\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 75 }{ 5 } \)=15
d'=\(\frac { x-\bar { x } }{ c } =\frac { x-A }{ c } \)
A is assumed mean c is common factor.
Here A= 15, C = 5
σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x c
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
If 3 is added to each value, we get 8, 13, 18,23, 28 as new values.

∴ σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
S.D. doesn't change when a number is added or subtracted to the values.

(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

Total number of cards = 52; n(S) = 52
Let A be the event of getting a king card, n(A) = 4
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 4 }{ 52 } \)
Let B be the event of getting a heart card. n(B) = 13
P(B) = \(\\ \frac { n(B) }{ n(S) } =\frac { 15 }{ 52 } \)
Let C be the event of getting a red card. n(C) = 26
P(C) = \(\frac { n(C) }{ n(S) } =\frac { 26 }{ 52 } \)
P(A ∩ B) = P (getting heart king) = \(\frac{1}{52}\)
P(B ∩ C) = P (getting red and heart)) = \(\frac{13}{52}\)
P(A ∩ C) = P (getting red king) = \(\frac{2}{52}\)
P(A ∩ B ∩ C) = (getting heart, king which is red = \(\frac{1}{52}\)
Therefore, required probability is
P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(C ∩ A) + P(A ∩ B ∩ C)
=\(\frac { 4 }{ 52 } +\frac { 13 }{ 52 } +\frac { 26 }{ 52 } -\frac { 1 }{ 52 } -\frac { 13 }{ 52 } -\frac { 2 }{ 52 } +\frac { 1 }{ 52 } =\frac { 28 }{ 52 } =\frac { 7 }{ 13 } \).

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) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 12 } =\frac { 1 }{ 4 } \)

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 = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 7 } \).

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) = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \).

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

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

CV =\(\frac { \sigma }{ \bar { x } } \) x 100 ⇒ \(\bar { x } =\frac { \sigma }{ CV } \) x 100
\(\bar { x } =\frac { 15.6 }{ 6.9 } \) x 100  = 22.6

Total number of students n(S) = 50.
Let A and B be the events of students opted for NCC and NSS respectively.
n(A) = 28, n(B) = 30, n(A⋂B) = 18
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 28 }{ 50 } \)
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 30 }{ 50 } \)
P(A∩B) = \(\frac { n(A\cap B) }{ n(S) } =\frac { 18 }{ 50 } \)
(i) Probability of the students opted for NCC but not NSS
P(A ∩ \(\bar { B } \)) = P(A) - P(A ∩ B) = \(\frac { 28 }{ 50 } -\frac { 18 }{ 50 } =\frac { 1 }{ 5 } \)
(ii) Probability of the students opted for NSS but not NCC.
P(A ∩ \(\bar { B } \)) = P(B) - P(A ∩ B) = \(\frac { 30 }{ 50 } -\frac { 18 }{ 50 } =\frac { 6 }{ 25 } \)
(iii) Probability of the students opted for exactly one of them
= P[(A ∩ \(\bar { B } \)) U (\(\bar { A } \) ∩ B)]
= P(A ∩ \(\bar { B } \)) + P(\(\bar { A } \) ∩ B) =\(\frac { 1 }{ 5 } +\frac { 6 }{ 25 } =\frac { 11 }{ 25 } \)
(Note that (A ∩ \(\bar { B } \)),(\(\bar { A } \) ∩ B) are mutually exclusive events)

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

When two dice are rolled together, there will be 6 x 6 = 36 outcomes. Let S be the sample space. Then n(S) = 36.
Let A be the event of getting a doublet and B be the event of getting face sum 4.
Then A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}
B = {(1,3),(2,2),(3,1)
Therefore, A∩B = {(2,2)}
Then, n(A) = 6, n(B) = 3, n(A∩B) = 1
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 6 }{ 36 } \)
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 36 } \)
P(A∩B) = \(\frac { n(A\cap B) }{ n(S) } =\frac { 1 }{ 36 } \)
Therefore, P (getting a doublet or a total of 4) = P(AUB)
P(AUB) = P(A) + P(B) - P(A∩B)
=\(\frac { 6 }{ 36 } +\frac { 3 }{ 36 } -\frac { 1 }{ 36 } =\frac { 8 }{ 36 } =\frac { 2 }{ 9 } \)
Hence, the required probability is \(\frac { 2 }{ 9 } \).

