Variance and Standard Deviation Questions in English

(C) Let the five observations be $x_1, x_2, x_3, x_4, x_5$.
Given mean $\bar{X} = \frac{24}{5}$ and variance $\sigma^2 = \frac{194}{25}$.
Sum of five observations: $\sum_{i=1}^5 x_i = 5 \times \frac{24}{5} = 24$.
Mean of first four observations is $\frac{7}{2}$,so $\sum_{i=1}^4 x_i = 4 \times \frac{7}{2} = 14$.
Thus,$x_5 = 24 - 14 = 10$.
Using the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{X})^2$:
$\frac{194}{25} = \frac{\sum_{i=1}^5 x_i^2}{5} - (\frac{24}{5})^2$.
$\frac{194}{25} = \frac{\sum_{i=1}^5 x_i^2}{5} - \frac{576}{25}$ $\Rightarrow \frac{\sum_{i=1}^5 x_i^2}{5} = \frac{770}{25} = \frac{154}{5}$.
$\sum_{i=1}^5 x_i^2 = 154$.
Since $x_5 = 10$,$x_5^2 = 100$.
$\sum_{i=1}^4 x_i^2 = 154 - 100 = 54$.
Variance of first four observations: $\text{Var} = \frac{\sum_{i=1}^4 x_i^2}{4} - (\text{mean of first four})^2$.
$\text{Var} = \frac{54}{4} - (\frac{7}{2})^2 = \frac{27}{2} - \frac{49}{4} = \frac{54 - 49}{4} = \frac{5}{4}$.

(B) First,calculate the mean $\overline{x}$:

$x_i$	$f_i$	$f_ix_i$	$f_ix_i^2$
$0$	$3$	$0$	$0$
$1$	$2$	$2$	$2$
$5$	$3$	$15$	$75$
$6$	$2$	$12$	$72$
$10$	$6$	$60$	$600$
$12$	$3$	$36$	$432$
$17$	$3$	$51$	$867$
Total	$\sum f_i = 22$	$\sum f_ix_i = 176$	$\sum f_ix_i^2 = 2048$

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{176}{22} = 8$
Variance $\sigma^2 = \frac{1}{N} \sum f_i x_i^2 - (\overline{x})^2$
$\sigma^2 = \frac{2048}{22} - (8)^2$
$\sigma^2 = 93.09 - 64 = 29.09$
Rounding to the nearest integer,the correct option is $B$.

(C) Given observations: $a, b, 68, 44, 48, 60$.
Mean $\overline{x} = 55$,Variance $\sigma^2 = 194$.
Sum of observations: $a + b + 68 + 44 + 48 + 60 = 6 \times 55 = 330$.
$a + b + 220 = 330 \Rightarrow a + b = 110$ (Equation $1$).
Variance formula: $\frac{1}{n} \sum (x_i - \overline{x})^2 = 194$.
$(a - 55)^2 + (b - 55)^2 + (68 - 55)^2 + (44 - 55)^2 + (48 - 55)^2 + (60 - 55)^2 = 194 \times 6$.
$(a - 55)^2 + (b - 55)^2 + 13^2 + (-11)^2 + (-7)^2 + 5^2 = 1164$.
$(a - 55)^2 + (b - 55)^2 + 169 + 121 + 49 + 25 = 1164$.
$(a - 55)^2 + (b - 55)^2 = 1164 - 364 = 800$.
$a^2 - 110a + 3025 + b^2 - 110b + 3025 = 800$.
$a^2 + b^2 - 110(a + b) + 6050 = 800$.
Substitute $a + b = 110$: $a^2 + b^2 - 110(110) + 6050 = 800$.
$a^2 + b^2 - 12100 + 6050 = 800 \Rightarrow a^2 + b^2 = 6850$ (Equation $2$).
From $(a + b)^2 = a^2 + b^2 + 2ab$,we have $110^2 = 6850 + 2ab$.
$12100 - 6850 = 2ab$ $\Rightarrow 2ab = 5250$ $\Rightarrow ab = 2625$.
Since $a + b = 110$ and $ab = 2625$,$a$ and $b$ are roots of $t^2 - 110t + 2625 = 0$.
$(t - 75)(t - 35) = 0$.
Since $a > b$,$a = 75$ and $b = 35$.
Therefore,$a + 3b = 75 + 3(35) = 75 + 105 = 180$.

