For comparing two data, first we have to find their coefficient of variations
Mean \(\bar { { x }_{ 1 } } \) = 155 cm, variance σ₁² = 72.25 cm²
Therefore standard deviation σ₁ = 8.5
Coefficient of variation C.V₁ = \(\frac { { \sigma }_{ 1 } }{ \bar { { x }_{ 1 } } } \) x 100%
C.V₁ = \(\frac { 8.5 }{ 15.5 } \) x 100% = 5.48% (for heights)
Mean \(\bar { { x }_{ 2 } } \) = 155 cm, variance σ₂² = 72.25 kg²
Standard deviation σ₂ = 5.3 kg
Coefficient of variation CV₂ = \(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100%
C.V₂ = \(\frac { 5.3 }{ 46.50 } \) x 100% = 11.40% (for weights)
C.V₁ = 5.48% = and CV₂ = 11.40%
Since C.V₂ > C.V₁,  the weight of the students is more varying than the height.

Mean \(\bar { x } \) = 25.6, Coefficient of variation, C.V. = 18.75
Coefficient of variation, C.V. = \(\frac { \sigma }{ \bar { x } } \) x 100%
18.75 = \(\frac { \sigma }{ 25.6 } \) x 100; σ = 4.8

Standard deviation of a data = 3.6
When each value of the data is divided by fixed constant; then the new standard deviation is also get divided by the constant.
If each value is divided by 3, then 
New standard deviation = \(\frac{3.6}{3}=1.2 \)
New variance = \(\sigma^{2}=(1.2)^{2} \)
New variance = 1.44
New standard deviation = 1.2

The standard deviation of a given data is 4.5. If we subtract some fixed constant from all the data, the standard deviation will not change.
Each value of the data decreased by 5, the new standard deviation will not change.
New standard deviation = 4.5

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

63,89,98, 125,79,108, 117,68
Largest value L = 125
Smallest value S = 63
\(\therefore\) R = L - S = 125- 63 = 62
Co-efficient of range = \(\frac { L-S }{ L+S } \)
=\(​​​​\frac {125-63 }{ 125+63 } \)
\(=\frac{62}{188}=0.329=0.33\)
Range = 62; coefficient of range = 0.33

Range R = 13.67
Largest value L = 70.08
Range R = L - S
13.67 = 70.08-S
S = 70.08 - 13.67 = 56.41
Therefore, the smallest value is 56.41

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