Total number of cards remaining
= 11 + 11 + 13 + 11 = 46
n(S) = 46
(i) Let A be the event of getting a clavor card
n(A) = 13
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{13}{46}\)
(ii) Let B be the event of getting queen of red card
n(B) = 0 (Removed red queens)
\(\mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=0\)
(iii) Let C be the event of getting a king of black card
n(C) = 1
\(\mathrm{P}(\mathrm{C})=\frac{n(C)}{n(S)}=\frac{1}{46}\)

Number of non defective bulbs = 20.
Let x be the number of defective bulbs
Then total number of bulbs n(s) = 20 + x
Let 'A' be the event of getting defective bulbs
\(\mathrm{P}(\mathrm{A})=\frac{x}{20+x}=\frac{3}{8}\)
8x = 3 (20 + x)
8x = 60 + 3x
8x - 3x = 60
5x = 60
\(x=\frac{60}{5}\)
x = 12
Number of defective bulbs = 12

When three fair coins are tossed together, the sample space
S = {(HHH), (THH), (HTH),(HHT), (TTH), (THT), (HTT), (TTT)}
N(s) = 8
(i) Let A be the event of getting all heads
A = {HHH}
n(A) = 1
\(P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(\mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=\frac{7}{8}\)
(iii) Let C be the event of getting at most one head
C = {HTT, THT, TTH, TTT}
n(C) = 4
\(\mathrm{P}(\mathrm{C})=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}\)
(iv) Let D be the event of getting at most two tails
P = {HHH, HHT, HTH, HTT, THH, THT, TTH}
n(D) = 7
\(\mathrm{P}(\mathrm{D})=\frac{n(D)}{n(S)}=\frac{7}{8}\)

Total number of balls = blue balls + red balls
n(S) = 12 + x
(i) The probability that it will be a red ball:
Let A be the event of selecting a red ball
n(A) = x
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)} \)
\(\mathrm{P}(\mathrm{A})=\frac{x}{12+x} \)
(ii) If 8 more red balls are put in the bag then the number of red balls = x + 8
Total number of balls = 12 + x + 8
Probability of drawing a red ball
\(\mathrm{P}(\mathrm{B}) =\frac{n(B)}{n(S)} \)
\(=\frac{x+8}{12+x+8}=\frac{x+8}{x+20} \)
If the probability of drawing a red ball will be twice that of the probability (i), then
\(\frac{x+8}{x+20}=2 \times\left[\frac{x}{12+x}\right]\)
(x + 8)(12 + x) = 2x (x + 20)
12x + 96 + x² + 8x = 2x² + 40x
2x²  - x² + 40x - 12x - 8x - 96 = 0
x² + 20x - 96 = 0
(x - 4) (x + 24) = 0
x = 4 and x = - 24 (not possible)
By applying the value of x in P(A)
we get P(A) \(=\frac{4}{12+4}=\frac{4}{16}\)
\(P(A)=\frac{1}{4}\)
x = 4

Number of green balls is n(G) = 6
Let number of red balls is n(R) = x
Therefore, number of black balls is n(B) = 2x
Total number of balls n(S) = 6 + x + 2x = 6 + 3x
It is given that, P(G) = 3 x P(R)
\(\frac { 6 }{ 6+3x } =3\times \frac { x }{ 6+3x } \)
3x = 6 gives, x = 2
(i) Number of black balls = 2 x 2 = 4
(ii) Total number of balls = 6 + (3 x 2) = 12

Total number of possible outcomes n(S) = 5 + 4 = 9
(i) Let A be the event of getting a blue ball.
Number of favourable outcomes for the event A. Therefore, n(A) = 5
Probability that the ball drawn is blue. Therefore, P(A) = \(\frac { n(A) }{ n(S) } =\frac { 5 }{ 9 } \)
(ii) \(\bar { A } \) will be the event of not getting a blue ball. So P(\(\bar { A } \)) = 1 - P(A) = \(1-\frac { 5 }{ 9 } =\frac { 4 }{ 9 } \).