(A) Given $n=10$ observations $x_1, x_2, \ldots, x_{10}$.
From $\sum_{i=1}^{10}(x_i-\alpha)=2$,we have $\sum x_i - 10\alpha = 2$.
Given mean $\mu = \frac{\sum x_i}{10} = \frac{6}{5}$,so $\sum x_i = 12$.
Substituting $\sum x_i = 12$ into the first equation: $12 - 10\alpha = 2$ $\Rightarrow 10\alpha = 10$ $\Rightarrow \alpha = 1$.
Variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - \mu^2 = \frac{84}{25}$.
We know $\sum (x_i - \beta)^2 = \sum x_i^2 - 2\beta \sum x_i + 10\beta^2 = 40$.
From $\sigma^2 = \frac{\sum x_i^2}{10} - (\frac{6}{5})^2 = \frac{84}{25}$,we get $\frac{\sum x_i^2}{10} = \frac{84}{25} + \frac{36}{25} = \frac{120}{25} = \frac{24}{5}$,so $\sum x_i^2 = 48$.
Substituting $\sum x_i^2 = 48$ and $\sum x_i = 12$ into the equation $48 - 2\beta(12) + 10\beta^2 = 40$:
$10\beta^2 - 24\beta + 8 = 0 \Rightarrow 5\beta^2 - 12\beta + 4 = 0$.
Solving the quadratic: $(5\beta - 2)(\beta - 2) = 0$.
Since $\beta$ is a positive integer,$\beta = 2$.
Thus,$\frac{\beta}{\alpha} = \frac{2}{1} = 2$.

(C) The total frequency $N = 2 + 1 + 1 + 1 + 1 + 1 = 7$.
The mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{2c + 2c + 3c + 4c + 5c + 6c}{7} = \frac{22c}{7}$.
The variance $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2$.
$\sum f_i x_i^2 = 2(c)^2 + 1(2c)^2 + 1(3c)^2 + 1(4c)^2 + 1(5c)^2 + 1(6c)^2 = c^2(2 + 4 + 9 + 16 + 25 + 36) = 92c^2$.
$\sigma^2 = \frac{92c^2}{7} - \left(\frac{22c}{7}\right)^2 = \frac{92c^2}{7} - \frac{484c^2}{49} = \frac{644c^2 - 484c^2}{49} = \frac{160c^2}{49}$.
Given $\sigma^2 = 160$,we have $\frac{160c^2}{49} = 160$.
$c^2 = 49 \Rightarrow c = 7$ (since $c \in N$).

(D) The given sequence is an arithmetic progression with first term $a = 8$ and common difference $d = 13$.
Let $n$ be the number of terms. The $n$-th term is $a_n = a + (n-1)d = 320$.
$8 + (n-1)13 = 320 \implies 13(n-1) = 312 \implies n-1 = 24 \implies n = 25$.
The mean $\bar{x} = \frac{8 + 320}{2} = \frac{328}{2} = 164$.
The variance $\sigma^2$ of an arithmetic progression with $n$ terms is given by $\sigma^2 = \frac{(n^2-1)d^2}{12}$.
Substituting the values: $\sigma^2 = \frac{(25^2 - 1) \times 13^2}{12} = \frac{(625 - 1) \times 169}{12} = \frac{624 \times 169}{12}$.
$\sigma^2 = 52 \times 169 = 8788$.

(A) Given the initial mean $\overline{x} = 5.5$ for $n = 10$ values,the initial sum is $\sum_{i=1}^{10} x_i = 5.5 \times 10 = 55$.
Given $\sum_{i=1}^{10} x_i^2 = 371$.
Correcting the sum: $(\sum x_i)_{\text{new}} = 55 - (4 + 5) + (6 + 8) = 55 - 9 + 14 = 60$.
Correcting the sum of squares: $(\sum x_i^2)_{\text{new}} = 371 - (4^2 + 5^2) + (6^2 + 8^2) = 371 - (16 + 25) + (36 + 64) = 371 - 41 + 100 = 430$.
The variance $\sigma^2$ is given by $\frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2$.
$\sigma^2 = \frac{430}{10} - (\frac{60}{10})^2 = 43 - 6^2 = 43 - 36 = 7$.

(A) Given $\sum_{i=1}^{10}(x_i-2)=30 \implies \sum x_i - 20 = 30 \implies \sum x_i = 50$. The mean $\bar{x} = \frac{50}{10} = 5$.
Variance $\sigma_x^2 = \frac{\sum x_i^2}{10} - (\bar{x})^2 = \frac{4}{5} \implies \frac{\sum x_i^2}{10} - 25 = 0.8 \implies \sum x_i^2 = 258$.
Given $\sum_{i=1}^{10}(x_i-\beta)^2 = 98 \implies \sum x_i^2 - 2\beta \sum x_i + 10\beta^2 = 98$.
Substituting values: $258 - 2\beta(50) + 10\beta^2 = 98 \implies 10\beta^2 - 100\beta + 160 = 0 \implies \beta^2 - 10\beta + 16 = 0$.
$(\beta-8)(\beta-2) = 0$. Since $\beta > 2$,we have $\beta = 8$.
Let $y_i = 2(x_i-1) + 4\beta = 2x_i - 2 + 32 = 2x_i + 30$.
Mean $\mu = 2\bar{x} + 30 = 2(5) + 30 = 40$.
Variance $\sigma^2 = 2^2 \cdot \sigma_x^2 = 4 \cdot \frac{4}{5} = \frac{16}{5}$.
Then $\frac{\beta\mu}{\sigma^2} = \frac{8 \cdot 40}{16/5} = \frac{320 \cdot 5}{16} = 20 \cdot 5 = 100$.

