Arranging the distribution in ascending order, we get
1, 1, 4, 4, 6, 8, 10, 12
Here, n = 8 (even).
So, their will be two median terms
i.e. (\(\frac { n }{ 2 }\))th, (\(\frac { 1 }{ 2 }\) + 1)th
Here, \(\frac { n }{ 2 }\) = \(\frac { 8 }{ 2 }\) = 4
Median = \(\frac { 4^{th} \;term + 5^{th} \;term }{ 2 }\)
= \(\frac { 4 + 6 }{ 2 }\) = 5