Mean (\(\bar { x } \)) =\(A+\frac { \sum { fu } }{ N } =64+\frac { 0 }{ 100 } =64\)
\(S.D.(\sigma )=\frac { 1 }{ 100 } \sqrt { N\sum { { fu }^{ 2 }-{ (\sum { fu) } }^{ 2 } } } \)
\(=\frac { 1 }{ 100 } \sqrt { 100\times 286-{ (0) }^{ 2 } } \)
\(=\frac { 1 }{ 100 } \sqrt { 28600 } =\frac { 1 }{ 100 } \times 169.1=1.69.\)

Mean\(\left( \overline { x } \right) =\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 80 } \times 4000=50\)
Mean deviation about mean=\(\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-\overline { x } \right| } \)
\(=\frac{1}{80}\times1280=16\)

Mean of the given data is
\(\overline { x } =\frac { 38+70+48+40+42+55+63+46+54+44 }{ 10 } \)
\(=\frac { 500 }{ 10 } =50\)

M.D.aboutmean =\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-\overline { x } \right| } \)
\(=\frac{1}{10}\times84=8.4\)

Mean of the given data is
\(\overline { x } =\frac { 4+7+8+9+10+12+13+17 }{ 8 } =\frac { 80 }{ 8 } =10\)

M.D.about mean =\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-\overline { x } \right| } =\frac { 1 }{ 8 } \times 24=3\)

Let us make the following table from the given data.

Here, N=44
Now,\(\frac { N }{ 2 } =\frac { 44 }{ 2 } =22\)  , which lies in the cumulative frequency of 30, therefore median class is 20-30.
Thus,   l = 20, f = 16, cf =14 and h = 10
Hence, \(Median\left( M \right) =l+\frac { \frac { N }{ 2 } -cf }{ f } \times h\)  
\(​​=20+\frac { 22-14 }{ 16 } \times 10=20+\frac { 8 }{ 16 } \times 10\)
\(=30+\frac { 80 }{ 16 } =20+5=25\) 
and mean deviation about median
\(\frac { \sum _{ i=1 }^{ 6 }{ { f }_{ i }|{ x }_{ i }-M| } }{ \sum _{ i=1 }^{ 6 }{ { f }_{ i } } } =\frac { 420 }{ 44 } =9.55\).

For Firm A
Number of wages earners = 586
Mean of monthly wages,\(\overline { x } \) = Rs. 5253
Amount paid by firm A = Rs \(\left( 586\times 5253 \right) \) = Rs.3078258
Variance of distribution of wages = 100
\(\therefore \) Standard deviation, \(\sigma \) = \(\sqrt { variance } =\sqrt { 100 } =10\) 
Coefficient of variance \(=\frac { \sigma }{ x } \times 100\)  
\(=\frac { 10 }{ 5253 } \times 100=0.19\).
For Firm B
Number of wages earners = 648
Mean of monthly wages,\(\overline { x } \) = Rs. 5253
Amount paid by firm B =Rs.\(\left( 648\times 5253 \right) \) = Rs.3403944
Standard deviation \(=\sqrt { 121 } =11\)  
\(\therefore \)  Coefficient of variance\(=\frac { \sigma }{ x } \times 100\)\(=\frac { 10 }{ 5253 } \times 100=0.21\).
Value of motivating employees by giving financial incentives is addressed by the firm.

Let us first make the data continuous by making classes as 32.5-36.5,36.5-40.5,40.5-44.5,44.5-48.5,48.5-52.5.
Let assumed mean is 42.5 and h=4.
Then, \({ u }_{ i }=\frac { { x }_{ i }-a }{ h } =\frac { { x }_{ i }-42.5 }{ 4 } \)
Now, let us make the following table from the given data.

Now,Mean \({ \left( \overline { x } \right) }=a+\frac { \sum { { f }_{ i } } { u }_{ i } }{ \sum { { f }_{ i } } } \times h=42.5+4\times \frac { 25 }{ 100 } \)
[\(\because\)  by using step derivation method]
=42.5+1 = 43.5
Standard deviation,,\((\sigma )=\sqrt { { h }^{ 2 }\left[ \frac { 1 }{ N } \sum { { f }_{ i }{ u }_{ i }^{ 2 } } -\left( \frac { 1 }{ N } \sum { { f }_{ i }{ u }_{ i } } \right) ^{ 2 } \right] } \)
[by shortcut method]
\(=\sqrt { { (4) }^{ 2 }\left[ \frac { 1 }{ 100 } \times 199-{ \left( \frac { 1 }{ 100 } \times 25 \right) }^{ 2 } \right] } \\ =\sqrt { 16\left( 1.99-\frac { 1 }{ 16 } \right) } =\sqrt { 16\left( \frac { 31.84-1 }{ 16 } \right) } \\ =\sqrt { 30.84 } =5.55\)
Hence standard deviation and mean diameter of circles are 5.55 and 43.5

Let the assumed mean be a=98
Now, let us make the following table from the given data.

