Let 0<α<1. Show that for a regularly distributed variable 100(1-α)%of all possible observations lie within zα/2standard deviations to either side of the mean, that is, between μ-zF/2·σand μ+zσ/2·σ.

Short Answer

Expert verified

Proved that 100(1-α)%of all possible observations lie between μ-zα/2·σand μ+zα/2·σ.

Step by step solution

01

Given Information

Normal distribution, 0<α<1.

02

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-awill be included in the 100(1-a)percent confidence range.

Consider p|z|zα/2=1-α

p-zα/2zzα/2=1-α

p-zα/2x-μσzα/2=1-αz=x-μσ

pμ-zα/2·σxμ+zα/2·σ=1-α

So, that 100(1-α)%of all possible observations lie between μ-zα/2·σand μ+zα/2·σ.

Hence, proved.

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