Word problem -Statistics Questions in English

(A) Let the two positive integers be $x$ and $y$. Given $x + y = 100$,where $x, y \in \{1, 2, \dots, 99\}$.
Total number of possible pairs $(x, y)$ is $99$.
We want the product $xy > 1000$.
Substituting $y = 100 - x$,we get $x(100 - x) > 1000$,which simplifies to $100x - x^2 > 1000$,or $x^2 - 100x + 1000 < 0$.
Solving the quadratic equation $x^2 - 100x + 1000 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{100 \pm \sqrt{10000 - 4000}}{2} = \frac{100 \pm \sqrt{6000}}{2} = 50 \pm 10\sqrt{15}$.
Since $\sqrt{15} \approx 3.87$,$x \approx 50 \pm 38.7$.
So,$x$ must be in the range $(11.3, 88.7)$.
The possible integer values for $x$ are $12, 13, \dots, 88$.
The number of such integers is $88 - 12 + 1 = 77$.
Therefore,the probability is $\frac{77}{99} = \frac{7}{9}$.

(D) Let the probabilities of hitting the plane in the four shots be $p_1 = 0.4$,$p_2 = 0.3$,$p_3 = 0.2$,and $p_4 = 0.1$.
The probability that the gun hits the plane is equal to $1$ minus the probability that the gun misses the plane in all four shots.
Let $q_i = 1 - p_i$ be the probability of missing the shot $i$.
$q_1 = 1 - 0.4 = 0.6$
$q_2 = 1 - 0.3 = 0.7$
$q_3 = 1 - 0.2 = 0.8$
$q_4 = 1 - 0.1 = 0.9$
The probability of missing all four shots is $P(\text{miss}) = q_1 \times q_2 \times q_3 \times q_4 = 0.6 \times 0.7 \times 0.8 \times 0.9 = 0.3024$.
Therefore,the probability that the gun hits the plane is $P(\text{hit}) = 1 - P(\text{miss}) = 1 - 0.3024 = 0.6976$.

(D) The formula for the mean is $\text{Mean} = \frac{\sum (x_i \times f_i)}{\sum f_i}$.
Given values: $x = \{1, 2, 3, 4\}$ and $f = \{5, 4, 6, f\}$.
Mean = $\frac{(1 \times 5) + (2 \times 4) + (3 \times 6) + (4 \times f)}{5 + 4 + 6 + f} = 3$.
$\frac{5 + 8 + 18 + 4f}{15 + f} = 3$.
$\frac{31 + 4f}{15 + f} = 3$.
$31 + 4f = 3(15 + f)$.
$31 + 4f = 45 + 3f$.
$4f - 3f = 45 - 31$.
$f = 14$.

(D) Given that the sum of deviations of $20$ observations from $30$ is $20$.
Let the observations be $x_1, x_2, ..., x_{20}$.
Then,$\sum_{i=1}^{20} (x_i - 30) = 20$.
Expanding the summation,we get $\sum_{i=1}^{20} x_i - \sum_{i=1}^{20} 30 = 20$.
$\sum_{i=1}^{20} x_i - (20 \times 30) = 20$.
$\sum_{i=1}^{20} x_i - 600 = 20$.
$\sum_{i=1}^{20} x_i = 620$.
The mean is given by $\bar{x} = \frac{\sum_{i=1}^{20} x_i}{20}$.
$\bar{x} = \frac{620}{20} = 31$.

(B) The weighted mean is given by the formula: $\frac{\sum_{i=1}^{n} (i \cdot i^2)}{\sum_{i=1}^{n} i^2} = \frac{\sum_{i=1}^{n} i^3}{\sum_{i=1}^{n} i^2}$.
We know that $\sum_{i=1}^{n} i^3 = \left[ \frac{n(n + 1)}{2} \right]^2$ and $\sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6}$.
Substituting these values,we get:
Weighted mean = $\frac{\left[ \frac{n(n + 1)}{2} \right]^2}{\frac{n(n + 1)(2n + 1)}{6}}$
$= \frac{n^2(n + 1)^2}{4} \cdot \frac{6}{n(n + 1)(2n + 1)}$
$= \frac{6n(n + 1)}{4(2n + 1)} = \frac{3n(n + 1)}{2(2n + 1)}$.

