Variance and Standard Deviation Questions in English

(D) We know that the formula for the coefficient of variation $(CV)$ is given by:
$CV = \frac{\sigma}{\mu} \times 100$
where $\sigma$ is the standard deviation and $\mu$ is the arithmetic mean.
Given: $CV = 60$ and $\sigma = 21$.
Substituting these values into the formula:
$60 = \frac{21}{\mu} \times 100$
Rearranging to solve for $\mu$:
$\mu = \frac{21 \times 100}{60}$
$\mu = \frac{2100}{60}$
$\mu = 35$
Therefore,the arithmetic mean of the distribution is $35$.

(C) The variance of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{n^2 - 1}{12}$.
Given $\frac{n^2 - 1}{12} = 10$,we have $n^2 - 1 = 120$,so $n^2 = 121$,which gives $n = 11$.
The first $m$ even natural numbers are $2, 4, 6, \dots, 2m$. This is equivalent to $2 \times (1, 2, 3, \dots, m)$.
The variance of a set of numbers multiplied by a constant $k$ is $k^2$ times the original variance.
Thus,the variance of the first $m$ even natural numbers is $2^2 \times \frac{m^2 - 1}{12} = 4 \times \frac{m^2 - 1}{12} = \frac{m^2 - 1}{3}$.
Given $\frac{m^2 - 1}{3} = 16$,we have $m^2 - 1 = 48$,so $m^2 = 49$,which gives $m = 7$.
Therefore,$n: m = 11: 7$.

(D) Step $1$: Find the mid-points $(x_i)$ of each class interval:
$x_1 = 1, x_2 = 3, x_3 = 5, x_4 = 7, x_5 = 9$.
Step $2$: Calculate the mean $(\bar{x})$:
$\sum f_i = 2+3+5+3+2 = 15$.
$\sum f_i x_i = (2 \times 1) + (3 \times 3) + (5 \times 5) + (3 \times 7) + (2 \times 9) = 2 + 9 + 25 + 21 + 18 = 75$.
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{75}{15} = 5$.
Step $3$: Calculate the variance $(\sigma^2)$:
$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.
$\sum f_i (x_i - 5)^2 = 2(1-5)^2 + 3(3-5)^2 + 5(5-5)^2 + 3(7-5)^2 + 2(9-5)^2$.
$= 2(16) + 3(4) + 5(0) + 3(4) + 2(16) = 32 + 12 + 0 + 12 + 32 = 88$.
$\sigma^2 = \frac{88}{15}$.

(B) Step $1$: Find the midpoints $(x_i)$ of each class interval:
$x_1 = \frac{0+4}{2} = 2$,$x_2 = \frac{4+8}{2} = 6$,$x_3 = \frac{8+12}{2} = 10$,$x_4 = \frac{12+16}{2} = 14$.
Step $2$: Calculate the mean $(\bar{x})$:
Total frequency $N = \sum f_i = 1+2+2+1 = 6$.
Sum of $f_i x_i = (1 \times 2) + (2 \times 6) + (2 \times 10) + (1 \times 14) = 2 + 12 + 20 + 14 = 48$.
Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{48}{6} = 8$.
Step $3$: Calculate the variance $(\sigma^2)$:
$\sigma^2 = \frac{1}{N} \sum f_i (x_i - \bar{x})^2$.
$\sigma^2 = \frac{1}{6} [1(2-8)^2 + 2(6-8)^2 + 2(10-8)^2 + 1(14-8)^2]$.
$\sigma^2 = \frac{1}{6} [1(-6)^2 + 2(-2)^2 + 2(2)^2 + 1(6)^2]$.
$\sigma^2 = \frac{1}{6} [36 + 8 + 8 + 36] = \frac{88}{6} = \frac{44}{3}$.
Thus,the variance is $\frac{44}{3}$.

(A) Given: $n_1 = 8$,$\bar{x}_1 = 25$,$\sigma_1 = 5$.
Sum of items: $\Sigma x_i = n_1 \times \bar{x}_1 = 8 \times 25 = 200$.
Variance: $\sigma_1^2 = \frac{\Sigma x_i^2}{n_1} - (\bar{x}_1)^2 = 25$.
$\frac{\Sigma x_i^2}{8} - 625 = 25 \Rightarrow \Sigma x_i^2 = 8 \times 650 = 5200$.
New data set: $n_2 = 8 + 2 = 10$.
New sum: $\Sigma x_{new} = 200 + 15 + 25 = 240$.
New mean: $\bar{x}_2 = \frac{240}{10} = 24$.
New sum of squares: $\Sigma x_{new}^2 = 5200 + 15^2 + 25^2 = 5200 + 225 + 625 = 6050$.
New variance: $\sigma_2^2 = \frac{\Sigma x_{new}^2}{n_2} - (\bar{x}_2)^2 = \frac{6050}{10} - (24)^2 = 605 - 576 = 29$.

(B) Let the original data be $x_i = \{6, 7, 8, 9, 10, 11\}$ with variance $\sigma^2$.
The new data set is $y_i = \{12, 14, 16, 18, 20, 22\}$,which can be written as $y_i = 2x_i$.
Using the property of variance,if each observation is multiplied by a constant $k$,the new variance becomes $k^2 \times \text{original variance}$.
Here,$k = 2$,so the new variance is $2^2 \times \sigma^2 = 4 \sigma^2$.

