(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-\(\frac { 5 }{ 8 } =\frac { 3 }{ 8 } \).

Arranging the numbers in ascending order
Rs. 280, Rs. 280, Rs. 290,Rs. 290, Rs. 300, Rs. 310, Rs. 310, Rs. 320, Rs. 320
\( \text { Mean } =\frac{280+280+290+290+300+310+310+320+320}{9} \)
\(\bar{x} =\frac{2700}{9}=300\)

Standard Deviation
\( \sigma =\sqrt{\frac{\Sigma d_{i}^{2}}{n}} \)
\(\sigma =\sqrt{\frac{2000}{9}} \)
\(\sigma=\sqrt{222.222 \ldots} =14.907=14.91 \)
\(\text { We have } \sigma =\sqrt{222.222} \)
\(\text { Variance } \sigma^{2} =222.22\)
Variance = 222.22;
Standard deviation = 14.91

Arranging the values in ascending order we get, 4, 7, 8, 10, 11 and n = 5

Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 350 }{ 5 } -\left( \frac { 40 }{ 5 } \right) ^{ 2 } } \)
σ = \(\sqrt { 6 } \) ⋍ 2.45
When we add 3 to all the values, we get the new values as 7, 10, 11, 13, 14.

 Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 635 }{ 5 } -\left( \frac { 55 }{ 5 } \right) ^{ 2 } } \)
σ = \(\sqrt { 6 } \) ⋍ 2.45

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 = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \).

Standard deviation
σ =\(\sqrt { \frac { \Sigma x_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
=\(\sqrt { \frac { 623 }{ 7 } -\left( \frac { 63 }{ 7 } \right) ^{ 2 } } \)
=\(\\ \sqrt { 89-81 } =\sqrt { 8 } \)
Hence, σ ≃ 2.83
(ii) Mean method:
Another convenient way of finding standard deviation is to use the following formula.
Standard deviation (by mean method) σ = \(\sqrt { \frac { \Sigma ({ x }_{ i }-\bar { x } )^{ 2 } }{ n } } \)
If di = xi - \(\bar { x } \) are the deviations, then σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } } \).

P(A) = 0.37, P(B) = 0.42, P(A∩B) = 0.09
P(AUB) = P(A) + P(B) - P(A∩B)
P(AUB) = 0.37 + 0.42 - 0.09 = 0.7