43.5, 13.6, 18.9,38.4,61.4,29.8
Largest value L= 61.4
Smallest value S = 13.6
R = L - S
= 61.4 - 13.6 = 4
Co-efficient of range = \(\frac { L-S }{ L+S } \)
= \(\frac { 47.8 }{ 75 } =0.64\)
Range = 47.8; co-efficient of range = 0.64.

(i) Given P(A) = 0.42
P(not A) = 1 - P(A)
\(\mathrm{P}(\bar{A})=1-0.42=0.58\)
(ii) Given P(B) = 0.48
P(not B) = 1 - P(B)
\(\mathrm{P}(\bar{B})=1-0.48=0.52\)
(iii) P(A or B) = \(P(A \cup B)\)
\(=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B)\)
= 0.42 + 0.48 - 015
= 0.90 - 0.16
P(A or B) = 0.74

Given \(P\left( A \right) :P\left( \bar { A } \right) =17:15\)
(i) \(\frac{P(A)}{P(\bar{A})} =\frac{17}{15} \)
\(\frac{P(A)}{1-P(A)} =\frac{17}{15} \quad[\therefore P(\bar{A})=1-P(A)] \)
15 P(A) = 17 [1- P(A)]
15 P(A) = 17 - 17 P(A)
\(32 \mathrm{P}(\mathrm{A})=17 ; \quad \mathrm{P}(\mathrm{A})=\frac{17}{32} \)
\(P(\bar{A})=1-\mathrm{P}(\mathrm{A})=1-\frac{17}{32}=\frac{32-17}{32}=\frac{15}{32} \)
(ii) \(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{17}{32}\)
n(A) = 17

Sample Space
s = {(1, 1) (1,2) (1,3) (1,4) (1, 5) (1,6)
(2, 1) (2,2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3,2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4,2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5,2) (5,3) (5,4) (5, 5) (5,6)
(6, 1) (6,2) (6, 3) (6, 4) (6, 5) (6, 6)}
Total number of outcomes = 36

When we toss three coins the outcome will be

The sample space = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Total number of outcomes = 8

Sample space
S = {1H,1T,2H,2T,3H,3T,4H,4T,5H,5T,6H,6T};
n(S) = 12
Let A be the event of getting an odd number and a head.
A = {1H, 3H, 5H}; n(A) = 3
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 12 } =\frac { 1 }{ 4 } \)

When two coins are tossed together, the sample space is
S = {HH, HT, TH, TT} n(S) = 4
Let A be the event of getting different faces on the coins.
A = {HT, TH}; n(A) = 2
Probability of getting different faces on the coins is P(A) = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \).

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%

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.