# what is difference between $95 \%$ CI of mean and 95% pdf of normal distribution?

We took sample mean $$\mu = 14$$, and $$\sigma = 0.45$$.

Calculate required area of normal distribution

We will apply $$\int_{13.19}^{15} f(x) \ dx = 95\%$$.

95 % CI of mean

But if we took sample size $$n = 9$$ and $$z_{95\%} = 1.96$$, hence

$$95 \%$$ CI = $$14 \pm 1.96(\frac{0.45}{\sqrt{9}})$$ = $$lower \ limit\ (13.7) \ and \ upper \ limit\ (14.2940)$$.

My question, If we apply these lower and upper limit in pdf function, then $$\int_{13.7}^{14.2940} f(x) \ dx =49 \%$$.

We are excited to know, how we can make difference $$95 \%$$ area of pdf function and $$95 \%$$ CI of mean.

• What do you mean by CI of mean when the population mean $\mu$ is known? Or do you mean the sample mean is 14? – StubbornAtom May 27 at 12:49
• Yes, the sample mean is $14$ – dtc348 May 27 at 13:20

You have a normal distribution, $$N(\mu,\sigma^2)$$, where $$\mu$$ is unknown and $$\sigma = 0.45$$. The central 95% of the distribution’s area is between $$\mu – 1.96\sigma$$ and $$\mu + 1.96\sigma$$. You took $$N = 9$$ iid samples from the distribution and computed their sample mean, $$m$$, as $$14$$. The sample mean is a sample from the sampling distribution of the mean, i.e., a sample from $$N(\mu,\sigma^2/N)$$, where $$N = 9$$. Thus the central 95% confidence interval for $$\mu$$ is $$m ± 1.96\sigma/√9$$. This is $$14 ± 1.96(0.45/3)$$, i.e., from 13.706 to 14.294. So you have two distributions: $$N(\mu,\sigma^2)$$ (from which you drew the $$N = 9$$ iid samples) and $$N(\mu,\sigma^2/N)$$ (from which m is a sample).
The interval $$\mu ± 1.96\sigma/√9$$, is only 48.6% of the area of $$N(\mu,\sigma^2)$$, but this is irrelevant: the interval applies to $$N(\mu,\sigma^2/9)$$.