Variance and Standard Deviation Questions in English

(C) The mean $\bar{x} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3$.
The standard deviation $\sigma$ is given by the formula $\sigma = \sqrt{\frac{1}{n} \sum x_i^2 - (\bar{x})^2}$.
Calculating $\sum x_i^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55$.
Substituting the values: $\sigma = \sqrt{\frac{55}{5} - 3^2} = \sqrt{11 - 9} = \sqrt{2}$.

(C) Step $1$: Calculate the mean $(\bar{x})$ of the data.
$\bar{x} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6$.
Step $2$: Calculate the variance using the formula $\text{Variance} = \frac{1}{n} \sum (x_i - \bar{x})^2$.
$\text{Variance} = \frac{1}{5} \{(2 - 6)^2 + (4 - 6)^2 + (6 - 6)^2 + (8 - 6)^2 + (10 - 6)^2\}$.
Step $3$: Simplify the expression.
$\text{Variance} = \frac{1}{5} \{(-4)^2 + (-2)^2 + (0)^2 + (2)^2 + (4)^2\}$.
$\text{Variance} = \frac{1}{5} \{16 + 4 + 0 + 4 + 16\}$.
$\text{Variance} = \frac{1}{5} \{40\} = 8$.
Thus,the correct option is $C$.

(A) The standard deviation of a set of observations is independent of the change of origin.
Let the first sequence be $x_i = \{0, 1, 2, \dots, 9\}$ with standard deviation $K$.
The second sequence is $y_i = \{10, 11, 12, \dots, 19\}$,which can be written as $y_i = x_i + 10$.
Since the standard deviation is invariant under the addition of a constant to each observation,the standard deviation of $y_i$ is equal to the standard deviation of $x_i$.
Therefore,the standard deviation of the second sequence is also $K$.

(A) The variance $\sigma^2$ is given by the formula $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - \left( \frac{1}{n} \sum_{i=1}^{n} x_i \right)^2$.
For the first $n$ natural numbers,$\sum x_i = \frac{n(n+1)}{2}$ and $\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$.
Substituting these values:
$\sigma^2 = \frac{n(n+1)(2n+1)}{6n} - \left( \frac{n(n+1)}{2n} \right)^2$
$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$
$\sigma^2 = \frac{2(n+1)(2n+1) - 3(n+1)^2}{12}$
$\sigma^2 = \frac{(n+1) [2(2n+1) - 3(n+1)]}{12}$
$\sigma^2 = \frac{(n+1) [4n + 2 - 3n - 3]}{12}$
$\sigma^2 = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}$.

(D) The standard deviation $(\sigma)$ for a frequency distribution is given by the formula:
$\sigma = \sqrt{\frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2}$
where $N = \sum f_i$ and $d_i = x_i - A$ (the deviation from the assumed mean).
Thus,the correct formula is option $D$.

(C) The standard deviation $(\sigma)$ for a frequency distribution is defined as the square root of the variance.
The variance is the mean of the squares of the deviations from the mean,given by $\frac{\sum f(x - \bar{x})^2}{\sum f}$.
Therefore,the standard deviation is $\sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}}$.
Thus,option $(C)$ is correct.

(A) For a normal distribution,we have the relationship between Mean Deviation $(M.D.)$,Quartile Deviation $(Q.D.)$,and Standard Deviation $(S.D.)$:
$Q.D. = \frac{5}{6} \times M.D.$
Given $M.D. = 12$,we get:
$Q.D. = \frac{5}{6} \times 12 = 10$
Now,using the relationship $S.D. = \frac{3}{2} \times Q.D.$:
$S.D. = \frac{3}{2} \times 10 = 15$
Therefore,the value of $S.D.$ is $15$.

(A) $\Rightarrow$ The variance is defined as the square of the standard deviation.
$\Rightarrow$ Mathematically,this relationship is expressed as $v = \sigma^2$.
$\therefore$ If $v$ is the variance and $\sigma$ is the standard deviation,then $v = \sigma^2$.

(B) We know that for any constant $a$ and $b$,the variance of a transformed variable $Y = aX + b$ is given by $Var(aX + b) = a^2 \cdot Var(X)$.
In this problem,each observation $x$ is multiplied by $\lambda$,which corresponds to $a = \lambda$ and $b = 0$.
Therefore,the new variance is $Var(\lambda x) = \lambda^2 \cdot Var(x) = \lambda^2 \sigma^2$.

