Word problem -Statistics Questions in English

(A) Given: $\sigma_{x}^{2} = 4$ and $\sigma_{y}^{2} = 5$ for two sets of size $n_1 = 5$ and $n_2 = 5$.
Means are $\bar{x} = 2$ and $\bar{y} = 4$.
Using the formula $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2$,we have:
$\Sigma x_i^2 = n(\sigma_x^2 + \bar{x}^2) = 5(4 + 2^2) = 5(8) = 40$.
$\Sigma y_i^2 = n(\sigma_y^2 + \bar{y}^2) = 5(5 + 4^2) = 5(21) = 105$.
Combined mean $\bar{z} = \frac{n_1\bar{x} + n_2\bar{y}}{n_1 + n_2} = \frac{5(2) + 5(4)}{10} = \frac{30}{10} = 3$.
Combined variance $\sigma_z^2 = \frac{\Sigma x_i^2 + \Sigma y_i^2}{n_1 + n_2} - (\bar{z})^2$.
$\sigma_z^2 = \frac{40 + 105}{10} - (3)^2$.
$\sigma_z^2 = \frac{145}{10} - 9 = 14.5 - 9 = 5.5 = \frac{11}{2}$.

(A) Given that the mean of $16$ observations is $16$,the sum of the observations is:
$\sum_{i=1}^{16} x_i = 16 \times 16 = 256$.
After deleting the observation with value $16$ and adding three new observations $(3, 4, 5)$,the new sum of the observations is:
$\text{New Sum} = 256 - 16 + 3 + 4 + 5 = 252$.
The new number of observations is $16 - 1 + 3 = 18$.
The new mean is:
$\text{New Mean} = \frac{252}{18} = 14$.

(C) The formula for combined mean is $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$.
Given,$\bar{x} = 500$,$\bar{x}_1 = 510$,$\bar{x}_2 = 460$.
Let the total number of employees be $100$,where $n_1$ is the number of male employees and $n_2$ is the number of female employees.
Thus,$n_2 = 100 - n_1$.
Substituting the values in the formula:
$500 = \frac{510n_1 + 460(100 - n_1)}{100}$
$50000 = 510n_1 + 46000 - 460n_1$
$50000 - 46000 = 50n_1$
$4000 = 50n_1$
$n_1 = \frac{4000}{50} = 80$.
Hence,the percentage of male employees in the factory is $80\%$.

(B) Given for the first set: $n_1 = 5$,$\bar{x}_1 = 8$,$\sigma_1^2 = 18$.
Given for the second set: $n_2 = 3$,$\bar{x}_2 = 8$,$\sigma_2^2 = 24$.
Combined mean $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{5 \times 8 + 3 \times 8}{5 + 3} = \frac{64}{8} = 8$.
Since $\bar{x}_1 = \bar{x}_2 = \bar{x} = 8$,the deviations $D_1 = \bar{x}_1 - \bar{x} = 0$ and $D_2 = \bar{x}_2 - \bar{x} = 0$.
The combined variance is given by $\sigma^2 = \frac{n_1(\sigma_1^2 + D_1^2) + n_2(\sigma_2^2 + D_2^2)}{n_1 + n_2}$.
Substituting the values: $\sigma^2 = \frac{5(18 + 0^2) + 3(24 + 0^2)}{5 + 3} = \frac{90 + 72}{8} = \frac{162}{8} = 20.25$.

(B) Let the two unknown observations be $x$ and $y$.
Given that the mean of $5$ observations is $4.4$,we have:
$\frac{1 + 2 + 6 + x + y}{5} = 4.4$
$9 + x + y = 22$
$x + y = 13$ ..... $(i)$
Given that the variance is $8.24$,we use the formula $\text{Variance} = \frac{\sum x_i^2}{n} - (\text{mean})^2$:
$8.24 = \frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - (4.4)^2$
$8.24 = \frac{1 + 4 + 36 + x^2 + y^2}{5} - 19.36$
$8.24 + 19.36 = \frac{41 + x^2 + y^2}{5}$
$27.6 = \frac{41 + x^2 + y^2}{5}$
$138 = 41 + x^2 + y^2$
$x^2 + y^2 = 97$ ..... $(ii)$
From $(i)$,$y = 13 - x$. Substituting into $(ii)$:
$x^2 + (13 - x)^2 = 97$
$x^2 + 169 - 26x + x^2 = 97$
$2x^2 - 26x + 72 = 0$
$x^2 - 13x + 36 = 0$
$(x - 9)(x - 4) = 0$
Thus,$x = 9$ or $x = 4$. If $x = 9$,then $y = 4$. If $x = 4$,then $y = 9$. The observations are $4$ and $9$.