(B) The formula for combined mean is $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$.
Given,$\bar{x} = 30$,$\bar{x}_1 = 32$ (mean age of men),and $\bar{x}_2 = 27$ (mean age of women).
Let the total number of people be $100$. Let $n_1$ be the number of men and $n_2$ be the number of women.
Then $n_2 = 100 - n_1$.
Substituting the values in the formula:
$30 = \frac{32n_1 + 27(100 - n_1)}{100}$
$3000 = 32n_1 + 2700 - 27n_1$
$3000 - 2700 = 5n_1$
$300 = 5n_1$
$n_1 = 60$.
Since $n_1$ is the number of men,the number of women is $n_2 = 100 - 60 = 40$.
Thus,the percentage of women in the group is $40\%$.

(D) Let the weight of the teacher be $w \ kg$.
Total weight of $35$ students = $35 \times 40 = 1400 \ kg$.
When the teacher is included,the total number of people becomes $35 + 1 = 36$.
The new average weight = $40 + 0.5 = 40.5 \ kg$.
Total weight of $36$ people = $36 \times 40.5 = 1458 \ kg$.
The weight of the teacher $w = \text{Total weight of } 36 \text{ people} - \text{Total weight of } 35 \text{ students}$.
$w = 1458 - 1400 = 58 \ kg$.

(D) Let $n_1$ and $n_2$ be the number of observations in two groups having means $\bar{x}_1$ and $\bar{x}_2$ respectively.
Then,the combined mean is given by $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$.
Now,consider $\bar{x} - \bar{x}_1 = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} - \bar{x}_1 = \frac{n_2(\bar{x}_2 - \bar{x}_1)}{n_1 + n_2}$.
Since $\bar{x}_2 > \bar{x}_1$ and $n_1, n_2 > 0$,we have $\bar{x} - \bar{x}_1 > 0$,which implies $\bar{x} > \bar{x}_1$ $(i)$.
Similarly,$\bar{x} - \bar{x}_2 = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} - \bar{x}_2 = \frac{n_1(\bar{x}_1 - \bar{x}_2)}{n_1 + n_2}$.
Since $\bar{x}_1 < \bar{x}_2$,we have $\bar{x} - \bar{x}_2 < 0$,which implies $\bar{x} < \bar{x}_2$ (ii).
Combining $(i)$ and (ii),we get $\bar{x}_1 < \bar{x} < \bar{x}_2$.

(A) Let the sum of all $n$ observations be $S$. Given that the $A.M.$ of $n$ observations is $M$,we have $S = nM$.
Let the sum of the remaining $4$ observations be $S_4$. We are given that the sum of $n - 4$ observations is $a$.
Therefore,$S = a + S_4$,which implies $S_4 = S - a = nM - a$.
The mean of the remaining $4$ observations is $\frac{S_4}{4} = \frac{nM - a}{4}$.

(C) The formula for the mean is given by $\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}$.
Given the distribution:
$\sum f_i x_i = (1 \times 4) + (2 \times 5) + (3 \times y) + (4 \times 1) + (5 \times 2) = 4 + 10 + 3y + 4 + 10 = 28 + 3y$.
$\sum f_i = 4 + 5 + y + 1 + 2 = 12 + y$.
Given $\text{Mean} = 2.6$,we have:
$2.6 = \frac{28 + 3y}{12 + y}$.
Multiplying both sides by $(12 + y)$:
$2.6(12 + y) = 28 + 3y$.
$31.2 + 2.6y = 28 + 3y$.
$31.2 - 28 = 3y - 2.6y$.
$3.2 = 0.4y$.
$y = \frac{3.2}{0.4} = 8$.

(C) Given,the number of items $n = 100$ and the initial mean $\bar{x} = 49$.
The initial sum of the items is $S_{initial} = 100 \times 49 = 4900$.
The sum of the correct items is $S_{correct} = 60 + 70 + 80 = 210$.
The sum of the wrongly read items is $S_{wrong} = 40 + 20 + 50 = 110$.
The correct sum of the items is $S_{new} = S_{initial} + S_{correct} - S_{wrong} = 4900 + 210 - 110 = 5000$.
The correct mean is $\bar{x}_{new} = \frac{S_{new}}{n} = \frac{5000}{100} = 50$.