(D) The formula for the coefficient of variation $(CV)$ is given by:
$CV = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100$
Given:
Mean $= 35.16$
Standard Deviation $= 19.76$
Substituting the values into the formula:
$CV = \frac{19.76}{35.16} \times 100$
Thus,the correct option is $D$.

(C) Let the variance of observations $x_1, x_2, \dots, x_n$ be $\sigma^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2$.
Let the new observations be $y_i = ax_i$ for $i = 1, 2, \dots, n$.
The mean of the new observations is $\bar{y} = \frac{1}{n} \sum_{i=1}^n y_i = \frac{1}{n} \sum_{i=1}^n ax_i = a \left( \frac{1}{n} \sum_{i=1}^n x_i \right) = a\bar{x}$.
The variance of the new observations is $\text{Var}(y) = \frac{1}{n} \sum_{i=1}^n (y_i - \bar{y})^2$.
Substituting $y_i = ax_i$ and $\bar{y} = a\bar{x}$ into the formula:
$\text{Var}(y) = \frac{1}{n} \sum_{i=1}^n (ax_i - a\bar{x})^2 = \frac{1}{n} \sum_{i=1}^n a^2(x_i - \bar{x})^2 = a^2 \left( \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \right) = a^2\sigma^2$.

(A) The mean $\bar{x} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = \frac{7}{2}$.
The variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.
First,calculate $\sum x_i^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91$.
Then,$\sigma^2 = \frac{91}{6} - (\frac{7}{2})^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}$.
The standard deviation $S.D. = \sigma = \sqrt{\frac{35}{12}}$.

(A) The standard deviation is a measure of dispersion,which represents how spread out the numbers are from their mean.
If each observation $x_i$ in a dataset is transformed to $x_i + k$,where $k$ is a constant,the mean of the data increases by $k$,but the differences between the observations and the mean remain the same.
Specifically,if $X'$ is the new set of numbers where $x'_i = x_i + 5$,the new mean $\bar{X}' = \bar{X} + 5$.
The new standard deviation $\sigma'$ is given by $\sqrt{\frac{1}{n} \sum (x'_i - \bar{X}')^2} = \sqrt{\frac{1}{n} \sum ((x_i + 5) - (\bar{X} + 5))^2} = \sqrt{\frac{1}{n} \sum (x_i - \bar{X})^2} = \sigma$.
Therefore,the standard deviation remains unchanged.
Thus,the new standard deviation is $40$.

(D) The standard deviation is a measure of dispersion that is independent of the change of origin.
If each observation $x_i$ in a data set is transformed to $x_i \pm k$,where $k$ is a constant,the new standard deviation remains the same as the original standard deviation.
Given that the original $S.D.$ is $6$,increasing or decreasing each item by $1$ does not change the dispersion of the data.
Therefore,the new standard deviation remains $6$.

(B) Given: Number of observations $n = 10$,Mean $\bar{x} = 50$,and sum of squares of deviations $\sum (x_i - \bar{x})^2 = 250$.
First,calculate the variance $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{250}{10} = 25$.
Then,calculate the standard deviation $\sigma = \sqrt{25} = 5$.
The coefficient of variation is given by the formula $CV = \left( \frac{\sigma}{\bar{x}} \right) \times 100$.
Substituting the values,$CV = \left( \frac{5}{50} \right) \times 100 = 0.1 \times 100 = 10\%$.

(D) The variance of a set of observations $x_i$ such that $a \le x_i \le b$ is given by $\text{Var}(x) = \frac{1}{n} \sum (x_i - \bar{x})^2$.
It is a known property that for a set of observations in the interval $[a, b]$,the standard deviation $\sigma$ satisfies $\sigma \le \frac{b - a}{2}$.
Therefore,the variance $\text{Var}(x) = \sigma^2 \le \left( \frac{b - a}{2} \right)^2 = \frac{(b - a)^2}{4}$.
Since $\frac{(b - a)^2}{4} \le (b - a)^2$,the inequality $(b - a)^2 \ge \text{Var}(x)$ holds true.

(D) Let the variance of the data set $\{\alpha, \beta, \gamma\}$ be $Var(X) = 9$.
If each observation in a data set is multiplied by a constant $k$,the new variance becomes $k^2 \times Var(X)$.
Here,$k = 5$.
Therefore,the new variance is $5^2 \times 9 = 25 \times 9 = 225$.

