Arranging the data in ascending order, we have
5, 7, 10, 12, 14, 19, 23, 26.
Here, n = 8 , which is even.
Therefore, Median = \(\cfrac { 1 }{ 2 } \left\{ \left( \cfrac { 8 }{ 2 } \right) ^{ th }term+\left( \cfrac { 8 }{ 2 } +1 \right) ^{ th }term \right\} \) 

= \(\cfrac { 1 }{ 2 } \left\{ { 4 }^{ th }term+5^{ th }term \right\} \) 
= \(\cfrac { 1 }{ 2 } \left\{ 12+14 \right\} \) 
= \(\cfrac { 26 }{ 2 } =13\) 
Therefore, the Median is 13.

As the number of observations is even, there are two middle values.
The average of those values must be 16.
We will now find any pair of numbers whose average is 16. Say 14 and 18.
Now an example of data with median 16 can be 2, 4, 7, 9, 12, 14, 18, 24, 28, 30, 45, 62.

Arranging the golf scores in ascending order, we have,
65, 68, 70, 73, 75, 78, 79
Here n = 7 , which is odd.
Therefore, Median = \(\left( \cfrac { n+1 }{ 2 } \right) ^{ th }term\) 
= \(\left( \cfrac { 7+1 }{ 2 } \right) ^{ th }term\) 
= \(\left( \cfrac { 8 }{ 2 } \right) ^{ th }term\) 
4^th term = 73
Hence, the Median is 73.

\(Mean=\cfrac { Total\quad ages\quad of\quad teachers }{ Number\quad of\quad teachers } \) 
\(42=\cfrac { 35+42+48+X+(X+8)+40+43+50+46+50+37+32+38+41+40 }{ 15 } \) 
\(\cfrac { 550+2X }{ 15 } =42\) 
550 + 2X = 42 X 15
= 630
2X = 630 - 550
2X = 80
\(X=\cfrac { 80 }{ 2 } \) 
X = 40
Therefore, the age of the teacher (X) is 40 and the age of the another teacher (X + 8) is 40 + 8 = 48.

\(Arthemetic\quad Mean=\cfrac { sum\quad of\quad observation }{ Number\quad of\quad observation } \) 
= \(\cfrac { 250+350+100+400+150+270+450+320+610+750 }{ 10 } \) 
= \(\cfrac { 3650 }{ 10 } =365\) 
Hence, the average income of the worker is Rs.365.