(C) The sum of $5$ numbers is $18 \times 5 = 90$.
After one number is excluded,the sum of the remaining $4$ numbers is $16 \times 4 = 64$.
Therefore,the excluded number is $90 - 64 = 26$.

(A) The class size is defined as the difference between two consecutive class marks.
Class size $= 10 - 6 = 4$.
Alternatively,Class size $= 14 - 10 = 4$.
Thus,the class size is $4$.

(D) We know that for any set of observations,the root mean square is greater than or equal to the arithmetic mean,i.e.,$RMS \ge AM$.
$\sqrt{\frac{\sum x_i^2}{n}} \ge \frac{\sum x_i}{n}$
Substituting the given values $\sum x_i^2 = 400$ and $\sum x_i = 80$:
$\sqrt{\frac{400}{n}} \ge \frac{80}{n}$
$\frac{20}{\sqrt{n}} \ge \frac{80}{n}$
$\frac{1}{\sqrt{n}} \ge \frac{4}{n}$
$\sqrt{n} \ge 4$
$n \ge 16$
Among the given options,the only value greater than or equal to $16$ is $18$. Therefore,the correct option is $D$.

(C) The formula for calculating the $k^{th}$ decile $(D_k)$ in a continuous frequency distribution is given by:
$D_k = l + \frac{(\frac{kN}{10} - C)}{f} \times i$
where:
$l$ is the lower limit of the decile class,
$N$ is the total frequency,
$C$ is the cumulative frequency of the class preceding the decile class,
$f$ is the frequency of the decile class,
$i$ is the class interval size.
For the $7^{th}$ decile $(k=7)$,the formula becomes:
$D_7 = l + \frac{(\frac{7N}{10} - C)}{f} \times i$.
Therefore,the correct option is $C$.

(D) Statement $(1)$ is correct: The mode of a grouped frequency distribution can be determined graphically using a histogram.
Statement $(2)$ is incorrect: Median is independent of the change of origin,but it is $NOT$ independent of the change of scale. If $y = a + bx$,then $Median(y) = a + b \times Median(x)$. Since it depends on $b$,it is not independent of the change of scale.
Statement $(3)$ is incorrect: Variance is independent of the change of origin,but it is $NOT$ independent of the change of scale. If $y = a + bx$,then $Var(y) = b^2 \times Var(x)$. Since it depends on $b^2$,it is not independent of the change of scale.
Therefore,only statement $(1)$ is correct.

(D) The measures of central tendency are statistical values that represent the center or typical value of a dataset.
Common measures of central tendency include the $Mean$,$Median$,and $Mode$.
$Range$ is a measure of dispersion (or variability),not central tendency,as it describes the spread of the data by calculating the difference between the maximum and minimum values.
Therefore,the correct option is $D$.

(C) From the given bar graph:
Mortality in the second quarter (April to June) = $250$.
Mortality in the third quarter (July to September) = $400$.
Increase in mortality = $400 - 250 = 150$.
Percentage increase = $\frac{\text{Increase}}{\text{Original Value}} \times 100$
Percentage increase = $\frac{150}{250} \times 100 = 0.6 \times 100 = 60\%$.

(B) In a histogram,the height of each bar corresponds to the frequency of the class interval.
Looking at the given data,the frequencies are:
- $150-300$: $300$
- $300-500$: $500$
- $500-800$: $900$
- $800-1200$: $1000$
- $1200-1800$: $1200$
The maximum frequency is $1200$,which corresponds to the income group $1200-1800$.
Therefore,the highest bar in the histogram will correspond to the class $1200-1800$.

(B) $Pie \ diagram$ (or pie chart) is the most effective way to represent the total expenditure of an industry distributed across different heads because it shows the proportion of each part to the whole.

(C) Total expenditure $= 560 + 420 + 180 + 160 + 120 = 1440$.
The angle of the sector for clothes is calculated as:
$\text{Angle} = \left( \frac{\text{Expenditure on clothes}}{\text{Total expenditure}} \right) \times 360^o$
$\text{Angle} = \left( \frac{180}{1440} \right) \times 360^o$
$\text{Angle} = \frac{1}{8} \times 360^o = 45^o$.

