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If the standard deviation of a normally distributed population is known to be 15, what size sample must be taken if 95% of the sample means are to differ from the population mean by less than 1?

I believe I can use $$\mathbb{P}\left(|\bar{X}-\mu|<\frac{2\sigma}{\sqrt{n}}=0.9544\right)$$ and I used $\frac{2\times 15}{\sqrt{n}}<1$ which yields $n=900$.

The correct answer is 865. Could anyone show me the correct way to solve this problem.

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    $\begingroup$ Just want to add that this answer (including the one from @Don Walpola below) is only an estimate. It is entire possible to collect that many samples and still not achieve the desired precision (i.e. need more). $\endgroup$ – SecretAgentMan Sep 7 '18 at 16:23
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Assuming your data are normally distributed, the formula you want to use is $n = \Big(\frac{z_{\alpha /2}\ \cdot \ \hat\sigma}{D} \Big)^{2}$, where $\alpha$ = $1 - \text{Confidence Level}$, $\hat\sigma$ is the estimate of the standard deviation, and $D$ is the desired level of accuracy for the estimate, i.e. the distance to the population mean. For your problem, $\alpha = 1 - 0.95 = 0.05$, $\hat\sigma = 15$, and $D = 1$. Then, after looking up the z-score for the $\alpha/2$ value of the probability in a z-table, will find that the z-score is $1.96$. Plugging these values into the formula will give you the desired result, after rounding up to the nearest whole number as fractional sample sizes are nonsensical.

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  • $\begingroup$ Thanks. I forgot about the error can be regarded as the standard error we talk about... $\endgroup$ – math101 Sep 8 '18 at 0:34

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