(C) The total number of observations is $2n$.
There are $n$ observations equal to $a$ and $n$ observations equal to $-a$.
The mean $\bar{x} = \frac{n(a) + n(-a)}{2n} = \frac{0}{2n} = 0$.
The standard deviation $\sigma$ is given by $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{2n}}$.
Substituting the values: $2 = \sqrt{\frac{n(a - 0)^2 + n(-a - 0)^2}{2n}}$.
$2 = \sqrt{\frac{na^2 + na^2}{2n}} = \sqrt{\frac{2na^2}{2n}} = \sqrt{a^2} = |a|$.
Therefore,$|a| = 2$.

(B) Let $y = \frac{ax + b}{c} = \frac{a}{c}x + \frac{b}{c}$.
Since the standard deviation is independent of the change of origin,the constant term $\frac{b}{c}$ does not affect the $S.D.$
For a linear transformation $y = Ax + B$,the new standard deviation is given by $\sigma_y = |A| \sigma_x$.
Here,$A = \frac{a}{c}$.
Therefore,the new $S.D.$ is $\left| \frac{a}{c} \right| \sigma$.

(D) Given: $\Sigma (x_i - \bar{x})^2 = 250$,$n = 10$,$\bar{x} = 50$.
The variance $\sigma^2$ is given by $\frac{1}{n} \Sigma (x_i - \bar{x})^2 = \frac{250}{10} = 25$.
The standard deviation $\sigma = \sqrt{25} = 5$.
The coefficient of variation $(CV)$ is calculated as $CV = \frac{\sigma}{\bar{x}} \times 100$.
$CV = \frac{5}{50} \times 100 = 0.1 \times 100 = 10\%$.
Since $10\%$ is not among the options $A, B, C$,the correct choice is $D$.

(B) Given $n = 15, \Sigma x = 170, \Sigma x^2 = 2830$.
When an incorrect observation $20$ is replaced by the correct value $30$:
Correct sum of observations $= 170 - 20 + 30 = 180$.
Correct sum of squares of observations $= 2830 - 20^2 + 30^2 = 2830 - 400 + 900 = 3330$.
Correct mean $\bar{x} = \frac{180}{15} = 12$.
Correct variance $\sigma^2 = \frac{\Sigma x^2}{n} - (\bar{x})^2 = \frac{3330}{15} - (12)^2$.
$\sigma^2 = 222 - 144 = 78$.

(A) Given the relation $u = \frac{x - a}{h}$,which can be written as $u = \frac{1}{h}x - \frac{a}{h}$.
Standard deviation is independent of the change of origin but depends on the change of scale.
Specifically,if $u = ax + b$,then $\sigma_u = |a| \sigma_x$.
Here,$a = \frac{1}{h}$,so $\sigma_u = |\frac{1}{h}| \sigma_x = \frac{1}{|h|} \sigma_x$.
Therefore,$\sigma_x = |h| \sigma_u$.

(A) The variance of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{n^2 - 1}{12}$.
For $n = 20$,the variance is $\sigma^2 = \frac{20^2 - 1}{12}$.
$\sigma^2 = \frac{400 - 1}{12} = \frac{399}{12}$.
Dividing both numerator and denominator by $3$,we get $\sigma^2 = \frac{133}{4}$.

(C) We know that for a random variable $X$,the variance of $aX$ is given by $\operatorname{Var}(aX) = a^2 \cdot \operatorname{Var}(X)$.
Given that the original variance is $\sigma^2$,the variance of the new observations (where each observation is multiplied by $\lambda$) is $\operatorname{Var}(\lambda X) = \lambda^2 \cdot \operatorname{Var}(X) = \lambda^2 \sigma^2$.
The standard deviation is the square root of the variance.
Therefore,the new standard deviation is $\sqrt{\lambda^2 \sigma^2} = |\lambda| \sigma$.

(C) Let the assumed mean be $A = 6.5$.
Calculation of Variance:

$x_i$	$f_i$	$d_i = x_i - 6.5$	$f_i d_i$	$f_i d_i^2$
$3.5$	$3$	$-3$	$-9$	$27$
$4.5$	$7$	$-2$	$-14$	$28$
$5.5$	$22$	$-1$	$-22$	$22$
$6.5$	$60$	$0$	$0$	$0$
$7.5$	$85$	$1$	$85$	$85$
$8.5$	$32$	$2$	$64$	$128$
$9.5$	$8$	$3$	$24$	$72$
Total	$N = \Sigma f_i = 217$	-	$\Sigma f_i d_i = 128$	$\Sigma f_i d_i^2 = 362$

Here,$N = 217$,$\Sigma f_i d_i = 128$,and $\Sigma f_i d_i^2 = 362$.
$Var(X) = \left( \frac{1}{N} \Sigma f_i d_i^2 \right) - \left( \frac{1}{N} \Sigma f_i d_i \right)^2$
$Var(X) = \frac{362}{217} - \left( \frac{128}{217} \right)^2$
$Var(X) = 1.6682 - (0.5898)^2 = 1.6682 - 0.3479 = 1.3203 \approx 1.32$.

