Ascending order -8, -3, -2, -1, 0, 7, 14
n = 8,(even)
\(\text { Median }=\frac{1}{2}\left\{\left(\frac{\mathrm{n}}{2}\right)^{\text {th }} \text { term }+\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }} \text { term }\right\}\)
\(=\frac{1}{2}\left\{\frac{8}{2}^{\text {th }} \text { term }+\left(\frac{8}{2}+1\right)^{\text {th }} \text { term }\right\} \)
\(=\frac{1}{2}\left\{4^{\text {th }} \text { term }+5^{\text {th }} \text { term }\right\} \)
\(=\frac{1}{2}\{-1+0\}=-\frac{1}{2} \)
Median = \(-\frac{1}{2}\)

\(\text { Mean marks } =\frac{\text { Sum of marks }}{\text { Number of students }} \)
\(8 =\frac{\text { Sum of marks }}{6} \)
Sum of marks = 8 x 6 = 48
Let x be the marks of a student to be added and the new mean is 8.
\(\text { New mean }=\frac{\text { Sum of marks }+\text { Mark of new student }}{\text { Number of students }} \)
\(8=\frac{48+\mathrm{x}}{7} \)
48 + x = 8 x 7
x = 56 - 48
Mark of one more student x = 8

(i) Mode
Arranging the scores in the ascending order
1, 1,2,2,2,3, 3, 4, 4, 6, 6
Here, 2 occurs maximum times
Mode = 2
(ii) Median
n = 11 (odd)
The median = \(\left( \cfrac { n+1 }{ 2 } \right) ^{ th }term\) 
= \(\left( \cfrac { 11+1 }{ 2 } \right) ^{ th }term\) 
= 6^th term
Median = 3
Hence Mode = 2, Median = 3.

Ascending order 1, 1,2,2,2,3,3,3,5,5,5
Here
1 occurs 2 times
2 occurs 3 times
3 occurs 3 times
5 occurs 3 times
Here 2, 3 and 5 occurs 3 times, maximum frequency
Mode = 2,3 and 5

 \(The\ mean\ of\ runs\ scored=\cfrac { Total\ run s\ scoredinlast\ 10\ innings }{ Number\ of\ innings } \) 
\(276=\cfrac { 235+400+351+x+100+315+410+260+284 }{ 10 } \) 
\(276=\cfrac { 2520+x }{ 10 } \) 
276 x 10 = 2520 + x
x = 2760 - 2520 = 240
The runs scored in the fourth innings = 240