# In practice, is there a $n$-value that we use to apply CLT?

For i.i.d. $$X_i$$ for $$i \in \mathbb N$$, denote $$\bar X_n := \frac{1}{n}\sum_{i=1}^n X_i$$. We know that $$\frac{\sqrt{n}\bar X_n}{\sigma}$$ converges (in distributional sense) to $$\mathcal N(0,1)$$, but we do not know how fast the convergence is.

In practice, is there a $$n$$-value that we use to apply CLT?

• note that for CLT to apply there are more conditions/assumptions that need to hold see (en.wikipedia.org/wiki/Central_limit_theorem) eg if $X_i$ are identically distributed you still need at least finite variance. Jan 17, 2023 at 19:50
• It depends on the distribution of the $X_i.$ You can learn much more about this by searching our site. Try stats.stackexchange.com/…
– whuber
Jan 17, 2023 at 19:53
• @James many questions and answers already on site relate to this question. With some population distributions, the distribution of sample means will be very close to normal at sample sizes like 3 or 5, and for other population distributions the distribution of sample means will still not be very close to normal at sample sizes like 3000 or 5 million. Some won't get there at any practicable sample size. Some will never get there. Unless you put some conditions on the distribution you sample from there's not really a relevant "n" you can rely on. Jan 17, 2023 at 21:34
• You missed $\mu$: it is $\frac{\sqrt{n}(\bar X_n- \mu)}{\sigma}$ which converges in distribution to a standard normal. A simple example to consider is a Bernoulli random variable so $1$ with probability $p$ or $0$ with probability $1-p$. The approximation is certainly poor if $n < \frac1p$ and for some larger $n$, but $\frac1p$ can be as large as you like if you choose a small enough $p$ Jan 18, 2023 at 0:38