(D) Given the mean $\overline{x} = \frac{\sum x_i}{5} = 5$,we have $1+3+a+7+b = 25$,which implies $a+b = 14$.
Given the variance $\sigma^2 = \frac{\sum x_i^2}{5} - (\overline{x})^2 = 10$,we have $\frac{1^2+3^2+a^2+7^2+b^2}{5} - 25 = 10$,so $1+9+a^2+49+b^2 = 175$,which gives $a^2+b^2 = 116$.
Since $a+b=14$ and $a^2+b^2=116$,we solve $(a+b)^2 - 2ab = 116$ $\Rightarrow 196 - 2ab = 116$ $\Rightarrow ab = 40$.
The values $a$ and $b$ are roots of $t^2 - 14t + 40 = 0$,so $(t-10)(t-4) = 0$. Given $a > b$,we have $a=10$ and $b=4$.
The new observations $y_n = n+x_n$ are $1+1=2, 2+3=5, 3+10=13, 4+7=11, 5+4=9$.
The new mean $\overline{y} = \frac{2+5+13+11+9}{5} = \frac{40}{5} = 8$.
The new variance is $\frac{2^2+5^2+13^2+11^2+9^2}{5} - 8^2 = \frac{4+25+169+121+81}{5} - 64 = \frac{400}{5} - 64 = 80 - 64 = 16$.

(A) To find the variance,we first calculate the mean and the sum of squares of the values.

$x_i$	$f_i$	$f_i x_i$	$f_i x_i^2$
$5$	$3$	$15$	$75$
$6$	$7$	$42$	$252$
$7$	$4$	$28$	$196$
$8$	$2$	$16$	$128$
$10$	$4$	$40$	$400$
Total	$N=20$	$\sum f_i x_i = 141$	$\sum f_i x_i^2 = 1051$

The formula for variance is $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \left(\frac{\sum f_i x_i}{N}\right)^2$.
Substituting the values:
$\sigma^2 = \frac{1051}{20} - \left(\frac{141}{20}\right)^2$
$\sigma^2 = 52.55 - (7.05)^2$
$\sigma^2 = 52.55 - 49.7025$
$\sigma^2 = 2.8475 \approx 2.85$.

(D) The formula for variance is given by $\sigma^2 = \frac{\sum X^2}{N} - \left(\frac{\sum X}{N}\right)^2$.
Given values are $N=60$,$\sum X^2=18000$,and $\sum X=960$.
Substituting these values into the formula:
$\sigma^2 = \frac{18000}{60} - \left(\frac{960}{60}\right)^2$.
$\sigma^2 = 300 - (16)^2$.
$\sigma^2 = 300 - 256$.
$\sigma^2 = 44$.
Thus,the variance of the data is $44$.

(B) Let the sixth test score be $x$.
Given the mean of six tests is $48$,we have:
$\frac{54+45+41+43+57+x}{6} = 48$
$240 + x = 288$
$x = 48$.
The marks are $54, 45, 41, 43, 57, 48$.
The mean $\bar{x} = 48$.
The standard deviation $\sigma = \sqrt{\frac{1}{n} \sum (x_i - \bar{x})^2}$.
$\sigma = \sqrt{\frac{(54-48)^2 + (45-48)^2 + (41-48)^2 + (43-48)^2 + (57-48)^2 + (48-48)^2}{6}}$
$\sigma = \sqrt{\frac{6^2 + (-3)^2 + (-7)^2 + (-5)^2 + 9^2 + 0^2}{6}}$
$\sigma = \sqrt{\frac{36 + 9 + 49 + 25 + 81 + 0}{6}}$
$\sigma = \sqrt{\frac{200}{6}} = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}$.

(A) The first $50$ even natural numbers are $2, 4, 6, \dots, 100$.
The mean $\bar{x} = \frac{2+4+\dots+100}{50} = \frac{2(1+2+\dots+50)}{50} = \frac{2 \times \frac{50 \times 51}{2}}{50} = 51$.
The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$\sum x_i^2 = 2^2 + 4^2 + \dots + 100^2 = 4(1^2 + 2^2 + \dots + 50^2)$.
Using the sum of squares formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$:
$\sum x_i^2 = 4 \times \frac{50 \times 51 \times 101}{6} = 2 \times 50 \times 17 \times 101 = 171700$.
$\sigma^2 = \frac{171700}{50} - (51)^2 = 3434 - 2601 = 833$.