(B) The combined variance $\sigma^{2}$ of two sets with sizes $n_{1}, n_{2}$,means $\bar{x}_{1}, \bar{x}_{2}$,and variances $\sigma_{1}^{2}, \sigma_{2}^{2}$ is given by:
$\sigma^{2} = \frac{n_{1} \sigma_{1}^{2} + n_{2} \sigma_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2}}{(n_{1} + n_{2})^{2}} (\bar{x}_{1} - \bar{x}_{2})^{2}$
Given $n_{1} = 6, \bar{x}_{1} = 11, \sigma_{1}^{2} = 24$ and $n_{2} = 3, \bar{x}_{2} = 14, \sigma_{2}^{2} = 36$:
$\sigma^{2} = \frac{6 \times 24 + 3 \times 36}{6 + 3} + \frac{6 \times 3}{(6 + 3)^{2}} (11 - 14)^{2}$
$\sigma^{2} = \frac{144 + 108}{9} + \frac{18}{81} \times (-3)^{2}$
$\sigma^{2} = \frac{252}{9} + \frac{18}{81} \times 9$
$\sigma^{2} = 28 + 2 = 30$

(C) The mean of $w_i$ is given by $\bar{w} = l \bar{y} + k$.
Given $\bar{y} = 48$ and $\bar{w} = 55$,we have $55 = 48l + k$ $(i)$.
The standard deviation of $w_i$ is given by $\sigma_w = |l| \sigma_y$.
Given $\sigma_y = 12$ and $\sigma_w = 15$,we have $15 = |l| \times 12$.
Thus,$|l| = \frac{15}{12} = 1.25$.
Assuming $l$ is positive,$l = 1.25$.
Substituting $l = 1.25$ into equation $(i)$:
$55 = 48(1.25) + k$
$55 = 60 + k$
$k = 55 - 60 = -5$.
Therefore,$l = 1.25$ and $k = -5$.

(B) Let the $10$ terms be $x_1, x_2, \dots, x_{10}$.
Given that the mean is $\frac{\sum_{i=1}^{10} x_i}{10} = 3$,which implies $\sum_{i=1}^{10} x_i = 30$.
When the first term is increased by $1$,the second by $2$,and the $i$-th term by $i$,the new sum of terms is:
$\sum_{i=1}^{10} (x_i + i) = \sum_{i=1}^{10} x_i + \sum_{i=1}^{10} i$.
Using the sum of the first $n$ natural numbers formula $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$,we get $\sum_{i=1}^{10} i = \frac{10 \times 11}{2} = 55$.
New sum $= 30 + 55 = 85$.
The new mean is $\frac{85}{10} = \frac{17}{2}$.

(B) Let the observations be $x_1, x_2, \dots, x_{30}$. The mean is given by $\bar{x} = \frac{1}{30} \sum_{i=1}^{30} x_i = 75$.
When each observation is multiplied by $\lambda$ and decreased by $25$,the new observations are $y_i = \lambda x_i - 25$.
The new mean is $\bar{y} = \frac{1}{30} \sum_{i=1}^{30} (\lambda x_i - 25) = \lambda \left( \frac{1}{30} \sum_{i=1}^{30} x_i \right) - 25 = 75\lambda - 25$.
According to the problem,the mean remains the same,so $\bar{y} = 75$.
Thus,$75\lambda - 25 = 75$.
$75\lambda = 100$.
$\lambda = \frac{100}{75} = \frac{4}{3}$.