\(\bar { { x }_{ 1 } } =\frac { { \Sigma x }_{ 1 } }{ n } \)
= \(\frac { 110 }{ 5 } \)
= 22
\({ \sigma }_{ 1 }=\sqrt { \frac { { \Sigma d }_{ 1 }^{ 2 } }{ n } } \)
= \(\sqrt { \frac { 40 }{ 5 } } \)
= \(\sqrt { 8 } \)
\(\bar { { x }_{ 2 } } =\frac { { \Sigma x }_{ 1 } }{ n } \)
= \(\frac { 75 }{ 5 } \)
= 15
\({ \sigma }_{ 2 }=\sqrt { \frac { { \Sigma d }_{ 2 }^{ 2 } }{ n } } \)
= \(\sqrt { \frac { 20 }{ 5 } } \)
= \(\sqrt { 4 } \)
= \(2\sqrt { 2 } \)
\(CV_{ 1 }=\frac { { { \sigma }_{ 1 } } }{ { x }_{ 1 } } \times 100\)
= \(\frac { 2.83 }{ 22 } \times 100\)
= 12.86
\({ CV }_{ 2 }=\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \times 100\)
= \(\frac { 2 }{ 15 } \times 100\)
= 13.33
\(\therefore\)Co-efficient of variation of City A is less than C.V of City B.
\(\therefore\) City A is more consistent.

First we find the coefficient of variation for guavas and oranges separately.
Number of guavas, n=7

Mean \(\bar { { x }_{ 1 } } \)=\(\frac { 30 }{ 7 } \)=4.29
Standard deviation σ₁=\(\sqrt { \frac { \Sigma { x }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
σ₁=\(\sqrt { \frac { 136 }{ 7 } -\left( \frac { 30 }{ 7 } \right) ^{ 2 } } =\sqrt { 19.43-18.40 } \) ≃ 1.01
Coefficient of variation for guavas
C.V₁=\(\frac { { \sigma }_{ 1 } }{ \bar { { x }_{ 1 } } } \times 100\)% = \(\frac { 1.01 }{ 4.29 } \) x 100% =23.54%

 Number of oranges n=7
Mean \(\bar { { x }_{ 2 } } \)=\(\frac { 30 }{ 7 } \)=4.29
Standard deviation σ₁=\(\sqrt { \frac { \Sigma { x }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
σ₂=\(\\ \sqrt { \frac { 184 }{ 7 } -\left( \frac { 30 }{ 7 } \right) ^{ 2 } } =\sqrt { 26.29-18.40 } \)=2.81
Coefficient of variation for oranges
C.V₁=\(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100% = \(\frac { 2.81 }{ 4.29 } \) x 100% = 65.50%
CV₁=23.54%, CV₂=65.50%. Since, C.V₁ < C.V₂,  we can conclude that the consumption of guavas is more consistent than oranges.

Mean = 8
Variance = 16.
Let a and b be the missing observations
\(\frac{\text { Sum of observations }}{\text { Number of observations }}=\text { Mean }\)
\(\frac{2+4+10+12+14+a+b}{7}=8\)
42 + a + b = 56
a + b = 56 - 42 = 14
b = 14 - a
Variance \(=\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}{n} \)
\(= \frac{\sum_{i=1}^{7}\left(x_{i}-\bar{x}\right)^{2}}{7} \)
\(=\frac{(2-8)^{2}+(4-8)^{2}+(10-8)^{2}+(12-8)^{2}+(14-8)^{2}+(a-8)^{2}+(b-8)^{2}}{7}\)
\(16=\frac{36+16+4+16+36+(a-8)^{2}+(b-8)^{2}}{7}\)
108 + (a - 8)² + (b - 8)² = 112
(a - 8)² + (b - 8)² = 4       ..(2)
Put b = 14 - a in (2)
(a - 8)² + (b - 8)² = 4 
(a - 8)² + (14 - a ) - 8)² = 4
(a - 8 )² + (6 - a)² = 4
a² + 64 - 16a + 36 + a² - 12a - 4 = 0
2a² - 28a + 96 = 0
Divided by 2,
a² - 14a + 48 = 0
(a - 6) (a - 8) = 0
a = 6 or a = 8
a = 6 , b = 14, a = 14 - 6 = 8
a = 6 and b = 8
Remaining numbers are 6 and 8.

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 \(\sigma=7.76\)

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

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

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