(A) Given: $n = 100$,$\bar{x} = 50$,and $\sigma = 5$.
We know that the mean $\bar{x} = \frac{\sum x_i}{n} = 50$.
Therefore,$\sum x_i = 50 \times 100 = 5000$.
The formula for standard deviation is $\sigma = \sqrt{\frac{\sum x_i^2}{n} - (\bar{x})^2}$.
Substituting the values: $5 = \sqrt{\frac{\sum x_i^2}{100} - (50)^2}$.
Squaring both sides: $25 = \frac{\sum x_i^2}{100} - 2500$.
$\frac{\sum x_i^2}{100} = 2500 + 25 = 2525$.
$\sum x_i^2 = 2525 \times 100 = 252500$.

(C) The mean of $a, b, c$ is $\bar{x} = \frac{a+b+c}{3}$.
Since $b = a + c$,we have $\bar{x} = \frac{b+b}{3} = \frac{2b}{3}$.
The standard deviation of $a+2, b+2, c+2$ is the same as the standard deviation of $a, b, c$,which is $d$.
Thus,$d^2 = \frac{a^2+b^2+c^2}{3} - (\bar{x})^2$.
$d^2 = \frac{a^2+b^2+c^2}{3} - \left(\frac{2b}{3}\right)^2$.
$d^2 = \frac{a^2+b^2+c^2}{3} - \frac{4b^2}{9}$.
$d^2 = \frac{3(a^2+b^2+c^2) - 4b^2}{9}$.
$9d^2 = 3(a^2+c^2+b^2) - 4b^2$.
$9d^2 = 3(a^2+c^2) + 3b^2 - 4b^2$.
$9d^2 = 3(a^2+c^2) - b^2$.
Therefore,$b^2 = 3(a^2+c^2) - 9d^2$.

(D) Let the observations be $x_1, x_2, \dots, x_{20}$ with variance $\sigma^2 = 5$.
If each observation is multiplied by a constant $a$,the new variance becomes $a^2 \sigma^2$.
Here,$a = 3$,so the new variance is $3^2 \times 5 = 9 \times 5 = 45$.
Adding a constant $b$ to each observation does not change the variance.
Therefore,the final variance remains $45$.

(D) Let the original observations be $x_1, x_2, \dots, x_6$ with variance $\sigma^2 = 12$.
When each observation is multiplied by a constant $k$,the new variance $\sigma'^2$ is given by $\sigma'^2 = k^2 \sigma^2$.
Here,$k = 3$ and $\sigma^2 = 12$.
Therefore,the new variance is:
$\sigma'^2 = 3^2 \times 12$
$\sigma'^2 = 9 \times 12$
$\sigma'^2 = 108$

(A) Given: $n = 15$,$\sum x = 170$,and $\sum x^2 = 2830$.
Corrected sum of observations: $\sum x = 170 - 20 + 30 = 180$.
Corrected sum of squares: $\sum x^2 = 2830 - (20)^2 + (30)^2 = 2830 - 400 + 900 = 3330$.
Corrected variance is given by the formula: $\sigma^2 = \frac{\sum x^2}{n} - \left(\frac{\sum x}{n}\right)^2$.
Substituting the values: $\sigma^2 = \frac{3330}{15} - \left(\frac{180}{15}\right)^2$.
$\sigma^2 = 222 - (12)^2 = 222 - 144 = 78$.

(C) First,we find the midpoints $(x_i)$ for each class interval ($C$.$I$.):
For $0-6$,$x_1 = \frac{0+6}{2} = 3$
For $6-12$,$x_2 = \frac{6+12}{2} = 9$
For $12-18$,$x_3 = \frac{12+18}{2} = 15$
Now,calculate $\sum f_i$,$\sum f_i x_i$,and $\sum f_i x_i^2$:
$\sum f_i = 2 + 4 + 6 = 12$
$\sum f_i x_i = (2 \times 3) + (4 \times 9) + (6 \times 15) = 6 + 36 + 90 = 132$
$\sum f_i x_i^2 = (2 \times 3^2) + (4 \times 9^2) + (6 \times 15^2) = (2 \times 9) + (4 \times 81) + (6 \times 225) = 18 + 324 + 1350 = 1692$
The variance $V(X)$ is given by:
$V(X) = \frac{\sum f_i x_i^2}{\sum f_i} - \left( \frac{\sum f_i x_i}{\sum f_i} \right)^2$
$V(X) = \frac{1692}{12} - \left( \frac{132}{12} \right)^2$
$V(X) = 141 - (11)^2$
$V(X) = 141 - 121 = 20$
Standard deviation $\sigma = \sqrt{V(X)} = \sqrt{20} = 2 \sqrt{5}$.