(B) The range of a set of observations is defined as the difference between the maximum value and the minimum value in the set.
Range $= X_{\max} - X_{\min}$
Given observations: $2, 3, 5, 9, 8, 7, 6, 5, 7, 4, 3$
Maximum value $(X_{\max}) = 9$
Minimum value $(X_{\min}) = 2$
Range $= 9 - 2 = 7$.

(B) The given observations are ${x_1}, -{x_1}, {x_2}, -{x_2}, \dots, {x_n}, -{x_n}, 0$. There are $(2n+1)$ observations.
$1$. Median $(M.D.)$: Arranging the observations in ascending order,the middle term is $0$. Thus,$M.D. = 0$.
$2$. Standard Deviation $(S.D.)$: The formula for $S.D.$ is $\sqrt{\frac{1}{N} \sum (x_i - \bar{x})^2}$.
Here,the mean $\bar{x} = \frac{x_1 - x_1 + x_2 - x_2 + \dots + x_n - x_n + 0}{2n+1} = 0$.
So,$S.D. = \sqrt{\frac{1}{2n+1} \sum_{i=1}^n (x_i^2 + (-x_i)^2 + 0^2)} = \sqrt{\frac{2 \sum x_i^2}{2n+1}}$.
Since all $x_i$ are distinct and non-zero,$\sum x_i^2 > 0$,which implies $S.D. > 0$.
Comparing the two,$S.D. > 0$ and $M.D. = 0$,therefore $S.D. > M.D.$

(C) Let the $n$ items be ${x_1}, {x_2}, \dots, {x_n}$. Then,the original mean is $\bar x = \frac{1}{n} \sum_{i=1}^{n} {x_i}$.
Let the new items be ${y_i} = {x_i} + i$ for $i = 1, 2, \dots, n$.
The new mean is $\bar y = \frac{1}{n} \sum_{i=1}^{n} {y_i} = \frac{1}{n} \sum_{i=1}^{n} ({x_i} + i)$.
This can be written as $\bar y = \frac{1}{n} \sum_{i=1}^{n} {x_i} + \frac{1}{n} \sum_{i=1}^{n} i$.
Since $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$,we have $\bar y = \bar x + \frac{1}{n} \cdot \frac{n(n+1)}{2}$.
Therefore,the new mean is $\bar x + \frac{n+1}{2}$.

(B) Given,number of observations $n = 100$ and mean $\bar{x} = 45$.
Sum of $100$ observations $= 100 \times 45 = 4500$.
Incorrect sum of observations $= 91 + 13 = 104$.
Correct sum of observations $= 19 + 31 = 50$.
Correct sum $= 4500 - 104 + 50 = 4446$.
Correct mean $= \frac{4446}{100} = 44.46$.

(D) First,calculate the total number of families:
$Total = 150 + 400 + 40 + 250 + 160 = 1000$.
The central angle for a category is calculated as:
$\text{Central Angle} = \left( \frac{\text{Value of category}}{\text{Total value}} \right) \times 360^\circ$.
For food and clothing,the value is $400$:
$\text{Central Angle} = \left( \frac{400}{1000} \right) \times 360^\circ = 0.4 \times 360^\circ = 144^\circ$.
Thus,the correct option is $D$.

(B) Given $n = 200$,incorrect mean $\bar{x} = 40$,incorrect $S.D. \sigma = 15$.
Incorrect $\Sigma x = n \times \bar{x} = 200 \times 40 = 8000$.
Corrected $\Sigma x = 8000 - 50 + 40 = 7990$.
Corrected mean $\bar{x}_{new} = \frac{7990}{200} = 39.95$.
Incorrect $\Sigma x^2 = n(\sigma^2 + \bar{x}^2) = 200(15^2 + 40^2) = 200(225 + 1600) = 200(1825) = 365000$.
Corrected $\Sigma x^2 = 365000 - 50^2 + 40^2 = 365000 - 2500 + 1600 = 364100$.
Corrected $\sigma_{new} = \sqrt{\frac{\Sigma x^2}{n} - (\bar{x}_{new})^2} = \sqrt{\frac{364100}{200} - (39.95)^2}$.
$\sigma_{new} = \sqrt{1820.5 - 1596.0025} = \sqrt{224.4975} \approx 14.98$.

