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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

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    Do you mean that it says n should be >30 to suitably estimate a distribution with the central limit theorem.2010-11-06

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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

$0.5 \sqrt{n} \ge 1.96.$

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    yay! This came somewhat handy in my paper today, which was SO awesum!! =)2010-11-08