(D) The variance of a set of observations $x_i$ is defined as $\sigma_x^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$,where $\bar{x}$ is the mean.
When each observation is multiplied by a constant $\lambda$,the new mean becomes $\bar{x}' = \lambda \bar{x}$.
The new variance is $\sigma_{new}^2 = \frac{1}{n} \sum_{i=1}^{n} (\lambda x_i - \lambda \bar{x})^2$.
$\sigma_{new}^2 = \frac{1}{n} \sum_{i=1}^{n} \lambda^2 (x_i - \bar{x})^2$.
$\sigma_{new}^2 = \lambda^2 \left[ \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \right] = \lambda^2 \sigma_x^2$.
Therefore,the new variance is $\lambda^2 \sigma_x^2$.

(A) The mean of the given numbers is $\overline{x} = \frac{2+3+11+x}{4} = \frac{16+x}{4}$.
Variance is given by $\sigma^2 = \frac{1}{n} \sum (x_i - \overline{x})^2$.
$\frac{49}{4} = \frac{1}{4} [(\frac{16+x}{4} - 2)^2 + (\frac{16+x}{4} - 3)^2 + (\frac{16+x}{4} - 11)^2 + (\frac{16+x}{4} - x)^2]$.
$49 = (\frac{8+x}{4})^2 + (\frac{4+x}{4})^2 + (\frac{x-28}{4})^2 + (\frac{16-3x}{4})^2$.
$49 \times 16 = (64 + x^2 + 16x) + (16 + x^2 + 8x) + (x^2 - 56x + 784) + (256 - 96x + 9x^2)$.
$784 = 12x^2 - 128x + 1120$.
$12x^2 - 128x + 336 = 0$.
Dividing by $4$,we get $3x^2 - 32x + 84 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{32 \pm \sqrt{1024 - 1008}}{6} = \frac{32 \pm 4}{6}$.
Thus,$x = \frac{36}{6} = 6$ or $x = \frac{28}{6} = \frac{14}{3}$.

(A) Let the original data be $X = \{2, 4, 5, 6, 8, 17\}$.
The variance of $X$ is given as $Var(X) = 23.33$.
The new data $Y = \{4, 8, 10, 12, 16, 34\}$ is obtained by multiplying each element of $X$ by $2$,i.e.,$Y = 2X$.
We know that if $Y = aX$,then $Var(Y) = a^2 \times Var(X)$.
Here,$a = 2$.
Therefore,$Var(Y) = (2)^2 \times 23.33 = 4 \times 23.33 = 93.32$.

(B) To determine which division has more uniformity,we calculate the Coefficient of Variation ($C$.$V$.) for each division. The formula for $C$.$V$. is $\text{C.V.} = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100$. $A$ lower $C$.$V$. indicates greater uniformity.\\
For division $A$: $\text{C.V.}_A = \frac{12}{80} = 0.15$ or $15\%$.\\
For division $B$: $\text{C.V.}_B = \frac{6}{75} = 0.08$ or $8\%$.\\
For division $C$: $\text{C.V.}_C = \frac{8}{70} \approx 0.114$ or $11.4\%$.\\
For division $D$: $\text{C.V.}_D = \frac{10}{72} \approx 0.139$ or $13.9\%$.\\
Since the $C$.$V$. is the least for division $B$,it has the most uniformity.

(D) Given $n = 16$ and $\Sigma x_i = 528$. \\ The mean $\overline{x} = \frac{\Sigma x_i}{n} = \frac{528}{16} = 33$. \\ The sum of the squares of deviations from the mean $33$ is given as $\Sigma(x_i - 33)^2 = 9158$. \\ The variance $\sigma^2$ is defined as $\frac{1}{n} \Sigma(x_i - \overline{x})^2$. \\ Therefore,$\sigma^2 = \frac{9158}{16} = 572.375$.

(D) The formula for the coefficient of variation is given by:
$\text{Coefficient of variation} = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100$
Given,$\text{Standard Deviation} = 12$ and $\text{Mean} = 72$.
Substituting the values:
$\text{Coefficient of variation} = \frac{12}{72} \times 100 \% = \frac{1}{6} \times 100 \% = 16.67 \%$

(A) Given: $n = 50$,$\Sigma x_i^2 = 3050$,and $\bar{x} = 6$.
The formula for standard deviation ($S$.$D$.) is $\sigma = \sqrt{\frac{1}{n} \Sigma x_i^2 - (\bar{x})^2}$.
Substituting the given values:
$\sigma = \sqrt{\frac{3050}{50} - (6)^2}$
$\sigma = \sqrt{61 - 36}$
$\sigma = \sqrt{25}$
$\sigma = 5$.
Thus,the standard deviation is $5$.

