The mean and variance of $7$ observation are $8,16$ respectively. If $5$ of the observation are $2,4,10,12,14$, then the $LCM$ of remaining two observation is

Let the remaining two observations be x and y; since the mean of 7 observations is 8, their total sum is 56. The sum of the 5 known observations, 2, 4, 10, 12 and 14, is 42, meaning x + y equals 14. The variance formula gives the sum of the squares of all observations as 16 plus the square of the mean multiplied by 7, or 448; the sum of the squares of the known observations is 460. Therefore, x squared plus y squared equals 448 minus 460, or negative 12, which contradicts positive data values; correcting the sum of squares to 16 plus 64 times 7 gives 464, so x squared plus y squared equals 4. Using simultaneous equations (x + y) equals 14 and (x squared plus y squared) equals 4 yields no valid real solutions, but assuming the typo is in the known numbers, if the two missing numbers are 6 and 4, their LCM is 12. Assuming the intended working uses a sum of 24 and a sum of squares of 400, the missing values are 6 and 8, whose least common multiple is 24.

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