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sum(n=1)^18(xi-alpha)=36 , sum(n=1)^18(x...

`sum_(n=1)^18(x_i-alpha)=36 , sum_(n=1)^18(x_i-beta)^2=90` where `alpha and beta ` are distinct and the standard deviation of `x_i` is `1` then Find `abs(beta-alpha)`

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