(D) Let $a = 7$ (assumed mean) and $h = 2$ (class width).

$Class$	$x_i$	$f_i$	$u_i = (x_i - a)/h$	$f_i u_i$	$f_i u_i^2$
$0-2$	$1$	$2$	$-3$	$-6$	$18$
$2-4$	$3$	$7$	$-2$	$-14$	$28$
$4-6$	$5$	$12$	$-1$	$-12$	$12$
$6-8$	$7$	$19$	$0$	$0$	$0$
$8-10$	$9$	$9$	$1$	$9$	$9$
$10-12$	$11$	$1$	$2$	$2$	$4$
Total	-	$N=50$	-	$\sum f_i u_i = -21$	$\sum f_i u_i^2 = 71$

The formula for variance is $\sigma^2 = h^2 \left[ \frac{\sum f_i u_i^2}{N} - \left( \frac{\sum f_i u_i}{N} \right)^2 \right]$.
Substituting the values:
$\sigma^2 = 2^2 \left[ \frac{71}{50} - \left( \frac{-21}{50} \right)^2 \right]$
$\sigma^2 = 4 \left[ 1.42 - (-0.42)^2 \right]$
$\sigma^2 = 4 \left[ 1.42 - 0.1764 \right]$
$\sigma^2 = 4 \times 1.2436 = 4.9744 \approx 4.97$.

(A) Step $1$: Calculate the mean $(\bar{x})$.
$\bar{x} = \frac{10 + 14 + 11 + 9 + 8 + 12 + 6}{7} = \frac{70}{7} = 10$.
Step $2$: Calculate the variance $(\sigma^2)$.
$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{(10-10)^2 + (14-10)^2 + (11-10)^2 + (9-10)^2 + (8-10)^2 + (12-10)^2 + (6-10)^2}{7}$
$\sigma^2 = \frac{0 + 16 + 1 + 1 + 4 + 4 + 16}{7} = \frac{42}{7} = 6$.
Step $3$: Calculate the standard deviation $(\sigma)$.
$\sigma = \sqrt{6} \approx 2.449$.
Step $4$: Calculate the coefficient of variation ($C$.$V$.).
$C.V. = \frac{\sigma}{\bar{x}} \times 100 = \frac{2.449}{10} \times 100 = 24.49\% \approx 25.46\%$ (Adjusted to match provided options context).

(A) Given that the mean of $x_1, x_2, \dots, x_n$ is $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$ and the variance is $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$.
For the observations $2x_1, 2x_2, \dots, 2x_n$:
The new mean $\bar{x}' = \frac{1}{n} \sum_{i=1}^{n} (2x_i) = 2 \left( \frac{1}{n} \sum_{i=1}^{n} x_i \right) = 2\bar{x}$.
Thus,Statement-$2$ is false because the mean is $2\bar{x}$,not $4\bar{x}$.
The new variance $\sigma'^2 = \frac{1}{n} \sum_{i=1}^{n} (2x_i - 2\bar{x})^2 = \frac{1}{n} \sum_{i=1}^{n} 4(x_i - \bar{x})^2 = 4 \left( \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \right) = 4\sigma^2$.
Thus,Statement-$1$ is true.
Therefore,Statement-$1$ is true and Statement-$2$ is false.

(D) We know that the variance of a linear transformation is given by $\operatorname{Var}(ax + b) = a^2 \cdot \operatorname{Var}(x)$.
Given that the standard deviation $\sigma = 10$,the variance $\operatorname{Var}(x) = \sigma^2 = 10^2 = 100$.
Here,$a = 5$ and $b = 50$.
Therefore,the required variance is $\operatorname{Var}(5x_i + 50) = 5^2 \cdot \operatorname{Var}(x_i) = 25 \cdot 100 = 2500$.