(D) The variance $\sigma^2$ of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - (\bar{x})^2$.
Here,$\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\bar{x} = \frac{n+1}{2}$.
Substituting these values:
$\sigma^2 = \frac{1}{n} \left( \frac{n(n+1)(2n+1)}{6} \right) - \left( \frac{n+1}{2} \right)^2$
$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$
Taking $\frac{n+1}{2}$ as a common factor:
$\sigma^2 = \frac{n+1}{2} \left( \frac{2n+1}{3} - \frac{n+1}{2} \right)$
$\sigma^2 = \frac{n+1}{2} \left( \frac{4n+2 - 3n-3}{6} \right)$
$\sigma^2 = \frac{n+1}{2} \left( \frac{n-1}{6} \right)$
$\sigma^2 = \frac{n^2-1}{12}$

(A) The standard deviation ($S$.$D$.) of the first $n$ natural numbers is given by the formula: $\sigma = \sqrt{\frac{n^2-1}{12}}$.
Given that $\sigma = 2$,we have:
$2 = \sqrt{\frac{n^2-1}{12}}$
Squaring both sides:
$4 = \frac{n^2-1}{12}$
$48 = n^2 - 1$
$n^2 = 49$
$n = 7$ (since $n$ must be a natural number).
Thus,the value of $n$ is $7$.

(C) Given $\sum x = 12$,$\sum x^2 = 16.9$,and $n = 10$.
The formula for standard deviation is $S.D. = \sqrt{\frac{\sum x^2}{n} - (\frac{\sum x}{n})^2}$.
Substituting the values:
$S.D. = \sqrt{\frac{16.9}{10} - (\frac{12}{10})^2}$
$S.D. = \sqrt{1.69 - (1.2)^2}$
$S.D. = \sqrt{1.69 - 1.44}$
$S.D. = \sqrt{0.25}$
$S.D. = 0.5$.

(C) We know that the Coefficient of Variation ($C$.$V$.) is given by the formula:
$C.V. = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100$
Calculating the $C$.$V$. for each subject:
$1. (C.V.)_{\text{Physics}} = \frac{3}{20} = 0.15$
$2. (C.V.)_{\text{Chemistry}} = \frac{2}{25} = 0.08$
$3. (C.V.)_{\text{Mathematics}} = \frac{4}{23} \approx 0.174$
$4. (C.V.)_{\text{Biology}} = \frac{5}{27} \approx 0.185$
Comparing the values,the highest variability is observed in Biology since it has the highest Coefficient of Variation $(0.185)$.

(B) The first six prime numbers greater than $5$ are $7, 11, 13, 17, 19, 23$.
The mean $\bar{x} = \frac{7+11+13+17+19+23}{6} = \frac{90}{6} = 15$.
The variance $\sigma^2$ is given by $\frac{1}{n} \sum (x_i - \bar{x})^2$.
$\sigma^2 = \frac{(7-15)^2 + (11-15)^2 + (13-15)^2 + (17-15)^2 + (19-15)^2 + (23-15)^2}{6}$
$\sigma^2 = \frac{(-8)^2 + (-4)^2 + (-2)^2 + (2)^2 + (4)^2 + (8)^2}{6}$
$\sigma^2 = \frac{64 + 16 + 4 + 4 + 16 + 64}{6} = \frac{168}{6} = 28$.

(D) Let the original observations be $x_1, x_2, \dots, x_6$ with variance $\sigma^2 = 16$.
When each observation is multiplied by a constant $\lambda$,the new variance $\sigma'^2$ is given by $\sigma'^2 = \lambda^2 \sigma^2$.
Here,$\lambda = 3$ and $\sigma^2 = 16$.
Therefore,the new variance is:
$\sigma'^2 = 3^2 \times 16$
$\sigma'^2 = 9 \times 16$
$\sigma'^2 = 144$

(B) The formula for variance is $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - \bar{x}^2$.
Here,$n = 4$ and variance $\sigma^2 = 5$.
The mean $\bar{x} = \frac{-1 + 0 + 1 + k}{4} = \frac{k}{4}$.
Substituting the values into the formula:
$5 = \frac{1}{4} [(-1)^2 + 0^2 + 1^2 + k^2] - (\frac{k}{4})^2$.
$5 = \frac{2 + k^2}{4} - \frac{k^2}{16}$.
Multiply the entire equation by $16$ to clear the denominators:
$80 = 4(2 + k^2) - k^2$.
$80 = 8 + 4k^2 - k^2$.
$80 - 8 = 3k^2$.
$72 = 3k^2$.
$k^2 = 24$.
Since $k > 0$,$k = \sqrt{24} = 2 \sqrt{6}$.

