If number of observations is even, then the median will be the average of \((\frac{N}2)^{th}\) and the \((\frac{N}2+1)^{th}\) observations.
In the given case, = 10 ⇒ \((\frac{N}2)^{th}\) = 5th and \((\frac{N}2+1)^{th}\) = 6th observation
Thus, 63 = \(\frac{x+(x+2)}2\)
⇒126 = 2x + 2
⇒ 124 = 2x
⇒ x = 62
Thus, the value of x is 62.