I interpret the question as the variance of 12 Y values (Y value minus the mean of Y=314.375). The variance is therefore:
(Y1-mean)^2+(Y2-mean)^2+...+(Y12-mean)^2=11008, where Y1 to Y12 make up the statistical set. Divide this number by n, giving 11008/12, and take the square root to get the standard error (standard deviation): 30.2875. This means that the standard error on the mean is +30.2875 so:
284.088<Y<344.663 gives range of possible values for Y, where the mean sits in the exact middle of the range.