(A) The range $r$ is defined as $r = \max |x_i - x_j|$ for $i \neq j$.
Given the variance $S^2 = \frac{1}{n - 1} \sum_{i = 1}^n (x_i - \bar{x})^2$.
It is a known statistical inequality that for any set of $n$ observations,the standard deviation $S$ and the range $r$ satisfy the relation $S \le r \sqrt{\frac{n}{n - 1}}$.
This is derived from the fact that $\sum (x_i - \bar{x})^2 \le n \frac{(n-1)}{n} r^2$ is not the correct bound,but rather $\sum (x_i - \bar{x})^2 \le \frac{n}{4} r^2$ for specific cases,however,the standard inequality for the sample standard deviation $S$ in terms of range $r$ is $S \le r \sqrt{\frac{n}{n-1}}$.

(A) The mean is given by the formula: $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$.
First,calculate $\Sigma f_i x_i = (1 \times 5) + (2 \times 4) + (3 \times f) + (4 \times 2) + (5 \times 3) = 5 + 8 + 3f + 8 + 15 = 3f + 36$.
Next,calculate $\Sigma f_i = 5 + 4 + f + 2 + 3 = f + 14$.
Given the mean is $2.6$,we have:
$2.6 = \frac{3f + 36}{f + 14}$
Multiply both sides by $(f + 14)$:
$2.6(f + 14) = 3f + 36$
$2.6f + 36.4 = 3f + 36$
Rearrange the terms to solve for $f$:
$36.4 - 36 = 3f - 2.6f$
$0.4 = 0.4f$
$f = 1$.

(A) Given $\Sigma f = 17 + f_1 + 32 + f_2 + 19 = 120$.
$\Rightarrow f_1 + f_2 + 68 = 120$ $\Rightarrow f_1 + f_2 = 52$ $(1)$
The class marks $(x)$ are $10, 30, 50, 70, 90$.
$\Sigma fx = (10 \times 17) + (30 \times f_1) + (50 \times 32) + (70 \times f_2) + (90 \times 19)$
$\Sigma fx = 170 + 30f_1 + 1600 + 70f_2 + 1710 = 30f_1 + 70f_2 + 3480$.
Mean $\bar{x} = \frac{\Sigma fx}{\Sigma f} = 50$.
$\frac{30f_1 + 70f_2 + 3480}{120} = 50$
$30f_1 + 70f_2 + 3480 = 6000$
$30f_1 + 70f_2 = 2520 \Rightarrow 3f_1 + 7f_2 = 252$ $(2)$
From $(1)$,$f_1 = 52 - f_2$. Substituting in $(2)$:
$3(52 - f_2) + 7f_2 = 252$
$156 - 3f_2 + 7f_2 = 252$
$4f_2 = 96 \Rightarrow f_2 = 24$.
Then $f_1 = 52 - 24 = 28$.
Thus,the missing frequencies are $28$ and $24$.

(B) Let the number of boys and girls be $n_1$ and $n_2$ respectively.
The combined mean is given by the formula:
$\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$
Substituting the given values:
$50 = \frac{52n_1 + 42n_2}{n_1 + n_2}$
$50(n_1 + n_2) = 52n_1 + 42n_2$
$50n_1 + 50n_2 = 52n_1 + 42n_2$
$8n_2 = 2n_1$
$n_1 = 4n_2$
The percentage of boys is given by:
$\text{Percentage} = \frac{n_1}{n_1 + n_2} \times 100$
Substituting $n_1 = 4n_2$:
$\text{Percentage} = \frac{4n_2}{4n_2 + n_2} \times 100 = \frac{4n_2}{5n_2} \times 100 = \frac{4}{5} \times 100 = 80\%$

(C) Let the original mean be $\bar{x} = \frac{x_1 + x_2 + \dots + x_n}{n}$.
The new values are $(x_1 + 1), (x_2 + 2), \dots, (x_n + n)$.
The new mean is $\bar{x}_{new} = \frac{(x_1 + 1) + (x_2 + 2) + \dots + (x_n + n)}{n}$.
$\bar{x}_{new} = \frac{(x_1 + x_2 + \dots + x_n) + (1 + 2 + \dots + n)}{n}$.
$\bar{x}_{new} = \frac{\sum x_i}{n} + \frac{\sum_{i=1}^{n} i}{n}$.
Since $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$,we have $\bar{x}_{new} = \bar{x} + \frac{n(n+1)}{2n}$.
$\bar{x}_{new} = \bar{x} + \frac{n+1}{2}$.

