\(Mean(\bar { x } )=A+\frac { \sum { fu } }{ N } \times h=25+\frac { 10 }{ 50 } \times 10=25+2=27\)
Variance \((\sigma ^{ 2 })\) = \(\frac { { h }^{ 2 } }{ { N }^{ 2 } } [N\sum { { fu }^{ 2 }-{ (\sum { fu) } }^{ 2 } } ]\)
\(=\frac { { (10) }^{ 2 } }{ { (50) }^{ 2 } } [50\times 68-{ (10) }^{ 2 }]=\frac { 100 }{ 2500 } [3400-100]=\frac { 1 }{ 25 } \times 3300=132\)

\(Mean(\bar { x } )=A+\frac { \sum { fu } }{ N } \times h=105+\frac { 2 }{ 30 } \times 30=107\)
\(Variance({ \sigma }^{ 2 })=\frac { { h }^{ 2 } }{ { N }^{ 2 } } [N\sum { { fu }^{ 2 }-(\sum { fu)^{ 2 } } ] } \)

Here x = 1, 2, 3, 4, ..., n
∴ Σx=1+2+3+4+ ... +n
=\(\frac{n(n+1)}{2}\)
\(\therefore \ Mean\left( \overline { x } \right) =\frac { n(n+1) }{ 2n } =\frac { (n+1) }{ 2 } \)
Σx² = (1)²+ (2)²+ (3)²+ (4)²+... + n²
Variance \({ \sigma }^{ 2 }=\frac { N\Sigma { x }^{ 2 }-\left( \Sigma x \right) ^{ 2 } }{ { N }^{ 2 } } \)
\(=\frac { n\times \frac { n(n+1)(2n+1) }{ 6 } -{ \left[ \frac { n(n+1) }{ 2 } \right] }^{ 2 } }{ { n }^{ 2 } } \)
\(=\frac { { n }^{ 2 }\left[ \frac { (n+1)(2n+1) }{ 6 } -\frac { { (n+1) }^{ 2 } }{ 4 } \right] }{ { n }^{ 2 } } \)
\(=\frac { 4n^{ 2 }+6n+2-3n^{ 2 }-6n-3 }{ 12 } =\frac { n^{ 2 }-1 }{ 12 } \) 

Here x = 6, 7, 10, 12, 13,4,8, 12
∴ Σx = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72
n = 8 \(\therefore \ \overline { x } =\frac { 72 }{ 8 } =9\)
Σx²= (6)²+ (7)²+ (10)²+ (12)²+ (13)² + (4)²+ (8)²+ (12)²
= 36 + 49 + 100 + 144 + 169
+ 16 + 64 + 144
= 722
∴  Variance =\({ \sigma }^{ 2 }=\frac { N\Sigma { x }^{ 2 }-\left( \Sigma x \right) ^{ 2 } }{ { N }^{ 2 } } \)
\(=\frac { 8\times 722-(72)^{ 2 } }{ { (8) }^{ 2 } } \)
\(=\frac { 5776-5184 }{ 64 } =\frac { 592 }{ 64 } =9.25\)

Let the remaining two observations be X and y.
\(Given,\bar { x } =8,{ x }_{ 1 }=2,{ x }_{ 2 }=4,{ x }_{ 3 }=10,{ x }_{ 4 }=12\ and\ { x }_{ 5 }=14.\)
\(\Rightarrow \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 }+{ x }_{ 5 }+{ x }+y }{ 7 } =8\)
\(\Rightarrow \frac { 2+4+10+12+14+x+y }{ 7 } =8\)
\(\Rightarrow 42+x+y=56\)
\(\Rightarrow x+y=14\)
\(Also,\ variance=16\Rightarrow { \sigma }^{ 2 }=16\)
\(\Rightarrow \frac { { x }_{ 1 }^{ 2 }+{ x }_{ 2 }^{ 2 }+{ x }_{ 3 }^{ 2 }+{ x }_{ 4 }^{ 2 }+{ x }_{ 5 }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 } }{ 7 } -(\bar { x } { ) }^{ 2 }=16\)
\(\Rightarrow\ { x }^{ 2 }+{ y }^{ 2 }=100\)
\(From\ Eq.(i),\quad Y=14-x\)
\(On\ putting\ this\ value\ of\ y\ in\ Eq.\quad (ii),\ we\ get\)
\( { x }^{ 2 }+(14-{ x) }^{ 2 }\ =100\)
\(\Rightarrow { x }^{ 2 }+196+{ x }^{ 2 }-28x=100\)
\(\Rightarrow 2{ x }^{ 2 }-28x+96=0\)
\(\Rightarrow { x }^{ 2 }-14x+48=0\) [divide both sides by 2]
\(\Rightarrow (x-6)(x-8)=0\Rightarrow x=6,8\)
\(If \ x=8,\ then\ y=14-8=6\)
Hence,the remaining two observations are 6 and 8