(C) Let the sum of the ages of the $25$ teachers be $S$.
Given that the mean age is $40 \text{ years}$,we have:
$\frac{S}{25} = 40 \Rightarrow S = 1000$.
Let the age of the new teacher be $A$.
After the retirement of a $60 \text{ year}$ old teacher and the appointment of the new teacher,the new sum of ages is $S - 60 + A$.
The new mean age is $39 \text{ years}$,so:
$\frac{S - 60 + A}{25} = 39$
$1000 - 60 + A = 39 \times 25$
$940 + A = 975$
$A = 975 - 940 = 35$.
Thus,the age of the newly appointed teacher is $35 \text{ years}$.

(C) Let the mean wage of the night shift workers be $x$.
The total number of workers is $70 + 30 = 100$.
The sum of wages for day shift workers is $70 \times 54 = 3780$.
The sum of wages for all workers is $100 \times 60 = 6000$.
The sum of wages for night shift workers is $30 \times x$.
We have the equation: $3780 + 30x = 6000$.
$30x = 6000 - 3780 = 2220$.
$x = \frac{2220}{30} = 74$.
Thus,the mean wage of the night shift workers is $Rs. 74$.

(A) Let the $2n$ distinct observations be $x_1 < x_2 < ... < x_{2n}$.
Since there are $2n$ observations,the median is the average of the $n^{th}$ and $(n+1)^{th}$ observations.
There are $n$ observations below the median (i.e.,$x_1, ..., x_n$) and $n$ observations above the median (i.e.,$x_{n+1}, ..., x_{2n}$).
The sum of the original observations is $S = \sum_{i=1}^{2n} x_i$.
The new sum $S'$ is obtained by adding $5$ to each of the first $n$ observations and subtracting $3$ from each of the remaining $n$ observations:
$S' = \sum_{i=1}^{n} (x_i + 5) + \sum_{i=n+1}^{2n} (x_i - 3)$
$S' = \sum_{i=1}^{n} x_i + 5n + \sum_{i=n+1}^{2n} x_i - 3n$
$S' = \sum_{i=1}^{2n} x_i + 2n = S + 2n$.
The new mean is $M' = \frac{S'}{2n} = \frac{S + 2n}{2n} = \frac{S}{2n} + 1$.
Since the original mean is $M = \frac{S}{2n}$,the new mean is $M' = M + 1$.
Thus,the mean increases by $1$.

(B) Given $d_i = -x_i - a$.
Statement $I$: The variance of a set of observations is invariant under change of origin and scale factor $-1$. Specifically,if $y_i = c x_i + k$,then $\text{Var}(y) = c^2 \text{Var}(x)$. Here,$c = -1$ and $k = -a$. Thus,$\text{Var}(d) = (-1)^2 \sigma^2 = \sigma^2$. So,Statement $I$ is true.
Statement $II$: The mean of $d_i$ is $\bar{d} = \frac{1}{n} \sum (-x_i - a) = -\bar{x} - a$. This is correct.
For the mode,if $M$ is the mode of $x_i$,then the mode of $d_i = -x_i - a$ is $-M - a$. This is also correct.
Therefore,both statements are true.

(A) Given,number of observations $n = 20$ and incorrect mean $\bar{x}_{incorrect} = 40$.
Incorrect sum of observations $= n \times \bar{x}_{incorrect} = 20 \times 40 = 800$.
Correct sum of observations $= \text{Incorrect sum} - \text{wrong value} + \text{correct value} = 800 - 33 + 53 = 820$.
Correct mean $= \frac{\text{Correct sum}}{n} = \frac{820}{20} = 41$.

(D) Given the condition $0 < y < x < 2y$,we can order the numbers as follows:
Since $y < x$ and $x < 2y$,we have $x - y < y < x < 2x + y$.
The median of the four numbers is the average of the two middle terms:
$\text{Median} = \frac{y + x}{2} = 10 \Rightarrow x + y = 20 \quad (i)$
The range is the difference between the largest and smallest values:
$\text{Range} = (2x + y) - (x - y) = x + 2y = 28 \quad (ii)$
Subtracting equation $(i)$ from $(ii)$:
$(x + 2y) - (x + y) = 28 - 20 \Rightarrow y = 8$.
Substituting $y = 8$ into $(i)$:
$x + 8 = 20 \Rightarrow x = 12$.
The four numbers are $(12 - 8), 8, 12, (2(12) + 8)$,which are $4, 8, 12, 32$.
The mean is $\frac{4 + 8 + 12 + 32}{4} = \frac{56}{4} = 14$.

