leap year has 366 days. So it has 52 full weeks and 2 days. 52 Saturdays must be in 52 full weeks.
The possible chances for the remaining two days will be the sample space.
S = {(Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun)}
n(S) = 7
Let A be the event of getting 53^rd Saturday.
Then A = {Fri-Sat, Sat-Sun}; n(A) = 2
Probability of getting 53 Saturdays in a leap year is P(A0 = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 7 } \).

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.

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

(i) P(A or B) = P(AUB)
= P(A) + P(B) - P(A∩B)
P(A or B) = \(\frac { 1 }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 8 } =\frac { 5 }{ 8 } \)
(ii) P(not A and not B) = \(P(\bar { A } \cap \bar { B } )\)
=\(P(\overline { A\cup B) } \)
= 1 - P(AUB)
P(not A and not B) = 1-\(\frac { 5 }{ 8 } =\frac { 3 }{ 8 } \).

Sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}; n(S) = 12
(i) Let A be the event of resting in 7. n(A) = 1
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 1 }{ 12 } \)
(ii) Let B be the event that the arrow will come to rest in a prime number.
B = {2 , 3, 5, 7, 11}; n(B) = 5
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 5 }{ 12 } \)
(iii) Let C be the event that arrow will come to rest in a composite number.
C = {4, 6, 8, 9, 10, 12}; n(C) = 6
P(C) = \(\frac { n(C) }{ n(S) } =\frac { 6 }{ 12 } =\frac { 1 }{ 2 } \)

When we roll two dice, the sample space is given by
S = \(\{ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \)
n(S) = 36
(i) Let A be the event of getting the sum of outcome values equal to 4.
Then A = {(1, 3),(2, 2),(3, 1)}; n(A) = 3.
Probability of getting the sum of outcomes equal to 4 is  P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
(ii) Let B be the event of getting the sum of outcome values greater than 10.
Then B = {(5,6),(6,5),(6,6)}; n(B) = 3
Probability of getting the sum of outcomes greater than 10 is P(B) = \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
(iii) Let C be the event of getting the sum of outcomes less than 13. Here all the outcomes have the sum value less than 13. Hence C = S
Therefore, n(C) = n(S) = 36
Probability of getting the sum value less than 13 is P(C) = \(\frac { n(C) }{ n(S) } =\frac { 36 }{ 36 } \) = 1

Given data are 38, 40, 43,44, 46, 47, 49, 53
Standard deviation \(\sigma=\sqrt{\frac{\Sigma d_{i}^{2}}{N}-\left(\frac{\Sigma d_{i}}{N}\right)^{2}}\)
Let us take the assumed mean A = 44

\(\sigma=\sqrt{\frac{\sum d_{i}^{2}}{N}-\left(\frac{\sum d_{i}}{N}\right)^{2}}\)
\(\sigma=\sqrt{\frac{172}{8}-\left(\frac{8}{8}\right)^{2}}=\sqrt{21.5-(1)^{2}} \)
\(\sigma=\sqrt{21.5-1}=\sqrt{20.5}=4.53 \)
\(\text { Mean } \bar{x}=\frac{\text { Sum of the given observations }}{\text { Number of observations }}\)
\(\bar{x}=\frac{38+40+43+44+46+47+49+53}{8} \)

\(\bar{x}=\frac{360}{8}=45 \)
Co-efficient of variation C.V \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{4.53}{45} \times 100 \%=\frac{453}{45}\)
Co-efficient of variation = 10.07%

N = 40
Standard deviation \(\sigma=\sqrt{\frac{\sum f_{i} d_{i}^{2}}{N}-\left(\frac{\sum f_{i} d_{i}}{N}\right)^{2}}\)
\(=\sqrt{\frac{2450}{40}-\left(\frac{40}{40}\right)^{2}}=\sqrt{61.25-1^{2}} \)
\(=\sqrt{61.25-1}=\sqrt{60.25} \)
Standard deviation \(\sigma=7.76\)

n = 15, \(\bar { x } \) = 10, σ = 5; \(\bar { x } =\frac { \Sigma x }{ n } \); Σx = 15 x 10 = 150
Wrong observation value = 8, Correct observation value = 23.
Correct total = 150 - 8 + 23 = 165
Correct mean \(\bar { x } \) = \(\frac { 165 }{ 15 } \) = 11
Standard deviation σ = \(\sqrt { \frac { \Sigma x^{ 2 } }{ n } -\left( \frac { \Sigma x }{ n } \right) ^{ 2 } } \)
Incorrect value of σ = 5 = \(\sqrt { \frac { \Sigma x^{ 2 } }{ 15 } -(10)^{ 2 } } \)
25 = \(\frac { \Sigma x^{ 2 } }{ 15 } \)-100 gives, \(\frac { \Sigma x^{ 2 } }{ 15 } \) = 125
Incorrect value of Σx² = 1875
Correct value of Σx² = 1875 - 8² + 23² = 2340
Correct standard deviation σ = \(\\ \sqrt { \frac { 2340 }{ 15 } -(11)^{ 2 } } \)
σ = \(\\ \sqrt { 156-121 } =\sqrt { 35 } \) σ ≃ 5.9