(A) The standard deviation is independent of the change of origin.
$\therefore$ $S$.$D$. of $x_{i} = \text{S.D. of } (x_{i}-2)$.
Let $y_{i} = x_{i}-2$. Then $\sum y_{i} = 20$ and $\sum y_{i}^2 = 100$ for $n = 20$.
$\text{S.D. of } y = \sqrt{\frac{\sum y_{i}^2}{n} - \left(\frac{\sum y_{i}}{n}\right)^2}$.
$\text{S.D. of } y = \sqrt{\frac{100}{20} - \left(\frac{20}{20}\right)^2}$.
$\text{S.D. of } y = \sqrt{5 - 1^2} = \sqrt{4} = 2$.
Thus,the standard deviation of $x$ is $2$.

(D) The terms are $x_i = (i-1)d$ for $i=1, 2, \ldots, n$.
$\bar{x} = \frac{1}{n} \sum_{i=1}^{n} (i-1)d = \frac{d}{n} \frac{(n-1)n}{2} = \frac{(n-1)d}{2}$.
$\sum x_i^2 = \sum_{i=1}^{n} (i-1)^2 d^2 = d^2 \sum_{k=0}^{n-1} k^2 = d^2 \frac{(n-1)n(2n-1)}{6}$.
$\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2 = \frac{d^2(n-1)(2n-1)}{6} - \frac{(n-1)^2 d^2}{4}$.
$\sigma^2 = \frac{d^2(n-1)}{2} \left[ \frac{2n-1}{3} - \frac{n-1}{2} \right] = \frac{d^2(n-1)}{2} \left[ \frac{4n-2-3n+3}{6} \right] = \frac{d^2(n-1)(n+1)}{12} = \frac{(n^2-1)d^2}{12}$.

(D) The variance $\sigma^2$ of the first $N$ natural numbers is given by $\sigma^2 = \frac{N^2-1}{12}$.
Here,$N = 2n$.
Substituting $N = 2n$ into the formula:
$\sigma^2 = \frac{(2n)^2-1}{12}$
$\sigma^2 = \frac{4n^2-1}{12}$

(A) Given $n = 15$,$\sum X^2 = 2830$,and $\sum X = 170$.
Incorrect observation $= 20$,Correct observation $= 30$.
Corrected $\sum X = 170 - 20 + 30 = 180$.
Corrected $\sum X^2 = 2830 - (20)^2 + (30)^2 = 2830 - 400 + 900 = 3330$.
Variance $\sigma^2 = \frac{\sum X^2}{n} - \left(\frac{\sum X}{n}\right)^2$.
$\sigma^2 = \frac{3330}{15} - \left(\frac{180}{15}\right)^2$.
$\sigma^2 = 222 - (12)^2$.
$\sigma^2 = 222 - 144 = 78$.

(C) Let the observations be $x_1, x_2, \dots, x_{15}$.
Given,mean $\bar{x} = 10$ and variance $\sigma^2 = 6$.
When each observation is increased by a constant $k=8$,the new observations are $x_i' = x_i + 8$.
The new mean $\bar{x}' = \bar{x} + 8 = 10 + 8 = 18$.
The variance is independent of the change of origin,so the new variance $\sigma'^2 = \sigma^2 = 6$.
Therefore,the new variance and new mean are $6$ and $18$ respectively.

(C) The first $10$ multiples of $3$ are $3, 6, 9, \ldots, 30$.
Let $x_i = 3i$ for $i = 1, 2, \ldots, 10$.
The variance $\sigma^2$ is given by $\frac{\sum x_i^2}{n} - (\bar{x})^2$.
$\bar{x} = \frac{3(1+2+\ldots+10)}{10} = \frac{3 \times 10 \times 11}{10 \times 2} = 16.5$.
$\sum x_i^2 = 3^2(1^2+2^2+\ldots+10^2) = 9 \times \frac{10(11)(21)}{6} = 9 \times 385 = 3465$.
$\sigma^2 = \frac{3465}{10} - (16.5)^2 = 346.5 - 272.25 = 74.25$.

(D) Given $n = 15$,$\sum x^2 = 2830$,and $\sum x = 170$.
The incorrect observation is $20$ and the correct observation is $30$.
Corrected $\sum x = 170 - 20 + 30 = 180$.
Corrected $\sum x^2 = 2830 - (20)^2 + (30)^2 = 2830 - 400 + 900 = 3330$.
Corrected Mean $\bar{x} = \frac{180}{15} = 12$.
Corrected Variance $\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 = \frac{3330}{15} - (12)^2$.
$\sigma^2 = 222 - 144 = 78$.