Here, N=\(\sum { { f }_{ i } } =22,\sum { { f }_{ i } } { d }_{ i }=44\quad and\quad \sum { { f }_{ i }{ d }_{ i }^{ 2 } } =728\)
\(Now,\ \overline { x } =a+\frac { \sum { { f }_{ i } } { d }_{ i } }{ \sum { { f }_{ i } } }\)
\(=98+\frac { 44 }{ 22 } =98+2=100\)
\(Variance,{ \sigma }^{ 2 }=\frac { 1 }{ N } (\sum { { f }_{ i } } { d }_{ i }^{ 2 })-\left( \frac { \sum { { f }_{ i }{ d }_{ i } } }{ N } \right) ^{ 2 }\)
\(=\frac { 1 }{ 22 } (728)-\left( \frac { 1 }{ 22 } \times 44 \right) ^{ 2 }=\frac { 1 }{ 22 } (728)-{ (2) }^{ 2 }\)
\(=\frac { 1 }{ 22 } (728-22\times { 2 }^{ 2 })=\frac { 1 }{ 22 } (728-88)\)
\(=\frac { 1 }{ 22 } \times 640=\frac { 320 }{ 11 } =29.09\)
\(and\ standard \ deviation, \sigma =\sqrt { 29.09 } =5.394\)

Here number of observations n = 6
By the given condition, \(\frac { 6+8+5+7+a+4 }{ 6 } =7\)
\(\Rightarrow 30+a=6\times 7 \Rightarrow a=42-30=12\)
Now, on arranging the observations in ascending order, we have 4,5,6,7,8,12
Now Median,
\(M=\frac { \left( \frac { n }{ 2 } \right) th\quad observation\quad +\left( \frac { n }{ 2 } +1 \right) th\quad observation\quad }{ 2 } \)
\(=\frac { \left( \frac { 6 }{ 2 } \right) th\quad observation\quad +\left( \frac { 6 }{ 2 } +1 \right) th\quad observation\quad }{ 2 }\)
\( \\ =\frac { 3rd\quad observation\quad +4th\quad observation\quad }{ 2 } \)
\(=\frac { 6+7 }{ 2 } =\frac { 13 }{ 2 } =6.5\)
Let us make the table for deviation and absolute deviation.

\(\therefore \) Mean deviation about median \(=\frac { \sum _{ i-1 }^{ 6 }{ \left| { x }_{ i }-M \right| } }{ 6 } =\frac { 12 }{ 6 } =2\)

Given, n = 100,
\(\bar { x } =20,\sigma =3\)
\(\therefore \ Mean,\bar { x } =20\)
\(\therefore \frac { \sum { { x }_{ i } } }{ 100 } =20\)
\(\Rightarrow \sum { { x }_{ i } } =100\times 20\Rightarrow \sum { { x }_{ i } } =2000\)
Now, as incorrect observations 21,21 and 18 are omitted, then correct sum is
\(\sum { { x }_{ i } } =2000-21-21-18=2000-60=1940\)
Therefore, correct mean of remaining 97 observations is
\(\bar { x } =\frac { 1940 }{ 97 } =20\)
\(\sigma =3\)
\(\sqrt { \sum { { x }_{ i }^{ 2 } } -(\bar { x } { ) }^{ 2 } } =3\)
On squaring both sides, we get
\(\Rightarrow \sum { { x }_{ i }^{ 2 } } =40900-441-441-324\)
\(=40900-441-441-324\)
\(=40900-1206=39694\)
Moe correct standard deviation for remaining 97 observations is
\(\sigma =\sqrt { \cfrac { 39694 }{ 97 } -(20{ ) }^{ 2 } }\)
\( =\sqrt { 409.2-(20{ ) }^{ 2 } } =\sqrt { 409.2-400= } \sqrt { 9.2 } =3.03\)