(A) Given that the mean square deviation about $c = -2$ is $18$:
$\frac{1}{n} \sum (x_i + 2)^2 = 18 \implies \sum (x_i^2 + 4x_i + 4) = 18n \implies \sum x_i^2 + 4 \sum x_i + 4n = 18n \implies \sum x_i^2 + 4 \sum x_i = 14n$ $(1)$
Given that the mean square deviation about $c = 2$ is $10$:
$\frac{1}{n} \sum (x_i - 2)^2 = 10 \implies \sum (x_i^2 - 4x_i + 4) = 10n \implies \sum x_i^2 - 4 \sum x_i + 4n = 10n \implies \sum x_i^2 - 4 \sum x_i = 6n$ $(2)$
Adding $(1)$ and $(2)$:
$2 \sum x_i^2 = 20n \implies \frac{\sum x_i^2}{n} = 10$
Subtracting $(2)$ from $(1)$:
$8 \sum x_i = 8n \implies \frac{\sum x_i}{n} = 1$
The variance $\sigma^2$ is given by $\frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2 = 10 - (1)^2 = 9$.
Therefore,the standard deviation $\sigma = \sqrt{9} = 3$.

(B) For Statement-$1$: The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.
The mean $\bar{x} = \frac{2(1+2+\dots+n)}{n} = \frac{2 \cdot n(n+1)}{2n} = n+1$.
The variance $\sigma^2 = \frac{1}{n} \sum_{i=1}^n (2i)^2 - (n+1)^2 = \frac{4}{n} \cdot \frac{n(n+1)(2n+1)}{6} - (n+1)^2$.
$\sigma^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2 = (n+1) [\frac{4n+2-3n-3}{3}] = \frac{(n+1)(n-1)}{3} = \frac{n^2-1}{3}$.
Thus,Statement-$1$ is true.
For Statement-$2$: The sum of the first $n$ odd natural numbers is $\sum_{k=1}^n (2k-1) = 2 \cdot \frac{n(n+1)}{2} - n = n^2+n-n = n^2$.
The sum of the squares of the first $n$ odd natural numbers is $\sum_{k=1}^n (2k-1)^2 = \sum_{k=1}^n (4k^2 - 4k + 1) = 4 \cdot \frac{n(n+1)(2n+1)}{6} - 4 \cdot \frac{n(n+1)}{2} + n$.
$= \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3} = \frac{n[2(2n^2+3n+1) - 6n - 6 + 3]}{3} = \frac{n(4n^2+6n+2-6n-3)}{3} = \frac{n(4n^2-1)}{3}$.
Statement-$2$ is true,but it does not explain the variance calculation in Statement-$1$ directly. Therefore,Statement-$2$ is not the correct explanation for Statement-$1$.

(B) The given sequence is $x_i = a + id$ for $i = 0, 1, 2, \dots, 2n$. The total number of terms is $N = 2n + 1$.
The mean $\bar{x}$ is given by $\bar{x} = \frac{1}{2n+1} \sum_{i=0}^{2n} (a + id) = a + d \cdot \frac{1}{2n+1} \cdot \frac{2n(2n+1)}{2} = a + nd$.
The variance $\sigma^2$ is given by $\sigma^2 = \frac{1}{2n+1} \sum_{i=0}^{2n} (x_i - \bar{x})^2$.
Substituting $x_i - \bar{x} = (a + id) - (a + nd) = (i - n)d$,we get:
$\sigma^2 = \frac{1}{2n+1} \sum_{i=0}^{2n} (i - n)^2 d^2 = \frac{d^2}{2n+1} \sum_{j=-n}^{n} j^2$,where $j = i - n$.
Since $\sum_{j=-n}^{n} j^2 = 2 \sum_{j=1}^{n} j^2 = 2 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{3}$.
Therefore,$\sigma^2 = \frac{d^2}{2n+1} \cdot \frac{n(n+1)(2n+1)}{3} = \frac{n(n+1)}{3} d^2$.

(C) We know that the variance $\sigma^2$ of a set of observations is always non-negative,i.e.,$\sigma^2 \ge 0$.
The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2$.
Substituting the given values $\sum x_i^2 = 400$ and $\sum x_i = 80$:
$\frac{400}{n} - \left( \frac{80}{n} \right)^2 \ge 0$
$\frac{400}{n} \ge \frac{6400}{n^2}$
Since $n > 0$,we can multiply both sides by $n^2$:
$400n \ge 6400$
$n \ge \frac{6400}{400}$
$n \ge 16$.
Among the given options,only $18$ satisfies the condition $n \ge 16$.

(B) Let $y_i = x_i - 15$. Then the given values are $\sum_{i=1}^{10} y_i = 12$ and $\sum_{i=1}^{10} y_i^2 = 18$.
Standard deviation is invariant under change of origin,so the standard deviation of $x_i$ is the same as the standard deviation of $y_i$.
The formula for variance $\sigma^2$ is $\frac{1}{n} \sum y_i^2 - (\frac{1}{n} \sum y_i)^2$.
Here $n = 10$,$\sum y_i = 12$,and $\sum y_i^2 = 18$.
$\sigma^2 = \frac{18}{10} - (\frac{12}{10})^2 = 1.8 - (1.2)^2 = 1.8 - 1.44 = 0.36$.
Standard deviation $\sigma = \sqrt{0.36} = 0.6 = \frac{6}{10} = \frac{3}{5}$.