(D) The modal class is the class interval with the highest frequency. Since the mode is $140$,which lies in the interval $100-150$,the modal class is $100-150$.
The formula for mode is:
$Mode = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
$L = 100$ (lower limit of modal class)
$f_1 = 37$ (frequency of modal class)
$f_0 = 33$ (frequency of class preceding modal class)
$f_2 = b$ (frequency of class succeeding modal class)
$h = 50$ (class size)
Substituting the values:
$140 = 100 + \left( \frac{37 - 33}{2(37) - 33 - b} \right) \times 50$
$40 = \left( \frac{4}{74 - 33 - b} \right) \times 50$
$40 = \frac{200}{41 - b}$
$40(41 - b) = 200$
$41 - b = 5$
$b = 41 - 5 = 36$

(A) The formula for the median of grouped data is:
$M = l + \left( \frac{\frac{N}{2} - F}{f} \right) \times C$
Where:
$l = 20$ (lower limit of the median class)
$N = 100$ (total number of observations)
$F = 45$ (cumulative frequency of the class preceding the median class)
$C = 30 - 20 = 10$ (width of the median class)
$M = 25$ (median)
$f$ is the frequency of the median class.
Substituting the values into the formula:
$25 = 20 + \left( \frac{\frac{100}{2} - 45}{f} \right) \times 10$
$25 - 20 = \left( \frac{50 - 45}{f} \right) \times 10$
$5 = \frac{5}{f} \times 10$
$5 = \frac{50}{f}$
$f = \frac{50}{5} = 10$

(B) Let the five observations be $1, 3, 8, x,$ and $y$.
Given mean $\mu = 5$,so $\frac{1 + 3 + 8 + x + y}{5} = 5$.
$12 + x + y = 25 \Rightarrow x + y = 13$ (Equation $1$).
Given variance $\sigma^2 = 9.20$,using the formula $\sigma^2 = \frac{\sum x_i^2}{N} - \mu^2$:
$9.2 = \frac{1^2 + 3^2 + 8^2 + x^2 + y^2}{5} - 5^2$.
$9.2 = \frac{1 + 9 + 64 + x^2 + y^2}{5} - 25$.
$34.2 = \frac{74 + x^2 + y^2}{5} \Rightarrow 171 = 74 + x^2 + y^2$.
$x^2 + y^2 = 97$ (Equation $2$).
We know $(x + y)^2 = x^2 + y^2 + 2xy$,so $13^2 = 97 + 2xy$.
$169 - 97 = 2xy$ $\Rightarrow 72 = 2xy$ $\Rightarrow xy = 36$.
Solving $x + y = 13$ and $xy = 36$,the quadratic equation $t^2 - 13t + 36 = 0$ gives $(t - 4)(t - 9) = 0$.
Thus,the two observations are $4$ and $9$.
The ratio is $\frac{4}{9}$ or $\frac{9}{4}$.

(A) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given $n = 5$,$\bar{x} = 4$,and $\sigma^2 = 5.2$.
Sum of observations: $\sum x_i = n \times \bar{x} = 5 \times 4 = 20$.
Given $x_1 = 3, x_2 = 4, x_3 = 4$,we have $3 + 4 + 4 + x_4 + x_5 = 20$,so $x_4 + x_5 = 9$ $(i)$.
Variance formula: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$5.2 = \frac{\sum x_i^2}{5} - 4^2$ $\Rightarrow 5.2 = \frac{\sum x_i^2}{5} - 16$ $\Rightarrow \sum x_i^2 = 5 \times 21.2 = 106$.
Sum of squares: $3^2 + 4^2 + 4^2 + x_4^2 + x_5^2 = 106$ $\Rightarrow 9 + 16 + 16 + x_4^2 + x_5^2 = 106$ $\Rightarrow x_4^2 + x_5^2 = 65$ $(ii)$.
We know $(x_4 - x_5)^2 = 2(x_4^2 + x_5^2) - (x_4 + x_5)^2$.
$(x_4 - x_5)^2 = 2(65) - (9)^2 = 130 - 81 = 49$.
Therefore,$|x_4 - x_5| = \sqrt{49} = 7$.

