# confusion regarding confidence interval of normal distribution

I am a bit confused when it comes to 68-95-99.7 rule of confidence interval of normal distribution. Normally I could see confidence interval of 95% for a sample statistic (s) with margin of error lets say e. So the confidence interval is s +- e. So 95% means that if we do the sampling 100 times and calculate the sample statistic s 100 times then for each we will have different s+-e. However 95 times the true population parameter p lies within s+-e.

This is what I have understood about confidence interval. The confidence interval is s+-e and the confidence level is 95%.

Now when it comes to normal distribution, how the above idea fits in. I mean here confidence interval is mui+-sigma and the confidence level is 68%?. According to wikipedia it is something like this

About 68% of values drawn from a normal distribution are within one standard deviation σ away from the mean; about 95% of the values lie within two standard deviations; and about 99.7% are within three standard deviations

I am confused where is the sample statistic here, it says the values drawn(which is directly the random variable).

• Where do you see that in the Wikipedia article? The central limit theorem part normalizes by multiplying by √n. Central limit theorem Main article: Central limit theorem The theorem states that under certain (fairly common) conditions, the sum of a large number of random variables will have an approximately normal distribution. For example if (x1, …, xn) is a sequence of iid random variables, each having mean μ and variance σ$^2$, then the central limit theorem states that – Michael R. Chernick Jun 6 '12 at 15:43
• When the provide the distribution for the sample mean they have Estimation of parameters It is often the case that we don't know the parameters of the normal distribution, but instead want to estimate them. That is, having a sample (x1, …, xn) from a normal N(μ, σ$^2$) population we would like to learn the approximate values of parameters μ and σ$^2$. – Michael R. Chernick Jun 6 '12 at 15:44
• The standard approach to this problem is the maximum likelihood method, which requires maximization of the log-likelihood function: Taking derivatives with respect to μ and σ$^2$ and solving the resulting system of first order conditions yields the maximum likelihood estimates:Estimator is called the sample mean, since it is the arithmetic mean of all observations. The statistic is complete and sufficient for μ, and therefore by the Lehmann–Scheffé theorem, is the uniformly minimum variance unbiased (UMVU) estimator.[29] In finite samples it is distributed normally: – Michael R. Chernick Jun 6 '12 at 15:44
• μ^ ~ N(μ,σ$^2$/n) – Michael R. Chernick Jun 6 '12 at 15:52