To find the coefficient of variation we need standard deviation
\(\sigma =\sqrt{\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}} \)
\(\frac{\Sigma x^{2}}{n} =\frac{765}{5}=153 \)
\(\left(\frac{\Sigma x}{n}\right)^{2} =(\bar{x})^{2}=6^{2}=36 \)
\(\sigma =\sqrt{(153)-36}=\sqrt{117} \)
\(=\sqrt{3 \times 3 \times 13} \)
\(\sigma =3 \sqrt{13} \)
Coefficient of variation \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{3 \sqrt{13}}{6} \times 100 \%=\frac{\sqrt{13}}{2} \times 100 \%=\frac{3.60555}{2} \times 100 \% \)
\(=1.80277 \times 100 \%=180.277 \% \)
Coefficient of variation = 180.28 %

Standard deviation \(\sigma=6.5\)
Mean \(\bar{x}=12.5\)
Coefficient of variation C.V  \(=\frac{\sigma}{x} \times 100 \% \)
\(=\frac{6.5}{12.5} \times 100 \% \)
\(=\frac{65}{125} \times 100 \% \)
\(=\frac{13}{25} \times 100 \% \)
= 52 %
Co-efficient of variation is 52%

Standard deviation of first n natural numbers
\(=\sqrt{\frac{n^{2}-1}{12}}\)
SD of first 21 natural numbers
\( =\sqrt{\frac{21^{2}-1}{12}} \)
\(=\sqrt{\frac{441-1}{12}}=\sqrt{\frac{440}{12}} \)
\(=\sqrt{36.6666}=6.05\)
Standard deviation of first 21 natural numbers = 6.05

If the range = 36.8 and
the smallest value =13.4
Range R = L - S
36.8 = L-13.4
= 36.8 + 13.4 = 50.2
The largest value L = 50.2

Radius of conical flask = 'r' units
Height of conical flask = 'h' units
Volume of conical flask = Volume of water
\(=\frac{1}{3} \pi r^{2} h \text { cu. units }\)
Since, water is poured into the cylindrical flask
Volume of cylinder = Volume of water
\(\pi(\mathrm{xr})^{2} H=\frac{1}{3} \pi r^{2} h\)
[xr - radius of cylinder, H - height]
\(\mathrm{X}^{2} \mathrm{r}^{2} \mathrm{H}=\frac{r^{2}}{3} h\)
Height of the water in cylinder flask
\(\mathrm{H}=\frac{h}{3 x^{2}}\)

Largest value L = 67; Smallest value S =18
Range R = L = S = 67 - 18 = 49
Coefficient of range = \(\frac { L-S }{ L+S } \)
Coefficient of range = \(\frac { 67-18 }{ 67+18 } =\frac { 49 }{ 85 } \) = 0.576

Given circumference = 484 cm
\(2 \pi r =484 \)
\(2 \times \frac{22}{7} \times r =484 \)
\(r =\frac{484 \times 7}{44}=77 \mathrm{~cm} \)
height h = 105 cm
Volume of cone \(=\frac{1}{3} \pi r^{2} h \text { cu. units } \)
\(=\frac{1}{3} \times \frac{22}{7} \times 77 \times 77 \times 105 \)
= 652190 cm³

Mathematics:
Mean \(\bar{x}=56 ; \mathrm{SD} \sigma=12\)
Co-efficient of variation C. V  \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{12}{56} \times 100 \%=\frac{1200}{56} \%=21.43 \%\)
Science:
Mean \(\bar{x}=65 ; \mathrm{SD} \sigma=14\)
Co-efficient of variation C. V \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{14}{65} \times 100 \%=\frac{1400}{65} \% \)
\(=21.538 \%=21.54 \% \)
Social Science:
Mean \(\bar{x}=60 ; \mathrm{SD} \sigma=10\)
Co-efficient of variation C. V \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{10}{60} \times 100 \%=\frac{100}{6} \%\)
= 16.666 %
= 16.67 %
The highest variation is in the subject science and the lowest variation is in the subject social science

(i) Sathya
Sum of marks in 5 subjects = 460
Mean marks of sathya \(\bar{x}=\frac{\sum x_{i}}{N}=\frac{460}{5}=92\)
Standard deviation \(\sigma=4.6\)
Co-efficient of variation C.V  \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{4.6}{92} \times 100 \%=\frac{460}{92} \% \)
\(=\frac{10}{2} \%=5 \% \)
C.V of Sathya = 5%
(ii) Vidhya
Sum of the marks in 5 subjects = 480
Mean marks of vidhya \(\bar{x}=\frac{\sum x_{i}}{N}=\frac{480}{5}=96\)
Standard deviation \(\sigma=2.4\)
Co-efficient of variation C.V. \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{2.4}{96} \times 100 \%=\frac{240}{96} \%\)
\(=\frac{10}{4} \%=2.5 \%\)
C. V. of Vidhya = 2.5%
C. V of Vidhya < C. V. of Sathya
Vidhya is more consistent in her performance.