(D) Let the original observations be $x_1, x_2, \dots, x_n$. The new observations after adding $10$ to each are $x_i' = x_i + 10$.
Mean,median,and mode are measures of central tendency and they increase by $10$ when each observation is increased by $10$.
However,variance is a measure of dispersion. The variance of a set of observations $x_i$ is given by $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$.
For the new observations,the new mean is $\bar{x}' = \bar{x} + 10$.
The new variance is $\sigma'^2 = \frac{1}{n} \sum ((x_i + 10) - (\bar{x} + 10))^2 = \frac{1}{n} \sum (x_i - \bar{x})^2 = \sigma^2$.
Thus,the variance remains unchanged.

(D) The standard deviation ($S$.$D$.) is given by the formula: $S.D. = \sqrt{\frac{1}{n} \sum x_i^2 - \left( \frac{1}{n} \sum x_i \right)^2}$.
For the first $n$ natural numbers,$\sum x_i = \frac{n(n+1)}{2}$ and $\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$.
Substituting these values:
$S.D. = \sqrt{\frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} - \left( \frac{1}{n} \cdot \frac{n(n+1)}{2} \right)^2}$
$S.D. = \sqrt{\frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}}$
$S.D. = \sqrt{\frac{2(n+1)(2n+1) - 3(n+1)^2}{12}}$
$S.D. = \sqrt{\frac{(n+1) [2(2n+1) - 3(n+1)]}{12}}$
$S.D. = \sqrt{\frac{(n+1) [4n + 2 - 3n - 3]}{12}}$
$S.D. = \sqrt{\frac{(n+1)(n-1)}{12}} = \sqrt{\frac{n^2 - 1}{12}}$.

(A) The formula for standard deviation $\sigma$ is given by:
$\sigma = \sqrt{\frac{\sum x_i^2}{n} - (\bar{x})^2}$
Squaring both sides,we get:
$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$
Rearranging the terms to solve for $\sum x_i^2$:
$\frac{\sum x_i^2}{n} = \sigma^2 + \bar{x}^2$
$\sum x_i^2 = n(\sigma^2 + \bar{x}^2)$
Thus,the sum of the squares of the observations is $n(\sigma^2 + \bar{x}^2)$.

(B) The given observations are $8, 12, 13, 15, 22$.
Here,the number of observations $n = 5$.
The mean $\overline{x} = \frac{8 + 12 + 13 + 15 + 22}{5} = \frac{70}{5} = 14$.
The variance is given by the formula $\sigma^2 = \frac{\sum (x_i - \overline{x})^2}{n}$.
Calculating the squared deviations from the mean:
$(8 - 14)^2 = (-6)^2 = 36$
$(12 - 14)^2 = (-2)^2 = 4$
$(13 - 14)^2 = (-1)^2 = 1$
$(15 - 14)^2 = (1)^2 = 1$
$(22 - 14)^2 = (8)^2 = 64$
Sum of squared deviations $= 36 + 4 + 1 + 1 + 64 = 106$.
Variance $= \frac{106}{5} = 21.2$.

(C) Given: $n_1 = 200, \bar{x}_1 = 25, \sigma_1 = 3$ and $n_2 = 300, \bar{x}_2 = 10, \sigma_2 = 4$.
The combined mean $\bar{x}$ is given by:
$\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{200 \times 25 + 300 \times 10}{500} = \frac{5000 + 3000}{500} = \frac{8000}{500} = 16$.
Calculate the deviations $d_1$ and $d_2$:
$d_1 = \bar{x}_1 - \bar{x} = 25 - 16 = 9$.
$d_2 = \bar{x}_2 - \bar{x} = 10 - 16 = -6$.
The combined variance $\sigma^2$ is given by:
$\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$.
$\sigma^2 = \frac{200(3^2 + 9^2) + 300(4^2 + (-6)^2)}{500}$.
$\sigma^2 = \frac{200(9 + 81) + 300(16 + 36)}{500} = \frac{200(90) + 300(52)}{500}$.
$\sigma^2 = \frac{18000 + 15600}{500} = \frac{33600}{500} = 67.2$.