(B) The first $50$ even natural numbers are $2, 4, 6, \ldots, 100$.
Mean,$\bar{x} = \frac{\sum_{i=1}^{50} x_i}{50} = \frac{2(1+2+3+\ldots+50)}{50} = \frac{2 \times \frac{50 \times 51}{2}}{50} = 51$.
Variance,$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$\sum x_i^2 = 2^2 + 4^2 + \ldots + 100^2 = 4(1^2 + 2^2 + \ldots + 50^2)$.
Using the sum of squares formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$\sum x_i^2 = 4 \times \frac{50(51)(101)}{6} = \frac{2 \times 50 \times 51 \times 101}{3} = 2 \times 50 \times 17 \times 101 = 171700$.
$\sigma^2 = \frac{171700}{50} - (51)^2 = 3434 - 2601 = 833$.

(D) Given observations are $10, 8, 5, a, b$.
Arithmetic Mean,$\bar{x} = \frac{10+8+5+a+b}{5} = 6$
$23 + a + b = 30 \implies a + b = 7$ ... $(i)$
Variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2 = 6.8$
$\frac{10^2 + 8^2 + 5^2 + a^2 + b^2}{5} - (6)^2 = 6.8$
$\frac{100 + 64 + 25 + a^2 + b^2}{5} - 36 = 6.8$
$\frac{189 + a^2 + b^2}{5} = 42.8$
$189 + a^2 + b^2 = 214$
$a^2 + b^2 = 25$ ... $(ii)$
We know that $(a + b)^2 = a^2 + b^2 + 2ab$
$(7)^2 = 25 + 2ab$
$49 = 25 + 2ab$
$2ab = 24 \implies ab = 12$

(B) Given,$\sum x = 170$ and $\sum x^2 = 2830$ for $n = 15$.
Replacing the wrong observation $20$ with the correct observation $30$:
Corrected $\sum x = 170 - 20 + 30 = 180$.
Corrected $\sum x^2 = 2830 - (20)^2 + (30)^2 = 2830 - 400 + 900 = 3330$.
We know that,$\text{Variance} = \frac{\sum x^2}{n} - \left(\frac{\sum x}{n}\right)^2$.
Substituting the values:
$\text{Variance} = \frac{3330}{15} - \left(\frac{180}{15}\right)^2$.
$\text{Variance} = 222 - (12)^2$.
$\text{Variance} = 222 - 144 = 78$.

(B) Given data: $6, 7, x-2, x, 18, 21$ (in ascending order).
Number of observations $n = 6$ (even).
Median = $\frac{(\frac{n}{2})^{\text{th}} \text{ observation} + (\frac{n}{2}+1)^{\text{th}} \text{ observation}}{2} = 16$.
$\frac{(x-2) + x}{2} = 16 \Rightarrow 2x - 2 = 32 \Rightarrow 2x = 34 \Rightarrow x = 17$.
The data set is $6, 7, 15, 17, 18, 21$.
Mean $\bar{x} = \frac{6+7+15+17+18+21}{6} = \frac{84}{6} = 14$.
Variance $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$.
Calculations:
$|\begin{array}{|c|c|c|} \hline x_i & x_i - \bar{x} & (x_i - \bar{x})^2 \\ \hline 6 & -8 & 64 \\ \hline 7 & -7 & 49 \\ \hline 15 & 1 & 1 \\ \hline 17 & 3 & 9 \\ \hline 18 & 4 & 16 \\ \hline 21 & 7 & 49 \\ \hline \text{Sum} & & 188 \\ \hline \end{array}|$
Variance = $\frac{188}{6} = \frac{94}{3} = 31 \frac{1}{3}$.

(C) Step $1$: Calculate the mean $(\bar{x})$ of the data.
$\bar{x} = \frac{2 + 12 + 3 + 11 + 5 + 10 + 6 + 7}{8} = \frac{56}{8} = 7$.
Step $2$: Calculate the squared deviations from the mean $(x_i - \bar{x})^2$.
$(2-7)^2 = 25$
$(12-7)^2 = 25$
$(3-7)^2 = 16$
$(11-7)^2 = 16$
$(5-7)^2 = 4$
$(10-7)^2 = 9$
$(6-7)^2 = 1$
$(7-7)^2 = 0$
Step $3$: Calculate the variance $(\sigma^2)$ using the formula $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$.
$\sigma^2 = \frac{25 + 25 + 16 + 16 + 4 + 9 + 1 + 0}{8} = \frac{96}{8} = 12$.

(A) Let $y_i = x_i - 5$. Then the given sums are $\sum_{i=1}^9 y_i = 9$ and $\sum_{i=1}^9 y_i^2 = 45$.
The variance of the observations $x_i$ is the same as the variance of $y_i$ because shifting the data by a constant does not change the variance.
The variance $\sigma^2$ is given by the formula $\sigma^2 = \frac{1}{n} \sum y_i^2 - (\bar{y})^2$.
Here,$n = 9$,$\sum y_i = 9$,and $\sum y_i^2 = 45$.
First,calculate the mean of $y$: $\bar{y} = \frac{1}{9} \sum_{i=1}^9 y_i = \frac{9}{9} = 1$.
Now,calculate the variance: $\sigma^2 = \frac{45}{9} - (1)^2 = 5 - 1 = 4$.
The standard deviation $\sigma$ is the square root of the variance: $\sigma = \sqrt{4} = 2$.

(D) Given the mean of the data $7, 8, 9, 7, 8, 7, \lambda, 8$ is $8$.
Mean $= \frac{7+8+9+7+8+7+\lambda+8}{8} = 8$
$\Rightarrow \frac{54+\lambda}{8} = 8$
$\Rightarrow 54+\lambda = 64$
$\Rightarrow \lambda = 10$
Now,the data set is $7, 8, 9, 7, 8, 7, 10, 8$.
Variance $(\sigma^2) = \frac{1}{n} \sum (x_i - \bar{x})^2$
Variance $= \frac{(7-8)^2 + (8-8)^2 + (9-8)^2 + (7-8)^2 + (8-8)^2 + (7-8)^2 + (10-8)^2 + (8-8)^2}{8}$
Variance $= \frac{(-1)^2 + 0^2 + 1^2 + (-1)^2 + 0^2 + (-1)^2 + 2^2 + 0^2}{8}$
Variance $= \frac{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0}{8} = \frac{8}{8} = 1$.

(C) Let the observations be $x_1, x_2, \dots, x_{20}$.
Given,variance $\sigma^2 = 5$.
We know that if each observation is multiplied by a constant $k$,the new variance $\sigma'^2$ is given by $\sigma'^2 = k^2 \times \sigma^2$.
Here,$k = 2$ and $\sigma^2 = 5$.
Therefore,the new variance $\sigma'^2 = (2)^2 \times 5 = 4 \times 5 = 20$.

(A) The variability of a data set is measured by its variance or standard deviation.
Since the average marks for both sections are equal,we compare their variances.
Variance of section $A = 64$.
Variance of section $B = 81$.
Since $81 > 64$,the variance of section $B$ is greater than the variance of section $A$.
Therefore,the variability of section $B >$ variability of section $A$.

(C) The first $10$ multiples of $4$ are $4, 8, 12, 16, 20, 24, 28, 32, 36, 40$.
Let $X = 4i$ where $i = 1, 2, \dots, 10$.
The standard deviation of the first $n$ natural numbers is $\sigma_n = \sqrt{\frac{n^2 - 1}{12}}$.
Since each term is multiplied by $4$,the standard deviation of the multiples is $4 \times \sigma_{10}$.
$\sigma_{10} = \sqrt{\frac{10^2 - 1}{12}} = \sqrt{\frac{99}{12}} = \sqrt{8.25}$.
Standard deviation $= 4 \times \sqrt{8.25} = \sqrt{16 \times 8.25} = \sqrt{132} \approx 11.489 \approx 11.5$.

(C) The given observations are $112, 116, 120, 125, 132$.
First,calculate the Arithmetic Mean $(\bar{x})$:
$\bar{x} = \frac{112 + 116 + 120 + 125 + 132}{5} = \frac{605}{5} = 121$.
Next,calculate the sum of the squares of the deviations from the mean:
$\sum(x_i - \bar{x})^2 = (112 - 121)^2 + (116 - 121)^2 + (120 - 121)^2 + (125 - 121)^2 + (132 - 121)^2$
$= (-9)^2 + (-5)^2 + (-1)^2 + (4)^2 + (11)^2$
$= 81 + 25 + 1 + 16 + 121 = 244$.
Finally,the variance $(\sigma^2)$ is given by:
$\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n} = \frac{244}{5} = 48.8$.

(C) Let the three angles of the triangle be $x, y, z$.
Since the sum of the three angles of a triangle is $180^{\circ}$,we have $x + y + z = 180^{\circ}$.
Given $x = 60^{\circ}$,then $y + z = 120^{\circ} \dots(1)$.
The variance of the angles is given by $\frac{x^2 + y^2 + z^2}{3} - (\frac{x+y+z}{3})^2 = 4614$.
Since $\frac{x+y+z}{3} = \frac{180^{\circ}}{3} = 60^{\circ}$,the variance is $\frac{60^2 + y^2 + z^2}{3} - 60^2 = 4614$.
$\frac{3600 + y^2 + z^2}{3} - 3600 = 4614$
$\frac{3600 + y^2 + z^2}{3} = 8214$
$3600 + y^2 + z^2 = 24642$
$y^2 + z^2 = 21042$.
We know $(y+z)^2 = y^2 + z^2 + 2yz$,so $120^2 = 21042 + 2yz$.
$14400 = 21042 + 2yz$
$2yz = -6642$ (Note: Variance formula $\frac{\sum x_i^2}{n} - \bar{x}^2$ is used).
Re-evaluating based on the provided variance value $4614$:
If $\frac{x^2+y^2+z^2}{3} = 4614$,then $3600 + y^2 + z^2 = 13842 \Rightarrow y^2 + z^2 = 10242$.
$(y+z)^2 - 2yz = 10242$ $\Rightarrow 14400 - 2yz = 10242$ $\Rightarrow 2yz = 4158$ $\Rightarrow yz = 2079$.
$(y-z)^2 = (y+z)^2 - 4yz = 14400 - 4(2079) = 14400 - 8316 = 6084$.
$y-z = \sqrt{6084} = 78^{\circ}$.
Solving $y+z=120$ and $y-z=78$,we get $2y = 198 \Rightarrow y = 99^{\circ}$ and $z = 21^{\circ}$.

(A) The variance measures the spread of data points around the mean. The formula for sample variance is $s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$.
For option $A$ $(1, 2, 3, 4, 5)$: Mean $\bar{x} = 3$. Variance $= \frac{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2}{5-1} = \frac{4+1+0+1+4}{4} = \frac{10}{4} = 2.5$.
For option $B$ $(1, 1, 2, 3, 6)$: Mean $\bar{x} = 2.6$. Variance $= \frac{(1-2.6)^2 + (1-2.6)^2 + (2-2.6)^2 + (3-2.6)^2 + (6-2.6)^2}{4} = \frac{2.56 + 2.56 + 0.36 + 0.16 + 11.56}{4} = \frac{17.2}{4} = 4.3$.
For option $C$ $(1, 1, 2, 3, 5)$: Mean $\bar{x} = 2.4$. Variance $= \frac{(1-2.4)^2 + (1-2.4)^2 + (2-2.4)^2 + (3-2.4)^2 + (5-2.4)^2}{4} = \frac{1.96 + 1.96 + 0.16 + 0.36 + 6.76}{4} = \frac{11.2}{4} = 2.8$.
For option $D$ $(1, 1, 2, 2, 5)$: Mean $\bar{x} = 2.2$. Variance $= \frac{(1-2.2)^2 + (1-2.2)^2 + (2-2.2)^2 + (2-2.2)^2 + (5-2.2)^2}{4} = \frac{1.44 + 1.44 + 0.04 + 0.04 + 7.84}{4} = \frac{10.8}{4} = 2.7$.
Comparing the variances $(2.5, 4.3, 2.8, 2.7)$,the minimum variance is $2.5$ for option $A$.

(D) Given data: $a, b, 8, 5, 10$.
Mean $= \frac{a+b+8+5+10}{5} = 6$.
$\Rightarrow a+b+23 = 30$ $\Rightarrow a+b = 7$ ...$(i)$
Variance $= \frac{\sum x_i^2}{n} - (\text{Mean})^2 = 6.80$.
$\frac{a^2+b^2+8^2+5^2+10^2}{5} - 6^2 = 6.80$.
$\frac{a^2+b^2+64+25+100}{5} - 36 = 6.80$.
$\frac{a^2+b^2+189}{5} = 42.80$.
$a^2+b^2+189 = 214 \Rightarrow a^2+b^2 = 25$ ...(ii)
From $(i)$,$b = 7-a$. Substituting in (ii):
$a^2 + (7-a)^2 = 25$.
$a^2 + 49 - 14a + a^2 = 25$.
$2a^2 - 14a + 24 = 0 \Rightarrow a^2 - 7a + 12 = 0$.
$(a-3)(a-4) = 0$.
So,$a=3, b=4$ or $a=4, b=3$.

(B) Given,$n=10$,$\Sigma x_i=60$,and $\Sigma x_i^2=1000$.
The mean $\bar{x} = \frac{\Sigma x_i}{n} = \frac{60}{10} = 6$.
The formula for variance is $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2$.
Substituting the values:
$\sigma^2 = \frac{1000}{10} - (6)^2$
$\sigma^2 = 100 - 36$
$\sigma^2 = 64$.

(C) Let $d_i = x_i - 5$. We are given $\Sigma d_i = 3$ and $\Sigma d_i^2 = 43$ with $n = 18$.
The variance of a distribution is invariant under change of origin.
Therefore,the variance of $x_i$ is the same as the variance of $d_i$.
The formula for variance is $\sigma^2 = \frac{\Sigma d_i^2}{n} - \left( \frac{\Sigma d_i}{n} \right)^2$.
Substituting the values: $\sigma^2 = \frac{43}{18} - \left( \frac{3}{18} \right)^2$.
$\sigma^2 = \frac{43}{18} - \left( \frac{1}{6} \right)^2 = \frac{43}{18} - \frac{1}{36}$.
$\sigma^2 = \frac{86 - 1}{36} = \frac{85}{36} \approx 2.3611$.
Thus,the variance is approximately $2.36$.

(A, B) Given the numbers are $2, 3, 2x, 11$. The mean $\bar{x} = \frac{2+3+2x+11}{4} = \frac{16+2x}{4} = 4 + \frac{x}{2}$.
Given standard deviation $\sigma = 3.5 = \frac{7}{2}$,so $\sigma^2 = \frac{49}{4}$.
The variance formula is $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.
$\frac{49}{4} = \frac{2^2 + 3^2 + (2x)^2 + 11^2}{4} - (4 + \frac{x}{2})^2$.
$\frac{49}{4} = \frac{4 + 9 + 4x^2 + 121}{4} - (16 + 4x + \frac{x^2}{4})$.
$\frac{49}{4} = \frac{134 + 4x^2}{4} - \frac{64 + 16x + x^2}{4}$.
$49 = 134 + 4x^2 - 64 - 16x - x^2$.
$49 = 3x^2 - 16x + 70$.
$3x^2 - 16x + 21 = 0$.
$(3x - 7)(x - 3) = 0$.
Thus,$x = \frac{7}{3}$ or $x = 3$.

(C) Given: $\bar{x} = 50$,$n = 100$,and $\sigma = 5$.
We know that $\bar{x} = \frac{\Sigma x_i}{n}$,so $\Sigma x_i = n \cdot \bar{x} = 100 \cdot 50 = 5000$.
The formula for standard deviation is $\sigma = \sqrt{\frac{\Sigma x_i^2}{n} - (\bar{x})^2}$.
Squaring both sides,we get $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2$.
Substituting the given values: $5^2 = \frac{\Sigma x_i^2}{100} - (50)^2$.
$25 = \frac{\Sigma x_i^2}{100} - 2500$.
$\frac{\Sigma x_i^2}{100} = 2525$.
$\Sigma x_i^2 = 252500$.

(D) First,calculate the mean $\bar{x}$:
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(5 \times 2) + (15 \times 3) + (25 \times 4) + (35 \times 1)}{2 + 3 + 4 + 1} = \frac{10 + 45 + 100 + 35}{10} = \frac{190}{10} = 19$.
Next,calculate the variance $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N}$:
$\sigma^2 = \frac{2(5-19)^2 + 3(15-19)^2 + 4(25-19)^2 + 1(35-19)^2}{10}$
$\sigma^2 = \frac{2(-14)^2 + 3(-4)^2 + 4(6)^2 + 1(16)^2}{10}$
$\sigma^2 = \frac{2(196) + 3(16) + 4(36) + 1(256)}{10}$
$\sigma^2 = \frac{392 + 48 + 144 + 256}{10} = \frac{840}{10} = 84$.
Thus,the variance is $84$.

(B) Given,Coefficient of Variation $(CV) = 60$ and Standard Deviation $(\sigma) = 21$.
We know that $CV = \frac{\sigma}{\mu} \times 100$,where $\mu$ is the mean.
Substituting the values: $60 = \frac{21}{\mu} \times 100 \Rightarrow \mu = \frac{2100}{60} = 35$.
When a constant $k = 15$ is added to every observation,the standard deviation remains unchanged,so $\sigma' = \sigma = 21$.
The new mean becomes $\mu' = \mu + 15 = 35 + 15 = 50$.
The new coefficient of variation is $CV' = \frac{\sigma'}{\mu'} \times 100$.
$CV' = \frac{21}{50} \times 100 = 21 \times 2 = 42$.
Thus,the correct option is $B$.

(B) Given that the standard deviations of $x_i$ and $y_i$ are $a$ and $b$ respectively,we have $a^2 = \frac{1}{10} \sum x_i^2 - \bar{x}^2$ and $b^2 = \frac{1}{10} \sum y_i^2 - \bar{y}^2$.
We are given $\sum_{i=1}^{10} (x_i - \bar{x})(y_i - \bar{y}) = c$.
Let $d_i = x_i - y_i$. The mean of $d_i$ is $\bar{d} = \bar{x} - \bar{y}$.
The variance of $d_i$ is $\sigma_d^2 = \frac{1}{10} \sum (d_i - \bar{d})^2 = \frac{1}{10} \sum ((x_i - y_i) - (\bar{x} - \bar{y}))^2$.
$\sigma_d^2 = \frac{1}{10} \sum ((x_i - \bar{x}) - (y_i - \bar{y}))^2 = \frac{1}{10} \sum (x_i - \bar{x})^2 + \frac{1}{10} \sum (y_i - \bar{y})^2 - \frac{2}{10} \sum (x_i - \bar{x})(y_i - \bar{y})$.
Since $\frac{1}{10} \sum (x_i - \bar{x})^2 = a^2$ and $\frac{1}{10} \sum (y_i - \bar{y})^2 = b^2$,and $\sum (x_i - \bar{x})(y_i - \bar{y}) = c$,we get:
$\sigma_d^2 = a^2 + b^2 - \frac{2c}{10} = a^2 + b^2 - \frac{c}{5}$.
Therefore,the standard deviation is $\sqrt{a^2 + b^2 - \frac{c}{5}}$.

(C) The given data set is $9, 3, 11, 5, 7$.
First,calculate the mean $(\bar{x})$:
$\bar{x} = \frac{9+3+11+5+7}{5} = \frac{35}{5} = 7$.
Next,calculate the variance $(\sigma^2)$:
$\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2 = \frac{(9-7)^2 + (3-7)^2 + (11-7)^2 + (5-7)^2 + (7-7)^2}{5}$
$= \frac{2^2 + (-4)^2 + 4^2 + (-2)^2 + 0^2}{5} = \frac{4 + 16 + 16 + 4 + 0}{5} = \frac{40}{5} = 8$.
The standard deviation $(\sigma)$ is $\sqrt{8} = 2\sqrt{2}$.
The coefficient of variation $(CV)$ is given by:
$CV = \frac{\sigma}{\bar{x}} \times 100 = \frac{2\sqrt{2}}{7} \times 100 = \frac{200\sqrt{2}}{7}$.

(D) First,calculate the mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
$\sum f_i = 1+2+3+4+5+6+7+8+9+10 = 55$.
$\sum f_i x_i = 1(1) + 2(2) + 3(3) + 4(4) + 5(5) + 6(6) + 7(7) + 8(8) + 9(9) + 10(10) = 1+4+9+16+25+36+49+64+81+100 = 385$.
$\bar{x} = \frac{385}{55} = 7$.
The variance $\sigma^2$ is given by $\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.
$\sum f_i (x_i - 7)^2 = 1(1-7)^2 + 2(2-7)^2 + 3(3-7)^2 + 4(4-7)^2 + 5(5-7)^2 + 6(6-7)^2 + 7(7-7)^2 + 8(8-7)^2 + 9(9-7)^2 + 10(10-7)^2$.
$= 1(36) + 2(25) + 3(16) + 4(9) + 5(4) + 6(1) + 7(0) + 8(1) + 9(4) + 10(9)$.
$= 36 + 50 + 48 + 36 + 20 + 6 + 0 + 8 + 36 + 90 = 330$.
$\sigma^2 = \frac{330}{55} = 6$.
Thus,the correct option is $D$.

(B) First,we calculate the mean $\bar{x}$ of the distribution:
The midpoints $x_i$ are $2.5, 7.5, 12.5, 17.5, 22.5$.
The sum of frequencies $N = \sum f_i = 4 + 1 + 10 + 3 + 2 = 20$.
The sum $\sum f_i x_i = (4 \times 2.5) + (1 \times 7.5) + (10 \times 12.5) + (3 \times 17.5) + (2 \times 22.5) = 10 + 7.5 + 125 + 52.5 + 45 = 240$.
Mean $\bar{x} = \frac{240}{20} = 12$.
Next,we calculate the variance $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N}$:
$\sigma^2 = \frac{4(2.5-12)^2 + 1(7.5-12)^2 + 10(12.5-12)^2 + 3(17.5-12)^2 + 2(22.5-12)^2}{20}$
$\sigma^2 = \frac{4(-9.5)^2 + 1(-4.5)^2 + 10(0.5)^2 + 3(5.5)^2 + 2(10.5)^2}{20}$
$\sigma^2 = \frac{4(90.25) + 20.25 + 10(0.25) + 3(30.25) + 2(110.25)}{20}$
$\sigma^2 = \frac{361 + 20.25 + 2.5 + 90.75 + 220.5}{20} = \frac{695}{20} = 34.75 = \frac{139}{4}$.
Standard deviation $\sigma = \sqrt{\frac{139}{4}} = \frac{\sqrt{139}}{2}$.
Coefficient of variation $CV = \frac{\sigma}{\bar{x}} \times 100 = \frac{\sqrt{139}/2}{12} \times 100 = \frac{\sqrt{139}}{24} \times 100 = \frac{25 \sqrt{139}}{6}$.
Thus,option $B$ is correct.

(C) Given,for $n=9$,$\bar{x} = 13$ and $\sigma = 5$.
$\sum_{i=1}^9 x_i = 9 \times 13 = 117$.
Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 \Rightarrow 25 = \frac{\sum x_i^2}{9} - 169$.
$\sum x_i^2 = 9(25 + 169) = 9(194) = 1746$.
Now,a new item $x_{10} = 3$ is added.
New sum $\sum_{i=1}^{10} x_i = 117 + 3 = 120$.
New mean $\bar{x}' = \frac{120}{10} = 12$.
New sum of squares $\sum_{i=1}^{10} x_i^2 = 1746 + (3)^2 = 1746 + 9 = 1755$.
New variance $\sigma'^2 = \frac{\sum x_i^2}{10} - (\bar{x}')^2 = \frac{1755}{10} - (12)^2 = 175.5 - 144 = 31.5$.

(B) To find the variance,we first calculate the mid-interval values $(x)$ and the frequency $(f)$:

Marks	$x$	$f$	$f \cdot x$	$f \cdot x^2$
$1-3$	$2$	$40$	$80$	$160$
$3-5$	$4$	$30$	$120$	$480$
$5-7$	$6$	$20$	$120$	$720$
$7-9$	$8$	$10$	$80$	$640$
Total		$100$	$400$	$2000$

The mean $\bar{x} = \frac{\sum f \cdot x}{\sum f} = \frac{400}{100} = 4$.
The variance $\sigma^2 = \frac{\sum f \cdot x^2}{\sum f} - (\bar{x})^2$.
$\sigma^2 = \frac{2000}{100} - (4)^2 = 20 - 16 = 4$.

(D) For student $A$: Marks are $30, 20, 40$.
Mean $\bar{x}_A = \frac{30+20+40}{3} = \frac{90}{3} = 30$.
Standard deviation $\sigma_A = \sqrt{\frac{(30-30)^2 + (20-30)^2 + (40-30)^2}{3}} = \sqrt{\frac{0 + 100 + 100}{3}} = \sqrt{\frac{200}{3}} = 10 \sqrt{\frac{2}{3}}$.
Coefficient of variation $(CV)_A = \frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{10 \sqrt{2}}{\sqrt{3} \times 30} \times 100 = \frac{\sqrt{2}}{3 \sqrt{3}} \times 100$.
For student $B$: Marks are $70, 0, 5$.
Mean $\bar{x}_B = \frac{70+0+5}{3} = \frac{75}{3} = 25$.
Standard deviation $\sigma_B = \sqrt{\frac{(70-25)^2 + (0-25)^2 + (5-25)^2}{3}} = \sqrt{\frac{45^2 + (-25)^2 + (-20)^2}{3}} = \sqrt{\frac{2025 + 625 + 400}{3}} = \sqrt{\frac{3050}{3}} = \sqrt{\frac{25 \times 122}{3}} = 5 \sqrt{\frac{122}{3}}$.
Coefficient of variation $(CV)_B = \frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{5 \sqrt{122}}{\sqrt{3} \times 25} \times 100 = \frac{\sqrt{122}}{5 \sqrt{3}} \times 100$.
Ratio $\frac{(CV)_A}{(CV)_B} = \frac{\sqrt{2}}{3 \sqrt{3}} \div \frac{\sqrt{122}}{5 \sqrt{3}} = \frac{\sqrt{2}}{3 \sqrt{3}} \times \frac{5 \sqrt{3}}{\sqrt{122}} = \frac{5 \sqrt{2}}{3 \sqrt{2} \sqrt{61}} = \frac{5}{3 \sqrt{61}}$.
Thus,the ratio is $5 : 3 \sqrt{61}$.

(C) The given scores are $505, 510, 515, \ldots, 595$. This is an arithmetic progression with $a = 505$,$d = 5$,and $n = 19$ terms.
The mean $\bar{X} = \frac{505 + 595}{2} = 550$.
The standard deviation $\sigma$ is given by $\sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{X})^2}$.
Let $x_i = 550 + 5k$,where $k$ ranges from $-9$ to $9$.
Then $(x_i - \bar{X})^2 = (5k)^2 = 25k^2$.
$\sum (x_i - \bar{X})^2 = 25 \sum_{k=-9}^{9} k^2 = 25 \times 2 \times \sum_{k=1}^{9} k^2$.
Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get $\sum_{k=1}^{9} k^2 = \frac{9(10)(19)}{6} = 285$.
So,$\sum (x_i - \bar{X})^2 = 50 \times 285 = 14250$.
$\sigma = \sqrt{\frac{14250}{19}} = \sqrt{750} = \sqrt{25 \times 30} = 5 \sqrt{30}$.

(C) Let the first distribution be $X$ with values $x_i$ and frequencies $f_i$. The variance is given by $\sigma_X^2 = 45.8$.
The second distribution $Y$ has values $y_i$ where $y_i = 2x_i + 2$.
We know that if $Y = aX + b$,then the variance $\sigma_Y^2 = a^2 \sigma_X^2$.
In this case,$y_i = 2x_i + 2$,so $a = 2$.
Therefore,$\sigma_Y^2 = 2^2 \times \sigma_X^2 = 4 \times 45.8$.
$\sigma_Y^2 = 183.2$.

(D) We know that the formula for standard deviation $(S.D.)$ is:
$S.D. = \sqrt{\frac{\sum(x_i - \mu)^2}{N}}$
where $\mu$ is the mean and $N$ is the number of terms.
First,calculate the mean $(\mu)$:
$\mu = \frac{22 + 26 + 28 + 20 + 24 + 30}{6} = \frac{150}{6} = 25$
Now,calculate the squared deviations:

$x_i$	$(x_i - \mu)^2$
$22$	$(22 - 25)^2 = 9$
$26$	$(26 - 25)^2 = 1$
$28$	$(28 - 25)^2 = 9$
$20$	$(20 - 25)^2 = 25$
$24$	$(24 - 25)^2 = 1$
$30$	$(30 - 25)^2 = 25$

Sum of squared deviations: $\sum(x_i - \mu)^2 = 9 + 1 + 9 + 25 + 1 + 25 = 70$
Standard Deviation: $S.D. = \sqrt{\frac{70}{6}} = \sqrt{11.666...} \approx 3.4156 \approx 3.42$

(D) For student $A$: Marks are $30, 20, 40$.
Mean $\bar{x}_A = \frac{30+20+40}{3} = \frac{90}{3} = 30$.
Variance $\sigma_A^2 = \frac{1}{3} \{(30-30)^2 + (20-30)^2 + (40-30)^2\} = \frac{1}{3} \{0 + 100 + 100\} = \frac{200}{3}$.
Standard Deviation $\sigma_A = \sqrt{\frac{200}{3}} = 10 \sqrt{\frac{2}{3}}$.
Coefficient of Variation $CV_A = \frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{\sqrt{200/3}}{30} \times 100$.
For student $B$: Marks are $70, 0, 5$.
Mean $\bar{x}_B = \frac{70+0+5}{3} = \frac{75}{3} = 25$.
Variance $\sigma_B^2 = \frac{1}{3} \{(70-25)^2 + (0-25)^2 + (5-25)^2\} = \frac{1}{3} \{45^2 + (-25)^2 + (-20)^2\} = \frac{1}{3} \{2025 + 625 + 400\} = \frac{3050}{3}$.
Standard Deviation $\sigma_B = \sqrt{\frac{3050}{3}} = \sqrt{\frac{25 \times 122}{3}} = 5 \sqrt{\frac{122}{3}}$.
Coefficient of Variation $CV_B = \frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{\sqrt{3050/3}}{25} \times 100$.
Ratio $\frac{CV_A}{CV_B} = \frac{\sigma_A / \bar{x}_A}{\sigma_B / \bar{x}_B} = \frac{\sigma_A}{\sigma_B} \times \frac{\bar{x}_B}{\bar{x}_A} = \frac{\sqrt{200/3}}{\sqrt{3050/3}} \times \frac{25}{30} = \sqrt{\frac{200}{3050}} \times \frac{5}{6} = \sqrt{\frac{4}{61}} \times \frac{5}{6} = \frac{2}{\sqrt{61}} \times \frac{5}{6} = \frac{5}{3 \sqrt{61}}$.

(A) $1$. Calculate the midpoints $(x_i)$ of each class interval: $5, 15, 25, 35$.
$2$. Calculate the mean $(\bar{x})$: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(1 \times 5) + (3 \times 15) + (4 \times 25) + (2 \times 35)}{1 + 3 + 4 + 2} = \frac{5 + 45 + 100 + 70}{10} = \frac{220}{10} = 22$.
$3$. Calculate the variance $(\sigma^2)$: $\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{\sum f_i}$.
$\sigma^2 = \frac{1(5-22)^2 + 3(15-22)^2 + 4(25-22)^2 + 2(35-22)^2}{10}$.
$\sigma^2 = \frac{1(-17)^2 + 3(-7)^2 + 4(3)^2 + 2(13)^2}{10} = \frac{289 + 147 + 36 + 338}{10} = \frac{810}{10} = 81$.
$4$. The standard deviation $(\sigma)$ is $\sqrt{81} = 9$.

(D) The first ten multiples of $3$ are $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$.
This is an arithmetic progression with $n = 10$ terms,where the first term $a = 3$ and common difference $d = 3$.
The variance of the first $n$ terms of an arithmetic progression with common difference $d$ is given by the formula $\sigma^2 = \frac{(n^2 - 1)d^2}{12}$.
Substituting the values $n = 10$ and $d = 3$:
$\sigma^2 = \frac{(10^2 - 1) \times 3^2}{12}$
$\sigma^2 = \frac{(100 - 1) \times 9}{12}$
$\sigma^2 = \frac{99 \times 9}{12} = \frac{891}{12} = 74.25$.
Thus,the variance is $74.25$.

(C) To calculate the variance,we use the assumed mean method where $a = 18$.
The calculations are as follows:

$x_i$	$f_i$	$d_i = x_i - 18$	$f_i d_i$	$f_i d_i^2$
$6$	$2$	$-12$	$-24$	$288$
$10$	$4$	$-8$	$-32$	$256$
$14$	$7$	$-4$	$-28$	$112$
$18$	$12$	$0$	$0$	$0$
$24$	$8$	$6$	$48$	$288$
$28$	$4$	$10$	$40$	$400$
$30$	$3$	$12$	$36$	$432$
Total	$N = 40$	-	$\sum f_i d_i = 40$	$\sum f_i d_i^2 = 1776$

The formula for variance is $\sigma^2 = \frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2$.
Substituting the values:
$\sigma^2 = \frac{1776}{40} - \left(\frac{40}{40}\right)^2$
$\sigma^2 = 44.4 - (1)^2$
$\sigma^2 = 44.4 - 1 = 43.4$.

(D)

$x_i$	$f_i$	$f_i x_i$	$D = x_i - \mu$	$f_i D^2$
$4$	$1$	$4$	$2$	$4$
$3$	$3$	$9$	$1$	$3$
$1$	$5$	$5$	$-1$	$5$
Total	$9$	$18$	-	$12$

Mean $(\mu) = \frac{\sum f_i x_i}{\sum f_i} = \frac{18}{9} = 2$
Standard deviation $(\sigma) = \sqrt{\frac{\sum f_i (x_i - \mu)^2}{\sum f_i}} = \sqrt{\frac{12}{9}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$
Coefficient of variation $= \frac{\sigma}{\mu} \times 100 = \frac{2/\sqrt{3}}{2} \times 100 = \frac{100}{\sqrt{3}}$.

(D) The standard deviation measures the dispersion of a data set relative to its mean. $A$ smaller range or smaller differences between consecutive values indicate a smaller standard deviation.
For the given sets:
$A: 10, 20, 30, 40$ (Range $= 30$)
$B: 2, 4, 6, 8$ (Range $= 6$)
$C: 3, 6, 9, 12$ (Range $= 9$)
$D: 1, 2, 3, 4$ (Range $= 3$)
Since the set $1, 2, 3, 4$ has the smallest range and the values are closest to each other,it has the least standard deviation.

(B) We know that the coefficient of variation $(CV)$ is given by the formula:
$CV = \frac{\sigma}{|\bar{x}|} \times 100$
Given $CV = 25$ and mean $\bar{x} = 44$.
Substituting the values:
$25 = \frac{\sigma}{44} \times 100$
$\sigma = \frac{25 \times 44}{100} = \frac{1100}{100} = 11$
Now,the variance is defined as the square of the standard deviation:
$\text{Variance} = \sigma^2 = 11^2 = 121$

(A) Given that $\sum(x_i+2)^2 = 28n$
$\Rightarrow \sum x_i^2 + 4\sum x_i + 4n = 28n$
$\Rightarrow \sum x_i^2 + 4\sum x_i = 24n$ $... (i)$
Similarly,$\sum(x_i-2)^2 = 12n$
$\Rightarrow \sum x_i^2 - 4\sum x_i + 4n = 12n$
$\Rightarrow \sum x_i^2 - 4\sum x_i = 8n$ $... (ii)$
Adding $(i)$ and $(ii)$:
$2\sum x_i^2 = 32n \Rightarrow \sum x_i^2 = 16n$
Subtracting $(ii)$ from $(i)$:
$8\sum x_i = 16n \Rightarrow \sum x_i = 2n$
The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$
$\sigma^2 = \frac{16n}{n} - \left(\frac{2n}{n}\right)^2 = 16 - 4 = 12$

(B) Let the original observations be $x_1, x_2, \ldots, x_n$. The variance is given by $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$.
If each observation is increased or decreased by $k$,the new observations are $y_i = x_i \pm k$.
The new mean becomes $\bar{y} = \frac{1}{n} \sum (x_i \pm k) = \bar{x} \pm k$.
The new variance is $\sigma_y^2 = \frac{1}{n} \sum (y_i - \bar{y})^2 = \frac{1}{n} \sum ((x_i \pm k) - (\bar{x} \pm k))^2 = \frac{1}{n} \sum (x_i - \bar{x})^2 = \sigma^2$.
Thus,the variance remains unchanged.

(C) Given,number of observations $n = 20$,$\sum x_i = 1000$,and $\sum x_i^2 = 84000$.
The mean $\overline{x} = \frac{\sum x_i}{n} = \frac{1000}{20} = 50$.
The variance $\sigma^2$ is given by the formula:
$\sigma^2 = \frac{1}{n} \sum x_i^2 - (\overline{x})^2$.
Substituting the values:
$\sigma^2 = \frac{84000}{20} - (50)^2$.
$\sigma^2 = 4200 - 2500$.
$\sigma^2 = 1700$.

(A) Let the original numbers be $w, x, y, z$ with variance $\sigma^2 = 9$.
The variance of a set of numbers is defined as $\text{Var}(X) = E[X^2] - (E[X])^2$.
If each number is multiplied by a constant $k$,the new variance is given by the property $\text{Var}(kX) = k^2 \text{Var}(X)$.
Here,$k = 5$ and $\text{Var}(X) = 9$.
Therefore,the new variance is $5^2 \times 9 = 25 \times 9 = 225$.