Let us take the assumed mean A = 20 and C = 5

Ages	Mid value x1	fi	\(d_{1}=x_{1}-A \) \(d_{i}=x_{1}-20 \)	\(\mathrm{d}_{\mathrm{i}}=\frac{x_{i}-A}{C} \) \(\mathrm{~d}_{\mathrm{i}}=\frac{x_{i}-A}{5} \)	fidi	fidi2
0-10	5	3	-15	-3	-9	27
10-20	15	5	-5	-1	-5	5
20-30	25	16	5	1	16	16
30-40	35	18	15	3	54	162
40-50	45	12	25	5	60	300
50-60	55	7	35	7	49	343
60-70	65	4	45	9	36	324
		\(\Sigma f_{i}\) = N = 65			\(\Sigma f_{i} d_{i}\) = 201	\(\Sigma f_{i} d_{i}^{2}\) = 1177

Standard deviation \(\sigma=\mathrm{C} \times \sqrt{\frac{\Sigma f_{i} d_{i}^{2}}{N}-\left(\frac{\Sigma f_{i} d_{i}}{N}\right)^{2}} \)
\(\sigma=5 \times \sqrt{\frac{1177}{65}-\left(\frac{201}{65}\right)^{2}} \)
\(\sigma=5 \times \sqrt{\frac{1177}{65}-\frac{40401}{4225}} \)
\(\sigma=5 \times \sqrt{\frac{76505-40401}{4225}} \)
\(\sigma=5 \times \sqrt{8.54} \)
\(\sigma=5 \times 2.92=14.6 \)
Standard deviation  \(\sigma=14.6\)

Mean \(\bar { x } \) = \(\frac { Sum\ of\ all\ observations }{ Number\ of\ observation } \)
= \(\frac { \Sigma x_{ i } }{ n } =\frac { 1+2+3+...+n }{ n } =\frac { n(n+1) }{ 2\times n } \)
Mean \(\bar { x } \) = \(\frac { n+1 }{ 2 } \)
Variance σ² = \(\frac { \Sigma x_{ i }^{ 2 } }{ n } -\left( \frac { \Sigma x_{ i } }{ n } \right) ^{ 2 }\left[ \begin{matrix} \Sigma x_{ i }^{ 2 }={ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ n }^{ 2 } \\ (\Sigma x_{ i })^{ 2 }=(1+2+3+...+n)2 \end{matrix} \right] \)
= \(\frac { n(n+1)(2n+1) }{ 6\times n } -\left[ \frac { n(n+1) }{ 2\times n } \right] ^{ 2 }\)
= \(\frac { 2n^{ 2 }+3n+1 }{ 6 } -\frac { { n }^{ 2 }+2n+1 }{ 4 } \)
Variance σ² = \(\frac { 4n^{ 2 }+6n+2-3n^{ 2 }-6n-3 }{ 12 } =\frac { { n }^{ 2 }-1 }{ 12 } \).

Solid sphere
radius = 6 cm
Volume \(=\frac{4}{3} \pi r^{3} \text { cu. units } \)
\(=\frac{4}{3} \pi(6)^{3} \)
\(=\frac{4}{3} \pi(216)=288 \pi \mathrm{cm}^{3} \)
Hollow cylinder
Internal radius = 'r'
External radius = 'R' = 5 cm
Height h = 32 cm
Volume of Hollow Cylinder
\(=\pi h\left(\mathrm{R}^{2}-r^{2}\right)\ cu. units \)
\(=\pi(32)\left(25-r^{2}\right) \mathrm{cm}^{3} \)
Given that solid sphere is melted to form a hollow cylinder.
Volume of Hollow Cylinder = Volume of Sphere
\(32 \pi\left(25-r^{2}\right) =288 \pi \)
\(25-r^{2} =\frac{288}{32}=9 \)
r² = 25 - 9 = 16
Internal radius r = 4 cm
Thickness = External radius - Internal radius
= R - r = 5 - 4 = 1 cm.