(D) Let the $7$ observations be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
Given mean $\bar{x} = 8$,so $\sum_{i=1}^{7} x_i = 7 \times 8 = 56$.
Given variance $\sigma^2 = 16$,we use the formula $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.
$16 = \frac{1}{7} \sum_{i=1}^{7} x_i^2 - 8^2$.
$16 = \frac{1}{7} \sum_{i=1}^{7} x_i^2 - 64 \Rightarrow \sum_{i=1}^{7} x_i^2 = 7 \times 80 = 560$.
Given $5$ observations are $2, 4, 10, 12, 14$. Let the remaining two be $x_6$ and $x_7$.
Sum of $5$ observations: $2 + 4 + 10 + 12 + 14 = 42$.
$x_6 + x_7 = 56 - 42 = 14$.
Sum of squares of $5$ observations: $2^2 + 4^2 + 10^2 + 12^2 + 14^2 = 4 + 16 + 100 + 144 + 196 = 460$.
$x_6^2 + x_7^2 = 560 - 460 = 100$.
We know $(x_6 + x_7)^2 = x_6^2 + x_7^2 + 2x_6x_7$.
$14^2 = 100 + 2x_6x_7$.
$196 = 100 + 2x_6x_7$ $\Rightarrow 2x_6x_7 = 96$ $\Rightarrow x_6x_7 = 48$.

(C) Given the mean of $8$ numbers is $10$:
$\frac{3+7+9+12+13+20+x+y}{8} = 10$
$64+x+y = 80$
$x+y = 16$ (Equation $1$)
Given the variance is $25$:
$\frac{\sum x_i^2}{n} - (\text{mean})^2 = 25$
$\frac{3^2+7^2+9^2+12^2+13^2+20^2+x^2+y^2}{8} - 10^2 = 25$
$\frac{9+49+81+144+169+400+x^2+y^2}{8} = 125$
$852+x^2+y^2 = 1000$
$x^2+y^2 = 148$ (Equation $2$)
Using $(x+y)^2 = x^2+y^2+2xy$:
$16^2 = 148 + 2xy$
$256 = 148 + 2xy$
$2xy = 108$
$xy = 54$

(A) Given $n = 20$,$\text{mean} = 10$,and $\text{variance} = 4$.
$\frac{\sum x_i}{20} = 10 \implies \sum x_i = 200$.
$\frac{\sum x_i^2}{20} - (10)^2 = 4 \implies \frac{\sum x_i^2}{20} = 104 \implies \sum x_i^2 = 2080$.
Correct sum of observations $= 200 - 9 + 11 = 202$.
Correct mean $= \frac{202}{20} = 10.1$.
Correct sum of squares $= 2080 - 9^2 + 11^2 = 2080 - 81 + 121 = 2120$.
Correct variance $= \frac{\sum x_i^2}{n} - (\text{mean})^2 = \frac{2120}{20} - (10.1)^2$.
$= 106 - 102.01 = 3.99$.

The formula for the coefficient of variation $(C.V.)$ is given by $C.V. = \frac{\sigma}{\bar{x}} \times 100$,where $\sigma$ is the standard deviation and $\bar{x}$ is the arithmetic mean.
For the first distribution:
$C.V._{1} = 60$,$\sigma_{1} = 21$
$60 = \frac{21}{\bar{x}_{1}} \times 100$
$\bar{x}_{1} = \frac{21 \times 100}{60} = \frac{2100}{60} = 35$
For the second distribution:
$C.V._{2} = 70$,$\sigma_{2} = 16$
$70 = \frac{16}{\bar{x}_{2}} \times 100$
$\bar{x}_{2} = \frac{16 \times 100}{70} = \frac{1600}{70} \approx 22.86$
Thus,the arithmetic means are $35$ and $22.86$ respectively.