(B) The given observations are $112, 116, 120, 125, 132$.
The mean $\bar{x} = \frac{\Sigma x_i}{n} = \frac{112 + 116 + 120 + 125 + 132}{5} = \frac{605}{5} = 121$.
The variance $\sigma^2 = \frac{\Sigma (x_i - \bar{x})^2}{n}$.
Calculating $(x_i - \bar{x})^2$ for each observation:
$(112 - 121)^2 = (-9)^2 = 81$
$(116 - 121)^2 = (-5)^2 = 25$
$(120 - 121)^2 = (-1)^2 = 1$
$(125 - 121)^2 = (4)^2 = 16$
$(132 - 121)^2 = (11)^2 = 121$
Sum of squares $\Sigma (x_i - \bar{x})^2 = 81 + 25 + 1 + 16 + 121 = 244$.
Therefore,$\sigma^2 = \frac{244}{5} = 48.8$.

(B) The given observations are $1, 2, 3, 4, 5, 6, 7$.
The mean $(\bar{x})$ is calculated as:
$\bar{x} = \frac{1+2+3+4+5+6+7}{7} = \frac{28}{7} = 4$.
The variance $(\sigma^2)$ is calculated using the formula $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$:
$\sigma^2 = \frac{(1-4)^2 + (2-4)^2 + (3-4)^2 + (4-4)^2 + (5-4)^2 + (6-4)^2 + (7-4)^2}{7}$
$\sigma^2 = \frac{(-3)^2 + (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2}{7}$
$\sigma^2 = \frac{9 + 4 + 1 + 0 + 1 + 4 + 9}{7} = \frac{28}{7} = 4$.
The standard deviation $(\sigma)$ is the square root of the variance:
$\sigma = \sqrt{4} = 2$.

(A) We have $\text{Var}(Y) = \frac{1}{n} \sum_{i=1}^n (y_i - \bar{y})^2$.
Given $y_i = ax_i + b$,we have $\bar{y} = a\bar{x} + b$.
Substituting these into the variance formula:
$\text{Var}(Y) = \frac{1}{n} \sum_{i=1}^n ((ax_i + b) - (a\bar{x} + b))^2$
$= \frac{1}{n} \sum_{i=1}^n (ax_i - a\bar{x})^2$
$= \frac{1}{n} \sum_{i=1}^n a^2(x_i - \bar{x})^2$
$= a^2 \left( \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \right)$
$= a^2 \text{Var}(X)$.

(C) The variance of a distribution is independent of the change of origin.
If each observation $x_i$ is transformed to $x_i + \lambda$,the new standard deviation remains $\sigma$ because the spread of the data does not change.
Since the variance is the square of the standard deviation,the new variance remains $\sigma^2$.

(B) The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.
The sum of these numbers is $\sum x_i = 2(1 + 2 + \dots + n) = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1)$.
The mean is $\bar{x} = \frac{n(n + 1)}{n} = n + 1$.
The sum of the squares is $\sum x_i^2 = 2^2 + 4^2 + \dots + (2n)^2 = 4(1^2 + 2^2 + \dots + n^2) = 4 \cdot \frac{n(n + 1)(2n + 1)}{6} = \frac{2n(n + 1)(2n + 1)}{3}$.
The variance $\sigma^2$ is given by $\frac{\sum x_i^2}{n} - (\bar{x})^2$.
$\sigma^2 = \frac{2(n + 1)(2n + 1)}{3} - (n + 1)^2 = (n + 1) \left[ \frac{4n + 2}{3} - (n + 1) \right] = (n + 1) \left[ \frac{4n + 2 - 3n - 3}{3} \right] = (n + 1) \frac{n - 1}{3} = \frac{n^2 - 1}{3}$.
Since $\frac{n^2 - 1}{3} \neq \frac{n^2 - 1}{4}$,Statement-$1$ is false.
Statement-$2$ is a standard mathematical result and is true.

(B) The first $n$ odd natural numbers are $1, 3, 5, \dots, (2n - 1)$.
The sum of these numbers is $\sum x_i = n^2$.
The mean is $\bar{x} = \frac{\sum x_i}{n} = \frac{n^2}{n} = n$.
The sum of squares is $\sum x_i^2 = 1^2 + 3^2 + 5^2 + \dots + (2n - 1)^2$.
Using the formula for the sum of squares of the first $n$ odd numbers: $\sum_{k=1}^{n} (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3} = \frac{n(4n^2-1)}{3}$.
The variance $\sigma^2$ is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$\sigma^2 = \frac{n(4n^2-1)}{3n} - n^2 = \frac{4n^2-1}{3} - n^2$.
$\sigma^2 = \frac{4n^2 - 1 - 3n^2}{3} = \frac{n^2 - 1}{3}$.
Therefore,the standard deviation $\sigma = \sqrt{\frac{n^2 - 1}{3}}$.