Hollow Hemisphere 
Internal diameter = 6 cm
Internal radius 'r' = 3 cm
External diameter = 10 cm
External radius 'R' = 5 cm
\(\left.\begin{array}{l} \text { Volume of hemisphere (or) } \\ \text {Volume of material used } \end{array}\right\}=\frac{2}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \text { cu. units }\)
\(=\frac{2}{3} \pi\left(5^{3}-3^{3}\right) \)
\(=\frac{2}{3} \pi(125-27)=\frac{196 \pi}{3} \mathrm{~cm}^{3} \)
Cylinder
Diameter = 14 cm
radius = 7 cm
height = h
Volume of cylinder \(=\pi r^{2} h\ cu. units \)
\(=\pi(7)^{2} h \)
\(=49 \pi h \mathrm{~cm}^{3} \)
Given that hollow hemisphere is melted and cast into a solid cylinder
Volume of cylinder = volume of hollow hemisphere
\(49 \pi h =\frac{196 \pi}{3} \)
\(h =\frac{196}{3 \times 49}=\frac{4}{3}=1.33 \)
Height of the cylinder = 1.33 cm.

Diameter of cone = 14 cm
Radius of cone = 7 cm
Height of cone = 8 cm
Volume of cone  \(=\frac{1}{3} \pi r^{2} h \text { cu. units } \)
\(=\frac{1}{3} \times \pi \times 7 \times 7 \times 8 \)
\(=\frac{392 \pi}{3} \mathrm{~cm}^{3} \)
External diameter of sphere = 10 cm
External radius of sphere 'R' = 5 cm
Internal radius = 'r'
Given, Right Circular Cone is melted to form a hollow sphere.
Volume of hollow sphere = Volume of cone
\(\frac{4}{3} \pi\left(5^{3}-r^{3}\right) =\frac{392 \pi}{3} \)
\(5^{3}-r^{3} =\frac{392}{4}=98 \)
r³ = 125 - 98 = 27
Radius r = 3cm
Internal diameter = 2r = 6 cm

Radius of sphere = 12 cm
Volume of sphere = \(\frac{4}{3} \pi r^{3} cu. units \)
= \(\frac{4}{3} \pi(12)^{3} \)
\(=2304 \pi \mathrm{cm}^{3}\)
Radius of cylinder = 8 cm
height = h cm
Volume of cylinder = \(\pi r^{2} h \) cu. units
= \(\pi(8)^{2} h \)
= \(64 \pi \mathrm{h} \mathrm{cm}^{3}\)
Given that sphere is melted and cast into a cylinder
Volume of cylinder = Volume of sphere
\(64 \pi h =2304 \pi \)
\(h =\frac{2304 \pi}{64 \pi}=36 \)
Height of the cylinder = 36 cm.

Let h and r be the height and radius of the cylinder respectively.
Given that, h = 15 cm, r = 6 cm
Volume of the container V = \(\pi\)r²h cubic units.
Let, r₁ = 3 cm, h₁ = 9 cm be the radius and height of the cone.
Also, r₁ = 3 cm is the radius of the hemispherical cap.
Volume of one ice cream cone = (Volume of the cone + Volume of the hemispherical cap)
\(=\frac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }+\frac { 2 }{ 3 } \pi { r }_{ 1 }^{ 3 }\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 3\times 3\times 9+\frac { 2 }{ 3 } \times \frac { 22 }{ 7 } \times 3\times 3\times 3\)
\(=\frac { 22 }{ 7 } \times 9(3+2)=\frac { 22 }{ 7 } \times 45\)
\(Number\ of\ cones=\frac { volume\ of\ the\ cylinder }{ volume\ of\ one\ ice\ cream\ cone } \)
Number of ice cream cones needed \(=\frac { \frac { 22 }{ 7 } \times 6\times 6\times 15 }{ \frac { 22 }{ 7 } \times 45 } =12\)
Thus 12 ice cream cones are required to empty the cylindrical container.

Let the number of small spheres obtained be n.
Let r be the radius of each small sphere and R be the radius of metallic sphere.
Here, R = 16 cm, r = 2 cm
Now, n\(\times\)(Volume of a small sphere) = Volume of big metallic sphere
\(n\left( \frac { 4 }{ 3 } { \pi r }^{ 2 } \right) =\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
\(n\left( \frac { 4 }{ 3 } \pi \times { 2 }^{ 3 } \right) =\frac { 4 }{ 3 } \pi \times { 16 }^{ 3 }\)
8n = 4096 gives n = 512
Therefore, there will be 512 small spheres.