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We have a measurement which has mean $\mu$ and variance $\sigma^2$ = 25. Let $\bar{X}$ be average of $\textit{n}$ such independent measurements.

How large should $\textit{n}$ be in so that $P(|\bar{X} - \mu| < 1) = 0.95$

This is what I got so far:

$P(\frac{\sqrt{n} * (-1)}{\sqrt{25}} < \frac{\bar{X} - \mu}{\sqrt{25}} < \frac{\sqrt{n} * (1)}{\sqrt{25}}) = 0.95$

Yielding

$\Phi(\frac{\sqrt{n}}{\sqrt{25}}) - \Phi(-\frac{\sqrt{n}}{\sqrt{25}}) = 0.95$

And this is when I get lost. How do I remove the $\Phi$ so to find $n$? I tried to find the z-value corresponding to 0.95, which is 1.65, but that did not help.

The correct result is $n=96$ (it is an exercise from Rice, third edition, problem 5/17).

HonzaB
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  • Hint: you can solve for $\Phi\left(\sqrt{n/25}\right).$ – whuber Nov 26 '18 at 15:15
  • @whuber that's exactly what I am confused about. I tried something like $\frac{\sqrt{n}}{5} = \Phi^{-1}(0.95)$, but that led to nowhere. Or I am doing something wrong. – HonzaB Nov 26 '18 at 15:32
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    Draw a picture. I just searched (for "self-study normal tail") and found the same question with a good picture for the answer. The sole difference is that $0.95$ is changed to $0.98$ in the duplicate. – whuber Nov 26 '18 at 15:33
  • (haven't seen you edited the comment) All right, that helped. Solving $\Phi(\sqrt{n/25}) = 0.975$ does it. Although I am not entirely sure why that works. Am I not getting the probabilty =0.975, even though the question asks for 0.95? – HonzaB Nov 26 '18 at 16:20
  • Why it works should be clear from the picture. That it works can be established by plugging your solution back into the original equation and verifying that it holds. – whuber Nov 26 '18 at 17:35
  • Yes, the picture helped and it is clear now. Thank you very much for your time. – HonzaB Nov 26 '18 at 17:36

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