(B) Let the new observation be $x$.
The sum of $9$ observations is $9 \times 15 = 135$.
After adding the new observation $x$,the total number of observations becomes $10$ and the new mean is $16$.
So,the new sum is $10 \times 16 = 160$.
Therefore,$135 + x = 160$.
$x = 160 - 135 = 25$.

(D) The Harmonic Mean $(H.M.)$ of a set of $n$ observations $x_1, x_2, \dots, x_n$ is defined as:
$H.M. = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}$
Here,the number of terms $n = 5$.
The terms are $3, 7, 8, 10, 14$.
Therefore,the Harmonic Mean is:
$H.M. = \frac{5}{\frac{1}{3} + \frac{1}{7} + \frac{1}{8} + \frac{1}{10} + \frac{1}{14}}$

(B) Let the two data sets be $X$ and $Y$ with sizes $n_1 = 5$ and $n_2 = 5$.
Given: $\sigma_1^2 = 4$,$\sigma_2^2 = 5$,$\bar{x}_1 = 2$,$\bar{x}_2 = 4$.
For set $X$: $\sigma_1^2 = \frac{\Sigma x_i^2}{n_1} - \bar{x}_1^2 \implies 4 = \frac{\Sigma x_i^2}{5} - 4 \implies \Sigma x_i^2 = 40$.
For set $Y$: $\sigma_2^2 = \frac{\Sigma y_i^2}{n_2} - \bar{x}_2^2 \implies 5 = \frac{\Sigma y_i^2}{5} - 16 \implies \Sigma y_i^2 = 105$.
The combined mean $\bar{x}_{comb} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{5(2) + 5(4)}{10} = 3$.
The combined variance $\sigma^2 = \frac{\Sigma x_i^2 + \Sigma y_i^2}{n_1 + n_2} - (\bar{x}_{comb})^2$.
$\sigma^2 = \frac{40 + 105}{10} - (3)^2 = \frac{145}{10} - 9 = 14.5 - 9 = 5.5 = \frac{11}{2}$.

(B) The formula for the mean using the assumed mean method is given by:
$\text{Mean} = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$
Where:
$a$ is the assumed mean.
$f_i$ is the frequency of the $i$-th class.
$d_i = x_i - a$ is the deviation of the $i$-th observation from the assumed mean $a$.
Therefore,$a$ represents the assumed mean.

(C) The maximum value of the given data is $14$ and the minimum value is $6$.
The range is defined as the difference between the maximum and minimum values.
Range $= 14 - 6 = 8$.

(D) Given that the mean of the series is $\bar{X} = \frac{x_1 + x_2 + \dots + x_n}{n}$.
This implies the sum of the observations is $\sum_{i=1}^{n} x_i = n\bar{X}$.
When $x_2$ is replaced by $\lambda$,the new sum of observations becomes $S_{new} = (n\bar{X} - x_2 + \lambda)$.
The new mean is $\bar{X}_{new} = \frac{S_{new}}{n} = \frac{n\bar{X} - x_2 + \lambda}{n}$.

(B) The weighted mean is given by the formula: $\text{Weighted Mean} = \frac{\sum x_i w_i}{\sum w_i}$.
Here,$x_i = i$ and $w_i = i$ for $i = 1, 2, \ldots, n$.
So,$\text{Weighted Mean} = \frac{\sum_{i=1}^n i \cdot i}{\sum_{i=1}^n i} = \frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}$.
Using the standard summation formulas $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^n i = \frac{n(n+1)}{2}$,we get:
$\text{Weighted Mean} = \frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}} = \frac{2n+1}{3}$.