(A) The first $10$ natural numbers are $1, 2, 3, \ldots, 10$.
Adding $1$ to each,we get the new set of numbers: $2, 3, 4, \ldots, 11$.
We know that the variance of a set of numbers remains unchanged if a constant is added to each term.
The variance of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{n^2 - 1}{12}$.
Here,$n = 10$.
$\sigma^2 = \frac{10^2 - 1}{12} = \frac{100 - 1}{12} = \frac{99}{12} = 8.25$.

(D) The given observations are $a, b, c, d, e$ with mean $m$ and standard deviation $S$.
Standard deviation is defined as $S = \sqrt{\frac{\sum (x_i - m)^2}{n}}$.
When a constant $k$ is added to each observation,the new observations are $x_i' = x_i + k$.
The new mean $m'$ is $\frac{\sum (x_i + k)}{n} = \frac{\sum x_i}{n} + k = m + k$.
The new standard deviation $S'$ is $\sqrt{\frac{\sum (x_i' - m')^2}{n}}$.
Substituting the values,$S' = \sqrt{\frac{\sum ((x_i + k) - (m + k))^2}{n}} = \sqrt{\frac{\sum (x_i - m)^2}{n}}$.
Thus,$S' = S$.
Adding a constant to each observation does not change the standard deviation.

(A) We know that the formula for standard deviation $\sigma$ is given by:
$\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2$
Given:
$n = 100$
$\bar{x} = 50$
$\sigma = 5$
Substituting the values:
$5^2 = \frac{\Sigma x_i^2}{100} - (50)^2$
$25 = \frac{\Sigma x_i^2}{100} - 2500$
$\frac{\Sigma x_i^2}{100} = 2500 + 25 = 2525$
$\Sigma x_i^2 = 2525 \times 100 = 252500$
Thus,the sum of squares of all observations is $252500$.

(C) The given numbers are $31, 32, 33, \ldots, 47$.
Subtracting $30$ from each term,we get the sequence $1, 2, 3, \ldots, 17$.
The standard deviation remains unchanged when a constant is subtracted from each term.
The formula for the standard deviation of the first $n$ natural numbers is $SD = \sqrt{\frac{n^{2}-1}{12}}$.
Here,$n = 17$.
$SD = \sqrt{\frac{17^{2}-1}{12}} = \sqrt{\frac{289-1}{12}} = \sqrt{\frac{288}{12}}$.
$SD = \sqrt{24} = 2 \sqrt{6}$.

(D) Given,the numbers are $-1, 0, 1, k$.
Standard deviation,$\sigma = \sqrt{5}$.
We know that the variance $\sigma^2$ is given by $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$.
Here,$n = 4$.
$\sigma^2 = 5 = \frac{(-1)^2 + 0^2 + 1^2 + k^2}{4} - \left(\frac{-1 + 0 + 1 + k}{4}\right)^2$.
$5 = \frac{2 + k^2}{4} - \left(\frac{k}{4}\right)^2$.
$5 = \frac{2 + k^2}{4} - \frac{k^2}{16}$.
Multiply the entire equation by $16$:
$80 = 4(2 + k^2) - k^2$.
$80 = 8 + 4k^2 - k^2$.
$72 = 3k^2$.
$k^2 = 24$.
Since $k > 0$,$k = \sqrt{24} = 2 \sqrt{6}$.

(C) Given: $n = 100$,$\bar{x} = 50$,and $\sigma = 4$.
The formula for standard deviation is $\sigma = \sqrt{\frac{\sum x_i^2}{n} - (\bar{x})^2}$.
Substituting the given values:
$4 = \sqrt{\frac{\sum x_i^2}{100} - (50)^2}$
Squaring both sides:
$16 = \frac{\sum x_i^2}{100} - 2500$
$\frac{\sum x_i^2}{100} = 2500 + 16 = 2516$
$\sum x_i^2 = 2516 \times 100 = 251600$.
Thus,the sum of the squares of all items is $251600$.

(A) Given data: $6, 7, 8, 9, 10$
Mean $(\bar{x})$ = $\frac{6+7+8+9+10}{5} = \frac{40}{5} = 8$
Standard Deviation $(SD)$ = $\sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}$
$SD$ = $\sqrt{\frac{(6-8)^2 + (7-8)^2 + (8-8)^2 + (9-8)^2 + (10-8)^2}{5}}$
$SD$ = $\sqrt{\frac{(-2)^2 + (-1)^2 + (0)^2 + (1)^2 + (2)^2}{5}}$
$SD$ = $\sqrt{\frac{4 + 1 + 0 + 1 + 4}{5}} = \sqrt{\frac{10}{5}} = \sqrt{2}$