Let \({ \sigma }_{ 1 }\) be the given standard deviation and \({ \sigma }_{ 2 }^{ 2 }\) be the required variance. Then ,We have \({ \sigma }_{ 1 }\) =5
We know that,\(c=\frac { 5 }{ 9 } (F-32)\)Where C and F represent temperature in degree Celsius and degree Fahrenheit respectively. Thus, we have \(F=\frac { 9 }{ 5 } c+32\)
So, the observation of data in \(^{0}F\) Can be obtained by multiplying each observation of data in  \(^{0}C\) by \(\frac { 9 }{ 5 } \) and by adding 32 in it. Hence\({ \sigma }_{ 2 }^{ 2 }=(\frac { 9 }{ 5 } { ) }^{ 2 }{ \sigma }_{ 1 }^{ 2 }\)
[ variance is dependent on change of scale, not on change of origin]
\(=(\frac { 9 }{ 5 } { ) }^{ 2 }\times { 5 }^{ 2 }=81\)

Let the observation be\({ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },{ x }_{ 4, }{ x }_{ 5 },{ x }_{ 6 }\)
Then, their mean,
\(\bar { x } =\frac { \sum _{ i=1 }^{ 6 }{ { x }_{ i } } }{ 6 } =8\)
\(\Rightarrow \sum _{ i=1 }^{ 6 }{ { x }_{ i } } =8\times 6=48\)
On multiplying each observation by 3,
we get the new observation as \({ 3x }_{ 1 },{ 3x }_{ 2 },{ 3x }_{ 3 },{ 3x }_{ 4 },\text{ and}{ 3x }_{ 6 }.\)
Now, their new mean,
\(\bar { x } =\frac { \sum _{ i=1 }^{ 6 }{ 3{ x }_{ i } } }{ 6 } =\frac { 3\sum _{ i=1 }^{ 6 }{ 3{ x }_{ i } } }{ 6 }\)
\(=\frac { 3\times 48 }{ 6 } =24\ [from\quad Eq\quad (i)]\)
and variance of new observation
\(=\frac { \sum _{ i=1 }^{ 6 }{ (3{ x }_{ i }-24{ ) }^{ 2 } } }{ 6 } =\frac { { 3 }^{ 2 }\sum _{ i=1 }^{ 6 }{ { (x }_{ i }-8{ ) }^{ 2 } } }{ 6 } \)
\(=\frac { 9 }{ 1 } \times \text{ Variance of old observation}\)
\(=\frac { 9 }{ 1 } \times ({ 4) }^{ 2 }=144 \quad [\because \text{SD of old observation is 4}]\)
\(\text{thus, SD of new observation}=\sqrt { variance=\sqrt { 144 } } =12\)

Given observations are 38,70,48,40,42,55,63,46,54,44
Here, number of observations, n = 10
\(\therefore \)  Mean,
\(\overline { x } =\frac { \left( 38+70+48+40+42+55+63+46+54+44 \right) }{ 10 } =\frac { 500 }{ 10 } =50\)
Let us make the table for deviation and absolute deviation

Now, \(MD=\frac { \sum _{ i-1 }^{ 10 }{ \left| { x }_{ i }-\overline { x } \right| } }{ 10 } =\frac { 84 }{ 10 } =8.4\)

We have ,mean
\(\bar { (x) } =\frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+...+{ x }_{ n } }{ n } \)
Now, mean of \( a{ x }_{ 1 },a{ x }_{ 2 },...,a{ x }_{ n }\)
\(=\frac { { ax }_{ 1 }+{ ax }_{ 2 }+...+a{ x }_{ n } }{ n } \)
\(=\frac { a({ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+...+{ x }_{ n } }{ n }\)
\(=a\bar { x } \quad [using\quad eq.(i)]\)
Also, we have variance.
\({ \sigma }^{ 2 }=\frac { \sum { ({ x }_{ i }-\bar { x } { ) }^{ 2 } } }{ n } ....(ii)\) 
\(\therefore Variance\ of\ a{ x }_{ 1 },a{ x }_{ 2 },a{ x }_{ 3 },...,a{ x }_{ n }=\frac { \sum { (a{ x }_{ 1 }-a\bar { x } { ) }^{ 2 } } }{ n } \)
\(=\frac { { a }^{ 2 }({ x }_{ 1 }-\bar { x } { ) }^{ 2 }+{ a }^{ 2 }({ x }_{ 2 }-\bar { x } { ) }^{ 2 }+...+{ a }^{ 2 }({ x }_{ n }-\bar { x } { ) }^{ 2 } }{ n } \)

\(=\frac { { a }^{ 2 }\sum { ({ x }_{ 1 }-\bar { x } { ) }^{ 2 } } }{ n } ={ a }^{ 2 }{ \sigma }^{ 2 }\)

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