The given data can be arranged in ascending order as 30,34,38,40,42,44,50,51,60,66.
Here, total number of observations are 10 i.e. n = 10, which is even
\(\therefore \) Median
\(M=\frac { \left( \frac { n }{ 2 } \right) th\quad observation\quad +\left( \frac { n }{ 2 } +1 \right) th\quad observation\quad }{ 2 } \)
\(=\frac { \left( \frac { 10 }{ 2 } \right) th\quad observation\quad +\left( \frac { 10 }{ 2 } +1 \right) th\quad observation\quad }{ 2 } \)
\(=\frac { 5th\quad observation\quad +6th\quad observation\quad }{ 2 } \)
\(=\frac { 42+44 }{ 2 } =\frac { 86 }{ 2 } =43\)
Let us make the table for absolute deviation

Now, mean deviation about the median.
\(MD=\frac { \sum _{ i-1 }^{ 10 }{ \left| { x }_{ i }-M \right| } }{ 10 } =\frac { 87 }{ 10 } =8.7\)

The first n natural numbers are 1,2,3...,n.
\(\because\) Standard deviation,
\(SD=\sqrt { \frac { \sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } }{ n } -({ \frac { \sum _{ i=1 }^{ n }{ { x }_{ i } } }{ n } ) }^{ 2 } } \)
\(\therefore \ SD=\sqrt { \frac { (n(n+1)(2n+1) }{ 6n } -({ \frac { n(n+1) }{ 2n } ) }^{ 2 } } \)
\([\because \ \sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } =\frac { n(n+1)(2n+1) }{ 6 } and\ \sum _{ i=1 }^{ n }{ { x }_{ i } } =\frac { n(n+1) }{ 2 } ]\)
\(=\sqrt { (n+1)(\frac { 2n+1 }{ 6 } -\frac { n+1 }{ 4 } ) } \)
\(=\sqrt { (n+1)(\frac { 4n+2-3n-3 }{ 12 } ) } \)
\(=\sqrt { \frac { (n+1)(n-1) }{ 12 } } =\sqrt { \frac { { n }^{ 2 }-1 }{ 12 } } \)
\(\therefore \ Variance={ (SD) }^{ 2 }=\frac { { n }^{ 2 }-1 }{ 12 } \)

We have known that, if any constant is added in each observation, then standard deviation remains same.
So, the standard deviation of the observations a + k,d + k,c + k,d + k,e + k is S

Given observations are 6,7,10,12,13,4,8,12
Number of observations =8
\(\therefore \ Mean(\overline { x } )=\frac { 6+7+10+12+13+4+8+12 }{ 8 } \)
\(=\frac { 72 }{ 8 } =9\)
Now, let us make the following table for deviation.

\(\therefore \) Sum of squares of deviations =\(\sum _{ i=1 }^{ 8 }{ { ({ x }_{ i }-\overline { x } ) }^{ 2 } } =74\)
Hence, variance, \({ \sigma }^{ 2 }=\frac { \sum _{ i=1 }^{ 8 }{ { ({ x }_{ i }-\overline { x } ) }^{ 2 } } }{ n } =\frac { 74 }{ 8 } \)=9.25
and standard deviation =\(\sqrt { \sigma } =\sqrt { 9.25 } \)
=3.04

Mean \((\overline { x } )=\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 22 } \times 2200=100\)
Variance = \({ \sigma }^{ 2 }=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }{ \left( { x }_{ i }-\overline { x } \right) }^{ 2 } } \)
\(=\frac { 1 }{ 22 } \times 640=29.09\)

\(\frac{N}{2}=\frac{50}{2}=25\)∴ Median class is 20 - 30
∴ Median=\(20+\frac { 25-14 }{ 14 } \times 10=20+7.86=27.86\)
M.D. about median =\(\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-M \right| } =\frac { 1 }{ 50 } \times 517.16=10.34\)

Mean\(\overline { x } =\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 50 } \times 17900=358\)
Mean deviation about mean\(=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-\overline { x } \right| } \)
\(=\frac{1}{50}\times7896 = 157.92\) 

Arrange the data in ascending order, we have
36,42,45,46,46,49,51,53,60,72
Here n = 10 (which is even)
So median is average of 5th and 6th observation
∴ Median=\(\frac{46+49}{2}=\frac{95}{2}=47.5\)

M.D. about median=\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-M \right| } \)
\(=\frac{1}{10}\times70=7\)

Arrange the data in ascending order, we have
10,11,12,13,13,14,16,16,17,17,18
Here n = 12 (which is even)
So median is average of 6th and 7th observations
∴ Median =\(\frac { 13+14 }{ 2 } =\frac { 27 }{ 2 } =13.5\)

M.D. about median =\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-M \right| } \)
\(=\frac{1}{12}\times28=2.33\)