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The normal distribution is given as

`N(x;a,b)=(1)/(\sqrt(2\pi b))e^(-((x-a)^(2))/(2b^(2)))`

Calculate the expectation value

`\mu =\int_(-\infty )^(\infty ) xN(x;a,b)dx`

and the variance

`\sigma ^(2)=\int_(-\infty )^(\infty ) (x-\mu )^(2)N(x;a,b)dx`