Question

Let $0<\alpha <1$. Show that for a regularly distributed variable $100(1-\alpha )\%$of all possible observations lie within $z_{\alpha /2}$standard deviations to either side of the mean, that is, between $\mu -z_{\mathrm{F}/2}·\sigma$and $\mu +z_{\sigma /2}·\sigma$.

Short answer

Proved that $100(1-\alpha )\%$of all possible observations lie between $\mu -z_{\alpha /2}·\sigma$and $\mu +z_{\alpha /2}·\sigma$.

Step-by-step solution

Given Information

Normal distribution, $0<\alpha <1$.

Explanation

$a$= the possibility. In a confidence interval, the population parameter will be left out.; $1-a$= the probability of the population parameter being included in the interval. With probability, the true value of a population parameter. $1-a$will be included in the $100(1-a)$percent confidence range.

Consider $p\left(\left|z\right|\le z_{\alpha /2}\right)=1-\alpha$

$p\left(-z_{\alpha /2}\le z\le z_{\alpha /2}\right)=1-\alpha$

$p\left(-z_{\alpha /2}\le \frac{x-\mu }{\sigma }\le z_{\alpha /2}\right)=1-\alpha \left(\because z=\frac{x-\mu }{\sigma }\right)$

$p\left(\mu -z_{\alpha /2}·\sigma \le x\le \mu +z_{\alpha /2}·\sigma \right)=1-\alpha$

So, that $100(1-\alpha )\%$of all possible observations lie between $\mu -z_{\alpha /2}·\sigma$and $\mu +z_{\alpha /2}·\sigma$.

Hence, proved.