sample mean, Xbar, takes on a value that is within one unit of the population mean "mu," so in other words i'm asked to find P(|Xbar - mu| <= 1).
a note, i'm pretty bad at Stats, so a strong explanation with nothing left out would be nice

Answer 1

Recall the definition of absolute value that:
\[|x| =\begin{cases} x& \text{if}&x\ge0 \\
-x& \text{if}&x<0\end{cases} \]
Thus,
\[\left|\bar{X}-\mu\right| =\begin{cases} \bar{X}-\mu & \text{if}&\bar{X}-\mu\ge0 \\
-\left( \bar{X}-\mu\right)& \text{if}&\bar{X}-\mu<0\end{cases} \]

Now, with an inequality, what does this mean: you have
\(\left| \bar{X}-\mu \right| \le 1\). From the definition of the absolute value , it means you have:

\(\bar{X}-\mu \le 1\) or
\(-(\bar{X}-\mu) \le 1\implies\bar{X}-\mu\ge -1\)

Combining both of these inequalities into "one" statement, you get:
\(-1 \le \bar{X}-\mu \le 1\)

Thus, \[ P\left(\left| \bar{X}-\mu\right| \le 1\right)=P\left(-1 \le \bar{X}-\mu \le 1
\right)\]

Now, recall the distribution of \(\bar{X}\): it has mean \(\mu\) and variance \(\sigma^2/n\), i.e. \[\bar{X}\sim\text{N}\left( \mu, \frac{\sigma^2}{n} \right)\]. So, you can standardize what you have above to get:

\[ P\left(\frac{-1}{\sigma/\sqrt{n}} \le \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \le \frac{1}{\sigma/\sqrt{n}}\right)=P\left(\frac{-1}{4/\sqrt{25}} \le Z \le \frac{1}{4/\sqrt{25}}\right)\\
=P(-1.25\le Z\le1.25)\], where \(Z\sim\text{N}(0,1)\)

Answer 2

Then you just you a normal table to find this probability which should be easy.

Also, if you haven't seen the result \[\bar{X}\sim\text{N}\left( \mu, \frac{\sigma^2}{n} \right)\], you should prove it (which is easy to do... or I could show it if you need to).