finding n with N $\left ( \mu ,\frac{\sigma^2}{n} \right ) $ with given probability This problem has me scared to my wits, mostly because there are more like it!
The mean of a random sample of n observations drawn from a N $ ( \mu ,\sigma^{2} ) $ distribution is denoted by $  \overline{\chi} $. Given that P $ (| \overline{\chi} - \mu| > 0.5\sigma ) \lt 0.05 $ 
Find the smallest value of n.
So I tried:
Since $\overline{\chi}$ ~ N $ \left ( \mu ,\frac{\sigma^2}{n} \right )  $
P $ (| \overline{\chi} - \mu| > 0.5\sigma ) \lt 0.05 $  Should be:
$$ \frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}} \lt 0.05$$
$$1-\phi\left(\frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}}\right) \lt 0.05 $$
($\phi$ is the normal distribution function, I lookup a value from the normal dist. table.)
$$ \frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}} \lt 1.645$$
Now where do I go from here!?
This is totally confusing me, especially that darn modulus. How do I use my result so far with this?
P $ (| \overline{\chi} - \mu| > 0.5\sigma )$
Thanks
Gideon

This is not a good question because it actually has nothing to do with sampling theory (nor with the Chebyshev inequality, for that matter): it's quite artificial.  Nevertheless, you have grasped its intent well.  A good way to work through it is to reason in rather basic terms, rather than relying solely on mathematical manipulation; to wit:


*

*The standard deviation of $\bar{\chi}$ is $1/\sqrt{n}$ times the SD of the parent distribution, $\sigma$.

*Therefore we should express the cutoff $0.5 \sigma$ in terms of the SD of $\bar{\chi}$ itself:
$$0.5 \sigma = 0.5 \sqrt{n} \left(\sigma / \sqrt{n}\right) = 0.5 \sqrt{n} SD_{\bar{\chi}}.$$

*We know (or can look up in a table) that $5$% of the time any normally distributed variate is further than $1.96$ times its standard deviation from its mean.  (The value $1.645$ corresponds to a probability of $10$%: the absolute value requires you to look at both tails, not just the right hand tail.)

*Thus the question is merely asking you to solve the equation
$$0.5 \sqrt{n} \ge 1.96.$$