(C) Given $n = 5$,mean $\bar{x} = 4$,and variance $\sigma^2 = 5.2$. Let the other two observations be $x_1$ and $x_2$.
The mean is given by $\frac{x_1 + x_2 + 1 + 2 + 6}{5} = 4$.
$x_1 + x_2 + 9 = 20 \implies x_1 + x_2 = 11$.
The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$5.2 = \frac{x_1^2 + x_2^2 + 1^2 + 2^2 + 6^2}{5} - 4^2$.
$5.2 = \frac{x_1^2 + x_2^2 + 1 + 4 + 36}{5} - 16$.
$21.2 = \frac{x_1^2 + x_2^2 + 41}{5}$.
$106 = x_1^2 + x_2^2 + 41 \implies x_1^2 + x_2^2 = 65$.
We have $x_1 + x_2 = 11$ and $x_1^2 + x_2^2 = 65$.
Using $(x_1 + x_2)^2 = x_1^2 + x_2^2 + 2x_1x_2$,we get $121 = 65 + 2x_1x_2 \implies 2x_1x_2 = 56 \implies x_1x_2 = 28$.
The quadratic equation $t^2 - (x_1+x_2)t + x_1x_2 = 0$ becomes $t^2 - 11t + 28 = 0$.
$(t - 4)(t - 7) = 0$.
Thus,the observations are $4$ and $7$.

(C) Let the total number of people in the group be $100$. Let $n_w$ be the number of women and $n_m$ be the number of men,such that $n_w + n_m = 100$.
The average age of the group is given by the formula:
$\frac{32n_m + 27n_w}{100} = 30$
Substituting $n_m = 100 - n_w$ into the equation:
$32(100 - n_w) + 27n_w = 3000$
$3200 - 32n_w + 27n_w = 3000$
$3200 - 5n_w = 3000$
$5n_w = 200$
$n_w = 40$
Therefore,the percentage of women in the group is $40\%$.

(C) Given: Total observations $N = 100$,so $\frac{N}{2} = 50$.
Cumulative frequency of the class preceding the median class $F = 45$.
Lower limit of the median class $l = 20$.
Class width $h = 30 - 20 = 10$.
Median $= 25$.
Let $f$ be the frequency of the median class.
The formula for median is: $\text{Median} = l + \left( \frac{\frac{N}{2} - F}{f} \right) \times h$.
Substituting the values: $25 = 20 + \left( \frac{50 - 45}{f} \right) \times 10$.
$25 - 20 = \left( \frac{5}{f} \right) \times 10$.
$5 = \frac{50}{f}$.
$f = \frac{50}{5} = 10$.

(B) Given observations are: $2, 3, 5, 9, 8, 7, 6, 5, 7, 4, 3$.
The largest value $(x_{max})$ in the given data is $9$.
The smallest value $(x_{min})$ in the given data is $2$.
The range of a data set is defined as the difference between the largest and the smallest observation.
$\text{Range} = x_{max} - x_{min} = 9 - 2 = 7$.

(D) The mode of the distribution is given as $140$. Since $140$ lies in the class interval $100-150$,the modal class is $100-150$.
The formula for mode is $Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$.
Here,$l = 100$ (lower limit of modal class),$f_1 = 37$ (frequency of modal class),$f_0 = 33$ (frequency of preceding class),$f_2 = b$ (frequency of succeeding class),and $h = 50$ (class size).
Substituting the values:
$140 = 100 + \left( \frac{37 - 33}{2(37) - 33 - b} \right) \times 50$
$40 = \left( \frac{4}{74 - 33 - b} \right) \times 50$
$40 = \frac{200}{41 - b}$
$40(41 - b) = 200$
$41 - b = 5$
$b = 41 - 5 = 36$.
Thus,the value of $b$ is $36$.

