Chevbyshev's theorem

[renatal1972]

I am in the middle of a alkes statistics class and am pulling my hair out because I must have missed something in algebra. The problem I am having is trying to figure out how

[100(1-1k^2)]% = 36% I have got all of the steps to get to the anser then I get to the last step ~~~~ k^2= 1/0.64and can not figure out how they are coming up with k= + or - 1.25

I have tried several diffrent variations, use excel and a sintific calculator. I am not comming up with the same answer. Can you enliten me.

 

[royhaas]

Since \(\displaystyle k^2 = 1/.64\), you have \(\displaystyle |k| = 1/\sqrt{.64} = 1/.8\).