Let the assumed mean, A = 8

Standard deviation
σ = \(\sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } -\left( \frac { \Sigma { f }_{ i }{ d }_{ i } }{ N } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 240 }{ 29 } -\left( \frac { 4 }{ 29 } \right) ^{ 2 } } =\sqrt { \frac { 240\times 29-16 }{ 29\times 29 } } \)
σ = \(\sqrt { \frac { 6944 }{ 29\times 29 } } \); σ ≃ 2.87
Calculation of Standard deviation for continuous frequency distribution
(i) Mean method:
Standard deviation σ = \(\sqrt { \frac { \Sigma { f }_{ i }({ x }_{ i }-\bar { x } )^{ 2 } }{ N } } \)
Where, x_i = Middle value of the i th class
f_i = Frequency of the i th class
(ii) Shortcut method (or) Step deviation method:
To make the calculation simple, we provide the following formula. Let A be the assumed mean, x_i be the middle value of the i^th class and c is the width of the class interval.
Let d_i = \(\frac { { x }_{ i }-A }{ c } \)
σ = \(\sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } -\left( \frac { \Sigma { f }_{ i }{ d }_{ i } }{ N } \right) ^{ 2 } } \).

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 = \(\frac { 4n^{ 2 }+6n+2-3n^{ 2 }-6n-3 }{ 12 } =\frac { { n }^{ 2 }-1 }{ 12 } \).

We note that all the observations are divisible by 5. Hence we can use the step deviation method. Let the Assumed mean A = 20, n = 8.

Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)xc
= \(\sqrt { \frac { 44 }{ 8 } -\left( \frac { 4 }{ 8 } \right) ^{ 2 } } \times 5=\sqrt { \frac { 11 }{ 2 } -\frac { 1 }{ 4 } \times 5 } \)
= \(\sqrt { 5.5-0.25 } \) = 2.29 x 5
σ ≃ 11.45

Arranging the numbers in ascending order we get, 11.4, 12.5, 12.8, 16.3, 17.8, 19.2 Number of observations
n = 6
Mean = \(\frac { 11.4+12.5+12.8+16.3+17.8+19.2 }{ 6 } =\frac { 90 }{ 6 } \)=15

Standard deviation σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } } \)
=\(\sqrt { \frac { 51.22 }{ 6 } } =\sqrt { 8.53 } \)
Hence, σ ≃2.9
Assumed Mean method:
When the mean value is not an integer (since calculations are very tedious in decimal form) then it is better to use the assumed mean method to find the standard deviation.
Ler x₁, x₂, x₃, .....xn be the given data values and let \(\bar { x } \) be their mean
Let di be the deviation of xi from the assumed mean A, which is usually the middle value or near the middle value of the given data.
d_i = x_i - A gives, x_i = d_i + A           ...(1)
Σd_i = Σ(x_i - A)
= Σx_i-(A + A + A + ..to n times)
Σd_i = Σxi - A x n
\(\frac { \Sigma { d }_{ i } }{ n } =\frac { \Sigma { x }_{ i } }{ n } \) - A
\(\bar { d } \) = \(\bar { x } \) - A (or) \(\bar { x } \) = \(\bar { d } \)+ A         ..(2)
Now, Standard deviation
σ = \(\\ \sqrt { \frac { \Sigma ({ x }_{ i }-\bar { x } )^{ 2 } }{ n } } =\sqrt { \frac { \Sigma ({ d }_{ i }+A-\bar { d } -A)^{ 2 } }{ n } } \) (using (1) and (2))
= \(\sqrt { \frac { \Sigma (d_{ i }-\bar { d } )^{ 2 } }{ n } } =\sqrt { \frac { \Sigma ({ d }_{ i }^{ 2 }+2d_{ i }-\bar { d } -\bar { d } ^{ 2 }) }{ n } } \)
= \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -2\bar { d } \frac { \Sigma { d }_{ i } }{ n } +\frac { { \bar { d } }^{ 2 } }{ n } (1+1+1+...to\quad n\quad times) } \)
\(=\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -2\bar { d } \times \bar { d } +\frac { { \bar { d } }^{ 2 } }{ n } \times n } \) (since \(\bar { d } \) is a constant)
= \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -{ \bar { d } }^{ 2 } } \)
Standard deviationσ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \).