Word problem -Statistics Questions in English

(B) Let the unknown numbers be $x$ and $y$.
Given mean $\bar{x} = 8$.
$\frac{2+4+10+12+14+x+y}{7} = 8$
$42 + x + y = 56$
$x + y = 14$ $... (i)$
Given variance $\sigma^2 = 16$.
We know $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$16 = \frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7} - 8^2$
$16 + 64 = \frac{4+16+100+144+196+x^2+y^2}{7}$
$80 \times 7 = 460 + x^2 + y^2$
$560 = 460 + x^2 + y^2$
$x^2 + y^2 = 100$ $... (ii)$
From $(x+y)^2 = x^2 + y^2 + 2xy$,we have $14^2 = 100 + 2xy$.
$196 - 100 = 2xy$ $\Rightarrow 2xy = 96$ $\Rightarrow xy = 48$.
Now,$(x-y)^2 = (x+y)^2 - 4xy = 14^2 - 4(48) = 196 - 192 = 4$.
$|x-y| = \sqrt{4} = 2$.

(B) The total number of students is $N = 20$.
Sum of frequencies $\Sigma f_i = (x+1)^2 + (2x-5) + (x^2-3x) + x = 20$.
Expanding the terms: $(x^2+2x+1) + 2x - 5 + x^2 - 3x + x = 20$.
$2x^2 + 2x - 4 = 20$ $\Rightarrow 2x^2 + 2x - 24 = 0$ $\Rightarrow x^2 + x - 12 = 0$.
Factoring the quadratic: $(x+4)(x-3) = 0$.
Since $x \in R^{+} \cup \{0\}$,we have $x = 3$.
Now,calculate the frequencies for $x=3$:
Marks $2$: $(3+1)^2 = 16$.
Marks $3$: $2(3)-5 = 1$.
Marks $5$: $3^2-3(3) = 0$.
Marks $7$: $3$.
Check sum: $16+1+0+3 = 20$. Correct.
Mean $\bar{x} = \frac{\Sigma f_i x_i}{N} = \frac{2(16) + 3(1) + 5(0) + 7(3)}{20} = \frac{32 + 3 + 0 + 21}{20} = \frac{56}{20} = 2.8$.

(C) Given,$\text{Mean} = 5$ and $\text{S.D.} = 2$.
For the data $3, 5, 7, a, b$ with $n=5$:
$\text{Mean} = \frac{3+5+7+a+b}{5} = 5$
$\Rightarrow 15+a+b = 25$
$\Rightarrow a+b = 10$ ... $(i)$
Using the formula for variance,$\text{Var} = \text{S.D.}^2 = 2^2 = 4$:
$\text{Var} = \frac{\sum x_i^2}{n} - (\text{Mean})^2$
$4 = \frac{3^2+5^2+7^2+a^2+b^2}{5} - 5^2$
$4 = \frac{9+25+49+a^2+b^2}{5} - 25$
$29 = \frac{83+a^2+b^2}{5}$
$145 = 83 + a^2 + b^2$
$a^2 + b^2 = 62$ ... $(ii)$
We know that $(a+b)^2 = a^2 + b^2 + 2ab$.
$10^2 = 62 + 2ab$
$100 - 62 = 2ab$
$38 = 2ab \Rightarrow ab = 19$.
The quadratic equation with roots $a$ and $b$ is given by $x^2 - (a+b)x + ab = 0$.
Substituting the values,we get $x^2 - 10x + 19 = 0$.

(C) Given: $n_1 = 5, \sigma_1^2 = 4, \overline{x}_1 = 2$ and $n_2 = 5, \sigma_2^2 = 5, \overline{x}_2 = 4$.
Combined mean $\overline{x}_c = \frac{n_1\overline{x}_1 + n_2\overline{x}_2}{n_1 + n_2} = \frac{5(2) + 5(4)}{5 + 5} = \frac{30}{10} = 3$.
Calculate deviations: $d_1 = \overline{x}_1 - \overline{x}_c = 2 - 3 = -1$ and $d_2 = \overline{x}_2 - \overline{x}_c = 4 - 3 = 1$.
Combined variance $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$.
Substituting the values: $\sigma^2 = \frac{5(4 + (-1)^2) + 5(5 + 1^2)}{5 + 5} = \frac{5(5) + 5(6)}{10} = \frac{25 + 30}{10} = \frac{55}{10} = \frac{11}{2}$.

(C) Let the five observations be $1, 2, 6, a,$ and $b$.
Given that the mean $\bar{x} = 4$ and $n = 5$.
$\bar{x} = \frac{1 + 2 + 6 + a + b}{5} = 4$
$9 + a + b = 20 \implies a + b = 11 \implies b = 11 - a$ ... $(1)$
Given that the variance $\sigma^2 = 5.2$.
$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = 5.2$
$\frac{(1-4)^2 + (2-4)^2 + (6-4)^2 + (a-4)^2 + (b-4)^2}{5} = 5.2$
$(-3)^2 + (-2)^2 + (2)^2 + (a-4)^2 + (b-4)^2 = 26$
$9 + 4 + 4 + (a-4)^2 + (b-4)^2 = 26$
$17 + (a-4)^2 + (b-4)^2 = 26$
$(a-4)^2 + (b-4)^2 = 9$ ... $(2)$
Substitute $b = 11 - a$ into equation $(2)$:
$(a-4)^2 + (11 - a - 4)^2 = 9$
$(a-4)^2 + (7 - a)^2 = 9$
$(a^2 - 8a + 16) + (49 - 14a + a^2) = 9$
$2a^2 - 22a + 65 = 9$
$2a^2 - 22a + 56 = 0$
$a^2 - 11a + 28 = 0$
$(a-4)(a-7) = 0$
So,$a = 4$ or $a = 7$.
If $a = 4$,then $b = 11 - 4 = 7$.
If $a = 7$,then $b = 11 - 7 = 4$.
Thus,the other two observations are $4$ and $7$.

(B) Given: Mean $\bar{x} = 16$ and Standard Deviation $\sigma = 16$.
Let $y_i = x_i - 5$.
The mean of the new observations $y_i$ is $\bar{y} = \bar{x} - 5 = 16 - 5 = 11$.
The standard deviation remains unchanged when a constant is subtracted from each observation,so $\sigma_y = 16$.
We know that $\sigma_y^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2$.
Substituting the values: $16^2 = \frac{\sum (x_i-5)^2}{50} - 11^2$.
$256 = \frac{\sum (x_i-5)^2}{50} - 121$.
$\frac{\sum (x_i-5)^2}{50} = 256 + 121 = 377$.
Thus,the mean of $(x_i-5)^2$ is $377$.

(C) Given $n=50$,$\bar{x}=16$,and $\sigma_x^2=256$.
Since $\sigma_x^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$,we have $256 = \frac{1}{50} \sum x_i^2 - 16^2$.
$256 = \frac{1}{50} \sum x_i^2 - 256 \implies \frac{1}{50} \sum x_i^2 = 512 \implies \sum x_i^2 = 25600$.
We need the mean of $(x_i-5)^2$,which is $\frac{1}{50} \sum_{i=1}^{50} (x_i-5)^2$.
Expanding the sum: $\sum (x_i^2 - 10x_i + 25) = \sum x_i^2 - 10 \sum x_i + \sum 25$.
Since $\bar{x} = \frac{1}{50} \sum x_i = 16$,then $\sum x_i = 50 \times 16 = 800$.
Thus,$\sum (x_i-5)^2 = 25600 - 10(800) + 50(25) = 25600 - 8000 + 1250 = 18850$.
Required mean $= \frac{18850}{50} = 377$.

(A) The coefficient of variation $(CV)$ is given by the formula: $CV = \frac{\sigma}{\bar{x}} \times 100$,where $\sigma$ is the standard deviation and $\bar{x}$ is the mean.
For the first distribution: $60 = \frac{21}{\bar{x}_1} \times 100 \Rightarrow \bar{x}_1 = \frac{2100}{60} = 35$.
For the second distribution: $70 = \frac{16}{\bar{x}_2} \times 100 \Rightarrow \bar{x}_2 = \frac{1600}{70} \approx 22.86$.
Thus,the means are $35$ and $22.86$ respectively.

(A) The coefficient of variation $( CV )$ is defined by the formula:
$ CV = \frac{\sigma}{\bar{x}} \times 100 $
Where $ \sigma $ is the standard deviation and $ \bar{x} $ is the arithmetic mean.
Given $ CV = 60 $ and $ \sigma = 24 $.
Substituting these values into the formula:
$ 60 = \frac{24}{\bar{x}} \times 100 $
$ 60 = \frac{2400}{\bar{x}} $
$ \bar{x} = \frac{2400}{60} $
$ \bar{x} = 40 $
Therefore,the arithmetic mean is $ 40 $.

(D) Let $n_1 = 15$,$\bar{x} = 2$,and $\sigma_x^2 = 4$. The sum of observations is $\sum x_i = n_1 \bar{x} = 15 \times 2 = 30$. The sum of squares is $\sum x_i^2 = n_1(\sigma_x^2 + \bar{x}^2) = 15(4 + 2^2) = 15(8) = 120$.
Let $n_2 = 10$,$\bar{y} = 2$,and $\sigma_y^2 = 5$. The sum of observations is $\sum y_i = n_2 \bar{y} = 10 \times 2 = 20$. The sum of squares is $\sum y_i^2 = n_2(\sigma_y^2 + \bar{y}^2) = 10(5 + 2^2) = 10(9) = 90$.
For the combined set of $N = n_1 + n_2 = 25$ observations,the combined mean is $\bar{z} = \frac{\sum x_i + \sum y_i}{n_1 + n_2} = \frac{30 + 20}{25} = \frac{50}{25} = 2$.
The combined variance is $\sigma^2 = \frac{\sum x_i^2 + \sum y_i^2}{n_1 + n_2} - \bar{z}^2 = \frac{120 + 90}{25} - 2^2 = \frac{210}{25} - 4 = 8.4 - 4 = 4.4$.

(B) Given $\sum_{i=1}^{11}(x_i-4)=22$.
Dividing by $11$,we get $\bar{x}-4 = \frac{22}{11} = 2$,so $\bar{x} = \alpha = 6$.
Now,the variance $\beta$ is given by $\beta = \frac{1}{n} \sum (x_i-\bar{x})^2$.
Since $\bar{x}=6$,we have $x_i-\bar{x} = x_i-6 = (x_i-4)-2$.
Thus,$\sum (x_i-6)^2 = \sum ((x_i-4)-2)^2 = \sum (x_i-4)^2 - 4\sum (x_i-4) + \sum 4$.
Substituting the values: $154 - 4(22) + 11(4) = 154 - 88 + 44 = 110$.
So,$\beta = \frac{110}{11} = 10$.
The roots are $\frac{\alpha}{\beta} = \frac{6}{10} = \frac{3}{5}$ and $\frac{\beta}{\alpha} = \frac{10}{6} = \frac{5}{3}$.
The quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Sum of roots $= \frac{3}{5} + \frac{5}{3} = \frac{9+25}{15} = \frac{34}{15}$.
Product of roots $= \frac{3}{5} \times \frac{5}{3} = 1$.
The equation is $x^2 - \frac{34}{15}x + 1 = 0$,which simplifies to $15x^2 - 34x + 15 = 0$.

(A) The range of a data set is calculated as the difference between the highest value and the lowest value in the set.
Given data: $35, 12, 21, 24, 15, 7, 16, 12, 30, 32, 13, 17$.
Highest value = $35$.
Lowest value = $7$.
Range = $\text{Highest value} - \text{Lowest value} = 35 - 7 = 28$.

(C) The total frequency is given as $120$. So,$17 + f_1 + 32 + f_2 + 19 = 120$.
$f_1 + f_2 + 68 = 120 \implies f_1 + f_2 = 52$ (Equation $1$).
The class marks $(x_i)$ are $10, 30, 50, 70, 90$.
The mean is $\frac{\sum f_i x_i}{\sum f_i} = 50$.
$\frac{17(10) + f_1(30) + 32(50) + f_2(70) + 19(90)}{120} = 50$.
$170 + 30f_1 + 1600 + 70f_2 + 1710 = 6000$.
$30f_1 + 70f_2 + 3480 = 6000$.
$30f_1 + 70f_2 = 2520 \implies 3f_1 + 7f_2 = 252$ (Equation $2$).
From Equation $1$,$f_1 = 52 - f_2$. Substituting into Equation $2$:
$3(52 - f_2) + 7f_2 = 252$.
$156 - 3f_2 + 7f_2 = 252$.
$4f_2 = 96 \implies f_2 = 24$.
Then $f_1 = 52 - 24 = 28$.

(D) The total wage bill is calculated as the product of the number of workers and the average daily wage.
For mill $A$:
Total wage bill $= 500 \times 186 = 93000$.
For mill $B$:
Total wage bill $= 600 \times 175 = 105000$.
Comparing the two,$105000 > 93000$.
Therefore,the wage bill of mill $B$ is greater than that of mill $A$.

(B) Number of boys $= 25$
Mean marks of boys $= 61$
Number of girls $= 35$
Mean marks of girls $= 58$
Total mean of all $60$ students $= \frac{61 \times 25 + 35 \times 58}{60}$
$= \frac{1525 + 2030}{60} = \frac{3555}{60} = 59.25$

(A) Let the number of boys and girls in the class be $m$ and $n$ respectively. According to the given information:
Total marks of boys $= 40m$
Total marks of girls $= 45n$
Total marks of boys and girls combined $= 42(m + n)$
Equating the total marks:
$40m + 45n = 42(m + n)$
$40m + 45n = 42m + 42n$
$3n = 2m$
$\frac{m}{n} = \frac{3}{2}$
The percentage of boys in the class is given by:
$\frac{m}{m + n} \times 100 = \frac{3}{3 + 2} \times 100$
$= \frac{3}{5} \times 100 = 60 \%$
Therefore,the percentage of boys is $60 \%$.
Hence,option $A$ is correct.

(A) The coefficient of variation $(CV)$ is defined as $CV = \frac{\sigma}{\bar{x}} \times 100$,where $\sigma$ is the standard deviation and $\bar{x}$ is the arithmetic mean.
For the first distribution: $60 = \frac{21}{\bar{x}_1} \times 100 \implies \bar{x}_1 = \frac{2100}{60} = 35$.
For the second distribution: $70 = \frac{16}{\bar{x}_2} \times 100 \implies \bar{x}_2 = \frac{1600}{70} \approx 22.857$.
Thus,the arithmetic means are $35$ and $22.85$ approximately.

(A) Consistency is measured by the coefficient of variation $(CV)$. $A$ lower $CV$ implies higher consistency.
Batsman $A$ is more consistent than $B$,so $CV_{A} < CV_{B}$.
This implies $\frac{\sigma_{A}}{\bar{x}} < \frac{\sigma_{B}}{\bar{y}}$.
Rearranging this gives $\frac{\sigma_{A}}{\sigma_{B}} < \frac{\bar{x}}{\bar{y}}$.
Since $\sigma_{A}, \sigma_{B}, \bar{x}, \bar{y} > 0$,we have $0 < \frac{\sigma_{A}}{\sigma_{B}} < \frac{\bar{x}}{\bar{y}}$.
For $A$ to be a higher run scorer,we must have $\bar{x} > \bar{y}$,which implies $\frac{\bar{x}}{\bar{y}} > 1$.

(A) Let the other two observations be $x$ and $y$.
Given mean $\bar{x} = 4.4$,so $\frac{1+2+6+x+y}{5} = 4.4$ $\Rightarrow 9+x+y = 22$ $\Rightarrow x+y = 13 \dots (i)$.
Variance $\sigma^2 = 8.24$,so $\frac{1}{5}(1^2+2^2+6^2+x^2+y^2) - (4.4)^2 = 8.24$.
$\frac{1}{5}(1+4+36+x^2+y^2) - 19.36 = 8.24$.
$\frac{41+x^2+y^2}{5} = 27.6$ $\Rightarrow 41+x^2+y^2 = 138$ $\Rightarrow x^2+y^2 = 97 \dots (ii)$.
We know $(x+y)^2 = x^2+y^2+2xy$,so $13^2 = 97+2xy$ $\Rightarrow 169-97 = 2xy$ $\Rightarrow 2xy = 72$ $\Rightarrow xy = 36$.
Since $x+y = 13$ and $xy = 36$,the quadratic equation $t^2 - 13t + 36 = 0$ gives the values.
$(t-9)(t-4) = 0$,so $t = 9$ or $t = 4$.
Thus,the other two observations are $9$ and $4$.

(D) The expression $\sum_{i=1}^{n}(x_i - \bar{x})^2$ represents the sum of squared deviations from the mean.
If this sum is almost zero,it implies that each individual observation $x_i$ must be very close to the mean $\bar{x}$.
Consequently,the variance,which is proportional to this sum,is very small,indicating that the degree of dispersion of the data points around the mean is low.

(A) The mean deviation from the mean and the mean deviation from the median are not necessarily equal for a given data set because the mean and median can take different values. Therefore,the statement that they must always be equal is false.

(C) For set $A$: $\text{Mean} = \frac{\sum x_i}{5} = 2 \Rightarrow \sum x_i = 10$.
Variance $= \frac{1}{5} \sum x_i^2 - (\text{Mean})^2 = 4$ $\Rightarrow \frac{1}{5} \sum x_i^2 - 4 = 4$ $\Rightarrow \sum x_i^2 = 40$.
For set $B$: $\text{Mean} = \frac{\sum y_i}{5} = 4 \Rightarrow \sum y_i = 20$.
Variance $= \frac{1}{5} \sum y_i^2 - (\text{Mean})^2 = 5$ $\Rightarrow \frac{1}{5} \sum y_i^2 - 16 = 5$ $\Rightarrow \sum y_i^2 = 105$.
For $A \cup B$: Total elements $N = 10$.
Combined Mean $\bar{X} = \frac{\sum x_i + \sum y_i}{10} = \frac{10 + 20}{10} = 3$.
Combined Variance $= \frac{\sum x_i^2 + \sum y_i^2}{10} - (\bar{X})^2 = \frac{40 + 105}{10} - (3)^2 = 14.5 - 9 = 5.5$.

(C) Given the set of numbers: $2, 3, 2x, 11$. The number of observations $n = 4$. The standard deviation $(SD)$ is $3.5 = \frac{7}{2}$.
First,calculate the mean $(\bar{x})$:
$\bar{x} = \frac{2 + 3 + 2x + 11}{4} = \frac{16 + 2x}{4} = \frac{8 + x}{2}$.
The variance $(V)$ is given by $V = (SD)^2 = (3.5)^2 = 12.25$.
The formula for variance is $V = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.
$\sum x_i^2 = 2^2 + 3^2 + (2x)^2 + 11^2 = 4 + 9 + 4x^2 + 121 = 134 + 4x^2$.
$12.25 = \frac{134 + 4x^2}{4} - \left(\frac{8 + x}{2}\right)^2$.
$12.25 = \frac{134 + 4x^2}{4} - \frac{64 + 16x + x^2}{4}$.
$12.25 = \frac{134 + 4x^2 - 64 - 16x - x^2}{4}$.
$49 = 3x^2 - 16x + 70$.
$3x^2 - 16x + 21 = 0$.
Solving the quadratic equation $3x^2 - 9x - 7x + 21 = 0$:
$3x(x - 3) - 7(x - 3) = 0$.
$(3x - 7)(x - 3) = 0$.
Therefore,$x = 3$ or $x = \frac{7}{3}$.

(C) The range of a data set is defined as the difference between the maximum value and the minimum value.
Given data: $15, 14, k, 25, 30, 35$.
Range $= 23$.
Case $1$: If $35$ is the maximum value,then the minimum value must be $35 - 23 = 12$.
If $k = 12$,the data set becomes $12, 14, 15, 25, 30, 35$. The range is $35 - 12 = 23$. This is a valid case.
Case $2$: If $k$ is the maximum value,then $k - 14 = 23$,which gives $k = 37$.
Since we are looking for the least positive value of $k$,we compare $12$ and $37$.
The least positive value is $12$.

(B) The given observations are $20, 28, 40, 12, 30, 15, 50$.
To find the range,we identify the maximum and minimum values in the data set.
Maximum value $= 50$.
Minimum value $= 12$.
Range $= \text{Maximum value} - \text{Minimum value}$.
Range $= 50 - 12 = 38$.

(C) Given $n=100$,$\bar{x}_1=40$,and $\sigma_1=15$.
The sum of observations is $\sum x_i = n \times \bar{x}_1 = 100 \times 40 = 4000$.
The corrected sum is $\sum x_i^{\prime} = 4000 - 30 - 60 + 40 + 50 = 4000$.
Thus,the corrected mean $\bar{x}_2 = \frac{4000}{100} = 40$.
So,$\bar{x}_1 = \bar{x}_2$.
Now,$\sigma_1^2 = \frac{\sum x_i^2}{n} - (\bar{x}_1)^2$ $\Rightarrow 225 = \frac{\sum x_i^2}{100} - 1600$ $\Rightarrow \sum x_i^2 = 182500$.
The corrected sum of squares is $\sum x_i^{\prime 2} = 182500 - 30^2 - 60^2 + 40^2 + 50^2 = 182500 - 900 - 3600 + 1600 + 2500 = 182100$.
The corrected variance is $\sigma_2^2 = \frac{182100}{100} - (40)^2 = 1821 - 1600 = 221$.
Since $\sigma_1^2 = 225$ and $\sigma_2^2 = 221$,we have $\sigma_1^2 > \sigma_2^2$,which implies $\sigma_1 > \sigma_2$.
Therefore,$\bar{x}_1 = \bar{x}_2$ and $\sigma_1 > \sigma_2$.

(C) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given $x_1=1, x_2=3, x_3=4$. Let the other two be $a$ and $b$.
Mean $\bar{x} = \frac{1+3+4+a+b}{5} = 4 \implies 8+a+b = 20 \implies a+b = 12$.
Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 4$.
$\frac{1^2+3^2+4^2+a^2+b^2}{5} - 4^2 = 4$.
$\frac{1+9+16+a^2+b^2}{5} - 16 = 4 \implies \frac{26+a^2+b^2}{5} = 20$.
$26+a^2+b^2 = 100 \implies a^2+b^2 = 74$.
We know $(a+b)^2 = a^2+b^2+2ab$.
$12^2 = 74+2ab \implies 144 = 74+2ab$.
$2ab = 70 \implies ab = 35$.

(A) The arithmetic mean is given by $\frac{a + b + 8 + 5 + 10}{5} = 6$.
$a + b + 23 = 30$,so $a + b = 7$.
The variance is given by $\frac{\sum x_i^2}{n} - (\text{mean})^2 = 6.8$.
$\frac{a^2 + b^2 + 8^2 + 5^2 + 10^2}{5} - 6^2 = 6.8$.
$\frac{a^2 + b^2 + 64 + 25 + 100}{5} - 36 = 6.8$.
$\frac{a^2 + b^2 + 189}{5} = 42.8$.
$a^2 + b^2 + 189 = 214$,so $a^2 + b^2 = 25$.
Since $a + b = 7$,we have $b = 7 - a$.
Substituting into the second equation: $a^2 + (7 - a)^2 = 25$.
$a^2 + 49 - 14a + a^2 = 25$.
$2a^2 - 14a + 24 = 0$.
$a^2 - 7a + 12 = 0$.
$(a - 3)(a - 4) = 0$.
So,$a = 3$ or $a = 4$.
If $a = 3$,then $b = 4$. If $a = 4$,then $b = 3$.
The ordered pair $(a, b)$ can be $(3, 4)$ or $(4, 3)$.
Comparing with the options,$(3, 4)$ is correct.

(C) To find the coefficient of variation $(CV)$,we use the formula: $CV = \frac{\sigma}{\bar{x}} \times 100$,where $\sigma$ is the standard deviation and $\bar{x}$ is the mean.
First,calculate the mean $(\bar{x})$:
Midpoints $(x_i)$: $35, 45, 55, 65, 75, 85$
Frequencies $(f_i)$: $17, 28, 21, 15, 13, 6$
Total frequency $(N = \sum f_i)$ = $17 + 28 + 21 + 15 + 13 + 6 = 100$
Sum of $(f_i x_i)$: $(17 \times 35) + (28 \times 45) + (21 \times 55) + (15 \times 65) + (13 \times 75) + (6 \times 85) = 595 + 1260 + 1155 + 975 + 975 + 510 = 5470$
Mean $(\bar{x})$ = $\frac{\sum f_i x_i}{N} = \frac{5470}{100} = 54.7$
Given standard deviation $(\sigma)$ = $14.72$
$CV = \frac{14.72}{54.7} \times 100 \approx 26.91$
Thus,the correct option is $C$.

(D) The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.
The mean $\bar{x} = \frac{2(1+2+\dots+n)}{n} = \frac{2 \times n(n+1)}{2n} = n+1$.
The variance is given by $\sigma^2 = \frac{1}{n} \sum_{i=1}^n (2i)^2 - (n+1)^2$.
$\sigma^2 = \frac{4}{n} \times \frac{n(n+1)(2n+1)}{6} - (n+1)^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2$.
$\sigma^2 = (n+1) \left[ \frac{4n+2-3n-3}{3} \right] = \frac{(n+1)(n-1)}{3} = \frac{n^2-1}{3}$.
Thus,Statement-$I$ is false.
For $n=20$,the mean is $20+1 = 21$.
The variance is $\frac{20^2-1}{3} = \frac{399}{3} = 133$.
The difference is $133 - 21 = 112$.
Thus,Statement-$II$ is true.

(D) Given that the sum of squares of deviations from the mean is $n\bar{x}^2$.
The formula for the sum of squares of deviations is $\sum_{i=1}^n (x_i - \bar{x})^2 = n\bar{x}^2$.
Expanding the left side: $\sum_{i=1}^n (x_i^2 - 2x_i\bar{x} + \bar{x}^2) = n\bar{x}^2$.
This simplifies to $\sum x_i^2 - 2\bar{x}\sum x_i + n\bar{x}^2 = n\bar{x}^2$.
Since $\sum x_i = n\bar{x}$,we substitute this into the equation: $\sum x_i^2 - 2\bar{x}(n\bar{x}) + n\bar{x}^2 = n\bar{x}^2$.
$\sum x_i^2 - 2n\bar{x}^2 + n\bar{x}^2 = n\bar{x}^2$.
$\sum x_i^2 - n\bar{x}^2 = n\bar{x}^2$.
Therefore,$\sum_{i=1}^n x_i^2 = 2n\bar{x}^2$.

(D) First,we calculate the mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4(1) + 24(3) + 28(5) + 16(7) + 8(9)}{4 + 24 + 28 + 16 + 8} = \frac{4 + 72 + 140 + 112 + 72}{80} = \frac{400}{80} = 5$.
Next,the mean deviation about the mean $m = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{4|1-5| + 24|3-5| + 28|5-5| + 16|7-5| + 8|9-5|}{80} = \frac{4(4) + 24(2) + 28(0) + 16(2) + 8(4)}{80} = \frac{16 + 48 + 0 + 32 + 32}{80} = \frac{128}{80} = \frac{8}{5} = 1.6$.
The variance $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - (\bar{x})^2 = \frac{4(1)^2 + 24(3)^2 + 28(5)^2 + 16(7)^2 + 8(9)^2}{80} - 5^2 = \frac{4 + 216 + 700 + 784 + 648}{80} - 25 = \frac{2352}{80} - 25 = 29.4 - 25 = 4.4 = \frac{22}{5}$.
Finally,$m + \sigma^2 = \frac{8}{5} + \frac{22}{5} = \frac{30}{5} = 6$.

(D) Given: $n_1 = 20, \bar{x}_1 = 25, \sigma_1^2 = 9$ and $n_2 = 30, \bar{x}_2 = 10, \sigma_2^2 = 16$.
Combined mean $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{20 \times 25 + 30 \times 10}{20 + 30} = \frac{500 + 300}{50} = \frac{800}{50} = 16$.
Let $d_1 = \bar{x}_1 - \bar{x} = 25 - 16 = 9$ and $d_2 = \bar{x}_2 - \bar{x} = 10 - 16 = -6$.
Combined variance $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$.
$\sigma^2 = \frac{20(9 + 9^2) + 30(16 + (-6)^2)}{20 + 30} = \frac{20(9 + 81) + 30(16 + 36)}{50} = \frac{20(90) + 30(52)}{50} = \frac{1800 + 1560}{50} = \frac{3360}{50} = 67.2$.

(D) We are given $\sum_{i=1}^n x_i^2 = 400$ and $\sum_{i=1}^n x_i = 80$.
We know that the variance of a set of observations is always non-negative,i.e.,$\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 \geq 0$.
Substituting the given values:
$\frac{400}{n} - \left(\frac{80}{n}\right)^2 \geq 0$
$\frac{400}{n} - \frac{6400}{n^2} \geq 0$
Multiplying by $n^2$ (since $n > 0$):
$400n - 6400 \geq 0$
$400n \geq 6400$
$n \geq \frac{6400}{400}$
$n \geq 16$.
Thus,the least value of $n$ is $16$.

(A) Given observations: $2, 3, 5, 8, 12$
Mean $\bar{x} = \frac{2+3+5+8+12}{5} = \frac{30}{5} = 6$
Variance $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{(2-6)^2 + (3-6)^2 + (5-6)^2 + (8-6)^2 + (12-6)^2}{5}$
$\sigma^2 = \frac{(-4)^2 + (-3)^2 + (-1)^2 + 2^2 + 6^2}{5} = \frac{16 + 9 + 1 + 4 + 36}{5} = \frac{66}{5} = 13.2$
Median $m$ of the data $2, 3, 5, 8, 12$ is the middle term,so $m = 5$.
Mean deviation about median $M = \frac{\sum |x_i - m|}{n} = \frac{|2-5| + |3-5| + |5-5| + |8-5| + |12-5|}{5}$
$M = \frac{3 + 2 + 0 + 3 + 7}{5} = \frac{15}{5} = 3$
Therefore,$\sigma^2 - M = 13.2 - 3 = 10.2$.

(D) Given: $n = 100$,$\bar{x} = 40$,$\sigma = 5.1$.
We know that $\text{Variance} = \sigma^2 = (5.1)^2 = 26.01$.
The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
Substituting the values: $26.01 = \frac{\sum x_i^2}{100} - (40)^2$.
$26.01 = \frac{\sum x_i^2}{100} - 1600$.
$\frac{\sum x_i^2}{100} = 1626.01$.
$\sum x_i^2 = 162601$ (This is the incorrect sum of squares).
To find the correct sum of squares,we subtract the square of the incorrect observation and add the square of the correct observation:
$\text{Correct } \sum x_i^2 = 162601 - (50)^2 + (40)^2$.
$\text{Correct } \sum x_i^2 = 162601 - 2500 + 1600$.
$\text{Correct } \sum x_i^2 = 161701$.

(C) Given that $x_1, x_2, \ldots, x_n$ are negative numbers in ascending order. Since they are negative,we can write them as $-|x_1|, -|x_2|, \ldots, -|x_n|$.
Because the order is ascending,we have $-|x_1| < -|x_2| < \ldots < -|x_n|$,which implies $|x_1| > |x_2| > \ldots > |x_n|$.
After changing the signs of the first term $(x_1)$ and the last term $(x_n)$,the new set of observations becomes $|x_1|, -|x_2|, -|x_3|, \ldots, -|x_{n-1}|, |x_n|$.
In this new set,the largest value is $|x_1|$ and the smallest value is $-|x_2|$ (since $|x_1| > |x_2| > \ldots > |x_n|$).
The range is defined as $\text{Maximum} - \text{Minimum}$.
Range $= |x_1| - (-|x_2|) = |x_1| + |x_2|$.
Wait,re-evaluating the sequence: The original sequence is $x_1 < x_2 < \ldots < x_n < 0$. After changing signs,the new sequence is $-x_1, x_2, x_3, \ldots, x_{n-1}, -x_n$. Since $x_1$ is the most negative,$-x_1$ is the largest positive. The smallest value is $x_2$. Thus,the range is $-x_1 - x_2 = |x_1| - x_2$.

(A) Statement $(I)$: The range is defined as the difference between the maximum and minimum observations. Since intermediate observations do not affect the maximum or minimum values,the range remains unchanged. Thus,Statement $(I)$ is true.
Statement $(II)$: $A$ well-known property of mean deviation is that it is minimized when calculated about the median. Thus,Statement $(II)$ is true.
Statement $(III)$: For grouped data,the range is defined as the difference between the upper limit of the largest class and the lower limit of the smallest class. The statement provided swaps these,making it false.
Therefore,Statements $I$ and $II$ are true but statement $III$ is false.

(C) The given data set is $50, 70, 60, B, 20, 40$. The range is defined as the difference between the maximum and minimum values,which is given as $65$.
Case $1$: If $B$ is the maximum value,then $B - 20 = 65$,which implies $B = 85$.
Case $2$: If $B$ is the minimum value,then $70 - B = 65$,which implies $B = 5$.
The possible values for $B$ are $85$ and $5$.
The absolute difference between these values is $|85 - 5| = 80$.

(A) The coefficient of variation $(CV)$ is defined as $CV = \frac{\sigma}{\bar{x}} \times 100$.
For group $A$ to be more consistent than group $B$,the coefficient of variation of $A$ must be less than the coefficient of variation of $B$.
Thus,$CV_A < CV_B$.
Substituting the values,we get $\frac{\sigma_A}{\bar{x}} < \frac{\sigma_B}{\bar{y}}$.
Given $\sigma_A = 2$ and $\sigma_B = 3$,we have $\frac{2}{\bar{x}} < \frac{3}{\bar{y}}$.
Rearranging the inequality to solve for $\frac{\bar{y}}{\bar{x}}$,we get $\frac{\bar{y}}{\bar{x}} < \frac{3}{2}$.

(C) The first five prime numbers are $2, 3, 5, 7, 11$.
Mean $(\bar{x}) = \frac{2+3+5+7+11}{5} = \frac{28}{5} = 5.6$.
Mean deviation about mean $(\alpha) = \frac{\sum |x_i - \bar{x}|}{n} = \frac{|2-5.6| + |3-5.6| + |5-5.6| + |7-5.6| + |11-5.6|}{5} = \frac{3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac{13.6}{5} = 2.72$.
Variance $(\beta) = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{2^2 + 3^2 + 5^2 + 7^2 + 11^2}{5} - (5.6)^2 = \frac{4 + 9 + 25 + 49 + 121}{5} - 31.36 = \frac{208}{5} - 31.36 = 41.6 - 31.36 = 10.24$.
Thus,the ordered pair $(\alpha, \beta)$ is $(2.72, 10.24)$.

(B) Given,$n = 100$,$\bar{x} = 40$,and $\sigma = 5.1$.
The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
Substituting the values: $(5.1)^2 = \frac{\sum x_i^2}{100} - (40)^2$.
$26.01 = \frac{\sum x_i^2}{100} - 1600$.
$\sum x_i^2 = (26.01 + 1600) \times 100 = 162601$.
This sum includes the incorrect observation $50$. To find the correct sum of squares,we subtract the square of the incorrect value and add the square of the correct value:
Correct $\sum x_i^2 = 162601 - (50)^2 + (40)^2$.
Correct $\sum x_i^2 = 162601 - 2500 + 1600 = 161701$.

(B) Let $n_1 = 50$,$\bar{x}_1 = 10$,and $\sigma_1 = 2$ be the parameters for the first group. Let $n_2 = 50$,$\bar{x}_2$,and $\sigma_2$ be the parameters for the second group. The total number of observations is $N = 100$,with mean $\bar{x} = 8$ and standard deviation $\sigma = \sqrt{10.5}$.
First,find the mean of the second group:
$\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} \implies 8 = \frac{50(10) + 50(\bar{x}_2)}{100} \implies 800 = 500 + 50\bar{x}_2 \implies \bar{x}_2 = 6$.
Next,use the combined variance formula:
$\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$,where $d_1 = \bar{x}_1 - \bar{x} = 10 - 8 = 2$ and $d_2 = \bar{x}_2 - \bar{x} = 6 - 8 = -2$.
$10.5 = \frac{50(2^2 + 2^2) + 50(\sigma_2^2 + (-2)^2)}{100}$
$10.5 = \frac{50(8) + 50(\sigma_2^2 + 4)}{100} = \frac{400 + 50\sigma_2^2 + 200}{100}$
$1050 = 600 + 50\sigma_2^2$
$50\sigma_2^2 = 450 \implies \sigma_2^2 = 9 \implies \sigma_2 = 3$.

(C) $(a) \sum_{i=1}^n(x_i-\bar{x}) = \sum x_i - \sum \bar{x} = n\bar{x} - n\bar{x} = 0$. Thus,$(a)-(iii)$.
$(b) \text{ Variance } (\sigma^2) = \frac{1}{n} \sum_{i=1}^n (x_i-\bar{x})^2$,which is the mean of the squares of the deviations from the mean. Thus,$(b)-(v)$.
$(c) \text{ Mean deviation} = \frac{1}{n} \sum_{i=1}^n |x_i - A|$,where $A$ is a measure of central tendency. Thus,$(c)-(iv)$.
$(d) \text{ Coefficient of variation} = \frac{\sigma}{\bar{x}} \times 100$,used to compare the homogeneity of two series. Thus,$(d)-(ii)$.
Therefore,the correct matching is $a-(iii), b-(v), c-(iv), d-(ii)$.

(D) Let $\bar{x}_1$ and $\bar{x}_2$ be the means and $\sigma_1^2$ and $\sigma_2^2$ be the variances of the two distributions.
Given $\sigma_1^2 = 144$ and $\sigma_2^2 = 64$,we have $\sigma_1 = 12$ and $\sigma_2 = 8$.
The coefficient of variation $(CV)$ is given by $CV = \frac{\sigma}{\bar{x}} \times 100$.
For the first distribution: $\frac{12}{\bar{x}_1} \times 100 = 40 \Rightarrow \bar{x}_1 = \frac{1200}{40} = 30$.
For the second distribution: $\frac{8}{\bar{x}_2} \times 100 = 20 \Rightarrow \bar{x}_2 = \frac{800}{20} = 40$.
The mean of the arithmetic means is $\frac{\bar{x}_1 + \bar{x}_2}{2} = \frac{30 + 40}{2} = \frac{70}{2} = 35$.

(D) Let $\bar{x}_1$ and $\bar{x}_2$ be the means and $\sigma_1^2$ and $\sigma_2^2$ be the variances of two distributions.
Given $\sigma_1^2 = 144$ and $\sigma_2^2 = 64$.
Thus,$\sigma_1 = \sqrt{144} = 12$ and $\sigma_2 = \sqrt{64} = 8$.
The coefficient of variation $(CV)$ is given by $CV = \frac{\sigma}{\bar{x}} \times 100$.
For the first distribution: $\frac{12}{\bar{x}_1} \times 100 = 40 \Rightarrow \bar{x}_1 = \frac{1200}{40} = 30$.
For the second distribution: $\frac{8}{\bar{x}_2} \times 100 = 20 \Rightarrow \bar{x}_2 = \frac{800}{20} = 40$.
The mean of their arithmetic means is $\frac{\bar{x}_1 + \bar{x}_2}{2} = \frac{30 + 40}{2} = \frac{70}{2} = 35$.

(D) The given data is $1, 3, 5, 7, 11, 13, 17, 19, 23$. The number of observations $n = 9$.
The mean $\bar{x} = \frac{1+3+5+7+11+13+17+19+23}{9} = \frac{99}{9} = 11$.
The mean deviation from the mean $M = \frac{\sum |x_i - \bar{x}|}{n} = \frac{|1-11| + |3-11| + |5-11| + |7-11| + |11-11| + |13-11| + |17-11| + |19-11| + |23-11|}{9} = \frac{10+8+6+4+0+2+6+8+12}{9} = \frac{56}{9}$.
The variance $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{(-10)^2 + (-8)^2 + (-6)^2 + (-4)^2 + 0^2 + 2^2 + 6^2 + 8^2 + 12^2}{9} = \frac{100 + 64 + 36 + 16 + 0 + 4 + 36 + 64 + 144}{9} = \frac{464}{9}$.
Now,$3(\sigma^2 - M) = 3 \left( \frac{464}{9} - \frac{56}{9} \right) = 3 \left( \frac{408}{9} \right) = \frac{408}{3} = 136$.

(A) Let the common mean of both distributions be $\bar{x}$.
Given that the coefficients of variation $(CV)$ for $A$ and $B$ are $6$ and $2$ respectively.
The formula for the coefficient of variation is $CV = \frac{\sigma}{\bar{x}} \times 100$.
For distribution $A$: $\frac{\sigma_A}{\bar{x}} \times 100 = 6 \implies \bar{x} = \frac{100 \sigma_A}{6}$.
For distribution $B$: $\frac{\sigma_B}{\bar{x}} \times 100 = 2 \implies \bar{x} = \frac{100 \sigma_B}{2}$.
Equating the two expressions for $\bar{x}$:
$\frac{100 \sigma_A}{6} = \frac{100 \sigma_B}{2}$.
$\frac{\sigma_A}{6} = \frac{\sigma_B}{2}$.
$\sigma_A = \frac{6}{2} \sigma_B$.
$\sigma_A = 3 \sigma_B$.

(C) Let the minimum percentage of the population of the town that should fall sick in order to raise the threat level be $x \%$.
Given that the probability of a sick person being admitted to an $ICU$ is $10 \%$.
Therefore,the probability of any person in the town being admitted to an $ICU$ is $10 \% \text{ of } x \%$.
We are given that the threat level is raised if this probability exceeds $5 \%$.
So,we set up the equation: $\frac{10}{100} \times x = 5$.
Solving for $x$: $0.1x = 5 \Rightarrow x = \frac{5}{0.1} = 50$.
Thus,at least $50 \%$ of the population must fall sick to raise the threat level.
Hence,option $C$ is correct.