# finding sample size with known $\sigma$

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.

• 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). Sep 7, 2018 at 16:23

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.