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).


1 Answer 1


It sounds like the statement you are quoting is not about confidence intervals but rather about the probability content within k standard deviations of the mean for k=1, 2, 3. This result plays a role in constructing confidence intervals for the mean of a normal sample though, because if Xi iid N(μ,σ) the sample mean Xb is N(μ,σ/√n) So for known σ [Xb-2σ/√n,Xb+2σ/√n] is approximately a 95% confidence interval for μ (actually 95.4%). When σ is unknown replacing σ with the sample standard deviation s will still give an approximate 95% confidence interval for μ when n is large. For small n the exact distribution to construct the confidence interval is student t with n-1 degrees of freedom. So 2 should be replaced by the (larger) appropriate percentile from the t distribution to get the 95% confidence.

  • $\begingroup$ But they haven't included √n so it's like [Xb-2σ,Xb+2σ] for the confidence interval of 95% isn't it. I mean it's not [Xb-2σ/√n,Xb+2σ/√n] but [Xb-2σ,Xb+2σ] as given in en.wikipedia.org/wiki/Normal_distribution. Also when you say confidence interval for μ (actually 95.4%), does it mean that the true population parameter mean is within +-2σ of the calculated sample mean or is it that when a draw a sample from the distribution there is 95% chance that it is from sample mean +- 2σ. This is where I am confused $\endgroup$
    – user31820
    Commented Jun 6, 2012 at 14:18
  • $\begingroup$ 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 $\endgroup$ Commented Jun 6, 2012 at 15:43
  • $\begingroup$ 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$^2$. $\endgroup$ Commented Jun 6, 2012 at 15:44
  • $\begingroup$ 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: $\endgroup$ Commented Jun 6, 2012 at 15:44
  • $\begingroup$ μ^ ~ N(μ,σ$^2$/n) $\endgroup$ Commented Jun 6, 2012 at 15:52

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