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Can I calculate the confidence level of the average depending on size of sample following a normal distribution?
Let's say I have a sample with 100 occurrences averaging to 0.87, can I say its average is 0.87 with a confidence of y%? If so, how to compute confidence level?
If you have a random sample of $n=100$ observations from a normal distribution with mean $\bar X = 0.87$ and you know the population standard deviation $\sigma,$ then a 95% confidence interval for the population mean $\mu$ is of the form $$\bar X \pm 1.96\sigma/\sqrt{n}$$ or in your case
$$0.87 \pm 1.96\sigma/10\;\;\text{or}\;\; 0.87 \pm 0.196\sigma.$$
The number $1.96$ cuts 2.5% of the probability from the upper tail of the standard normal distribution and $-1.96$ cuts 2.5% from the lower tail (leaving 95% in the 'middle' part of the distribution).
If you don't know the population standard deviation $\sigma,$ you can estimate it with the sample standard deviation $S = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2$ and use the confidence interval $$\bar X \pm t^*S/\sqrt{n},$$ where $t^*$ cuts 2.5% from the probability from the upper tail of Student's t distribution with $n-1$ degrees of freedom. Most statistics books have tables of the t distribution so you can find the appropriate value $t^*.$ For $n = 100,$ there are 99 degrees of freedom and $t^* = 1.98.$
Sometimes people use the approximation $$\bar X \pm 2S/\sqrt{n},$$ for $n > 30.$
For smaller values of $n,$ values of $t^*$ are larger than $2$.
Note: You need some kind of information about the variability of the population in order to make a confidence interval. So just with the information you give in your problem, you can't make a confidence interval.