(D) Let the two series be $x_1, x_2, \dots, x_{n_1}$ and $y_1, y_2, \dots, y_{n_2}$ with sizes $n_1$ and $n_2$ respectively.
$G_1 = (x_1 \times x_2 \times \dots \times x_{n_1})^{1/n_1} \implies G_1^{n_1} = x_1 \times x_2 \times \dots \times x_{n_1} \dots (1)$
$G_2 = (y_1 \times y_2 \times \dots \times y_{n_2})^{1/n_2} \implies G_2^{n_2} = y_1 \times y_2 \times \dots \times y_{n_2} \dots (2)$
The combined geometric mean $G$ is given by:
$G = [(x_1 \times x_2 \times \dots \times x_{n_1}) \times (y_1 \times y_2 \times \dots \times y_{n_2})]^{\frac{1}{n_1 + n_2}}$
Substituting $(1)$ and $(2)$ into the expression for $G$:
$G = (G_1^{n_1} \times G_2^{n_2})^{\frac{1}{n_1 + n_2}}$
Taking the logarithm on both sides:
$\log G = \log [(G_1^{n_1} \times G_2^{n_2})^{\frac{1}{n_1 + n_2}}]$
Using the property $\log(a^b) = b \log a$ and $\log(ab) = \log a + \log b$:
$\log G = \frac{1}{n_1 + n_2} \log (G_1^{n_1} \times G_2^{n_2})$
$\log G = \frac{n_1 \log G_1 + n_2 \log G_2}{n_1 + n_2}$

(C) Let the number of boys be $n_1$ and the number of girls be $n_2$.
The combined mean weight is given by the formula:
$\frac{65n_1 + 55n_2}{n_1 + n_2} = 61$
Multiplying both sides by $(n_1 + n_2)$:
$65n_1 + 55n_2 = 61n_1 + 61n_2$
Rearranging the terms to group $n_1$ and $n_2$:
$65n_1 - 61n_1 = 61n_2 - 55n_2$
$4n_1 = 6n_2$
Therefore,the ratio is:
$\frac{n_1}{n_2} = \frac{6}{4} = \frac{3}{2}$
Thus,the ratio of the number of boys to girls is $3 : 2$.

(A) Let the number of individuals in the two groups be $n_1$ and $n_2$ respectively.
Given: $\bar{x}_1 = 400$,$\bar{x}_2 = 480$,and the combined mean $\bar{x} = 430$.
The formula for the combined mean is $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$.
Substituting the values: $430 = \frac{400n_1 + 480n_2}{n_1 + n_2}$.
$430(n_1 + n_2) = 400n_1 + 480n_2$.
$430n_1 + 430n_2 = 400n_1 + 480n_2$.
$430n_1 - 400n_1 = 480n_2 - 430n_2$.
$30n_1 = 50n_2$.
$\frac{n_1}{n_2} = \frac{50}{30} = \frac{5}{3}$.

(C) Let the number of individuals in the first group be $n_1 = 4k$ and in the second group be $n_2 = 3k$,where $k$ is a constant.
The combined average income is given by the formula:
$\text{Combined Average} = \frac{n_1\bar{x} + n_2\bar{y}}{n_1 + n_2}$
Substituting the values:
$\text{Combined Average} = \frac{(4k)\bar{x} + (3k)\bar{y}}{4k + 3k}$
$\text{Combined Average} = \frac{k(4\bar{x} + 3\bar{y})}{7k}$
$\text{Combined Average} = \frac{4\bar{x} + 3\bar{y}}{7}$

(A) The coefficient of variation $(CV)$ is given by the formula:
$CV = \left( \frac{\sigma}{\bar{x}} \right) \times 100$
where $\sigma$ is the standard deviation and $\bar{x}$ is the mean.
Given: $CV = 50\%$ and $\sigma = 20$.
Substituting the values into the formula:
$50 = \left( \frac{20}{\bar{x}} \right) \times 100$
$50 = \frac{2000}{\bar{x}}$
$\bar{x} = \frac{2000}{50}$
$\bar{x} = 40$.
Therefore,the mean is $40$.

(A) Let the number of boys be $x$ and the number of girls be $y$.
The total marks of boys is $52x$ and the total marks of girls is $42y$.
The combined average marks is given by $\frac{52x + 42y}{x + y} = 50$.
Multiplying both sides by $(x + y)$,we get $52x + 42y = 50(x + y)$.
Expanding the equation: $52x + 42y = 50x + 50y$.
Rearranging the terms: $52x - 50x = 50y - 42y$.
$2x = 8y$,which simplifies to $x = 4y$.
The total number of students is $x + y = 4y + y = 5y$.
The percentage of boys is $\frac{x}{x + y} \times 100 = \frac{4y}{5y} \times 100 = 80\%$.