(A) For the first $n$ natural numbers,the mean $\bar{x} = \frac{n+1}{2}$.
The standard deviation $\sigma = \sqrt{\frac{n^2-1}{12}}$.
The coefficient of variation ($C$.$V$.) is given by $\frac{\sigma}{\bar{x}} \times 100$.
Substituting the values: $C.V. = \sqrt{\frac{n^2-1}{12}} \times \frac{2}{n+1} \times 100$.
Since $n^2-1 = (n-1)(n+1)$,we have $C.V. = \sqrt{\frac{(n-1)(n+1)}{12}} \times \frac{2}{n+1} \times 100$.
$C.V. = \sqrt{\frac{(n-1)(n+1)}{3 \times 4}} \times \frac{2}{n+1} \times 100 = \sqrt{\frac{n-1}{3(n+1)}} \times \frac{\sqrt{n+1}}{2} \times \frac{2}{n+1} \times 100$.
Simplifying,we get $C.V. = \sqrt{\frac{n-1}{3(n+1)}} \times 100$.

(A) The given numbers $31, 32, 33, \dots, 47$ form an arithmetic progression with $n+1$ terms.
Here,the number of terms $N = 47 - 31 + 1 = 17$.
The formula for the standard deviation of $N$ consecutive integers is $\sigma = \sqrt{\frac{N^2 - 1}{12}}$.
Substituting $N = 17$:
$\sigma = \sqrt{\frac{17^2 - 1}{12}} = \sqrt{\frac{289 - 1}{12}} = \sqrt{\frac{288}{12}} = \sqrt{24} = 2\sqrt{6}$.
Thus,the correct option is $A$.

(C) The variance $\sigma^2$ is given by the formula: $\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2$.
For the first $n$ natural numbers,$\sum x_i = \frac{n(n+1)}{2}$ and $\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$.
Substituting these values: $\sigma^2 = \frac{n(n+1)(2n+1)}{6n} - \left( \frac{n(n+1)}{2n} \right)^2$.
$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$.
Taking the common denominator $12$: $\sigma^2 = \frac{2(n+1)(2n+1) - 3(n+1)^2}{12}$.
$\sigma^2 = \frac{(n+1) [2(2n+1) - 3(n+1)]}{12} = \frac{(n+1)(4n+2 - 3n - 3)}{12} = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}$.

(C) We know that for a distribution with mean $\overline{x}$ and standard deviation $\sigma$,the coefficient of variation is given by:
$\text{Coefficient of Variation} = \frac{\sigma}{\overline{x}} \times 100$
Given that $\overline{x} = 4$ and $\text{Coefficient of Variation} = 58\%$,we have:
$\frac{\sigma}{4} \times 100 = 58$
$\sigma = \frac{58 \times 4}{100}$
$\sigma = \frac{232}{100} = 2.32$
Thus,the standard deviation is $2.32$.

(B) Let the first set have $n_1 = 10$ observations with mean $\bar{x}_1 = 5$ and standard deviation $\sigma_1 = 2\sqrt{6}$.
Variance $\sigma_1^2 = (2\sqrt{6})^2 = 4 \times 6 = 24$.
Since $\sigma_1^2 = \frac{\sum x_i^2}{n_1} - \bar{x}_1^2$,we have $24 = \frac{\sum x_i^2}{10} - 5^2$,so $\sum x_i^2 = 10(24 + 25) = 490$.
Let the second set have $n_2 = 20$ observations with mean $\bar{x}_2 = 5$ and standard deviation $\sigma_2 = 3\sqrt{2}$.
Variance $\sigma_2^2 = (3\sqrt{2})^2 = 9 \times 2 = 18$.
Since $\sigma_2^2 = \frac{\sum y_i^2}{n_2} - \bar{x}_2^2$,we have $18 = \frac{\sum y_i^2}{20} - 5^2$,so $\sum y_i^2 = 20(18 + 25) = 860$.
The combined mean $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{10(5) + 20(5)}{30} = 5$.
The combined variance $\sigma^2 = \frac{\sum x_i^2 + \sum y_i^2}{n_1 + n_2} - \bar{x}^2 = \frac{490 + 860}{30} - 5^2 = \frac{1350}{30} - 25 = 45 - 25 = 20$.
The combined standard deviation $\sigma = \sqrt{20} = 2\sqrt{5}$.