(B) The total monthly wages paid by a firm is calculated as the product of the mean monthly wage and the number of wage earners.
For firm $A$:
Total amount $= 5253 \times 586 = 3,078,258$.
For firm $B$:
Total amount $= 5253 \times 648 = 3,403,944$.
Comparing the two,$3,403,944 > 3,078,258$.
Therefore,firm $B$ pays a larger total amount as monthly wages.

(A) Let the other two observations be $x$ and $y$.
The sum of the $5$ observations is $5 \times 4.4 = 22$.
Thus,$1 + 2 + 6 + x + y = 22$,which gives $x + y = 13$ (Equation $1$).
The variance is given by $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.
$8.24 = \frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - (4.4)^2$.
$8.24 = \frac{1 + 4 + 36 + x^2 + y^2}{5} - 19.36$.
$8.24 + 19.36 = \frac{41 + x^2 + y^2}{5}$.
$27.6 = \frac{41 + x^2 + y^2}{5} \implies 138 = 41 + x^2 + y^2 \implies x^2 + y^2 = 97$ (Equation $2$).
From $(x + y)^2 = x^2 + y^2 + 2xy$,we have $13^2 = 97 + 2xy$.
$169 = 97 + 2xy \implies 2xy = 72 \implies xy = 36$.
Since $x + y = 13$ and $xy = 36$,$x$ and $y$ are roots of $t^2 - 13t + 36 = 0$.
$(t - 9)(t - 4) = 0$,so $t = 9$ or $t = 4$.
The other two observations are $4$ and $9$.

(A) Given that number of observations $(n) = 100$.
Incorrect mean $(\bar{x}) = 40$.
Incorrect standard deviation $(\sigma) = 5.1$.
We know that $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$.
$40 = \frac{1}{100} \sum_{i=1}^{100} x_i \implies \sum_{i=1}^{100} x_i = 4000$.
Incorrect sum of observations $= 4000$.
Correct sum of observations $= 4000 - 50 + 40 = 3990$.
Correct mean $= \frac{3990}{100} = 39.9$.
Also,standard deviation $\sigma = \sqrt{\frac{1}{n} \sum x_i^2 - (\bar{x})^2}$.
$5.1 = \sqrt{\frac{1}{100} \sum x_i^2 - (40)^2}$.
$26.01 = \frac{1}{100} \sum x_i^2 - 1600$.
Incorrect $\sum x_i^2 = 100(26.01 + 1600) = 162601$.
Correct $\sum x_i^2 = 162601 - (50)^2 + (40)^2 = 162601 - 2500 + 1600 = 161701$.
Correct standard deviation $= \sqrt{\frac{161701}{100} - (39.9)^2} = \sqrt{1617.01 - 1592.01} = \sqrt{25} = 5$.

(A) Let the remaining two observations be $x$ and $y$.
The sum of the eight observations is $8 \times 9 = 72$.
The sum of the six given observations is $6 + 7 + 10 + 12 + 12 + 13 = 60$.
Thus,$x + y = 72 - 60 = 12$ ........... $(1)$
The variance is given by $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.
$9.25 = \frac{1}{8} (6^2 + 7^2 + 10^2 + 12^2 + 12^2 + 13^2 + x^2 + y^2) - 9^2$.
$9.25 + 81 = \frac{1}{8} (36 + 49 + 100 + 144 + 144 + 169 + x^2 + y^2)$.
$90.25 \times 8 = 642 + x^2 + y^2$.
$722 = 642 + x^2 + y^2 \Rightarrow x^2 + y^2 = 80$ ........... $(2)$
From $(1)$,$(x + y)^2 = 144 \Rightarrow x^2 + y^2 + 2xy = 144$.
$80 + 2xy = 144$ $\Rightarrow 2xy = 64$ $\Rightarrow xy = 32$.
Since $x + y = 12$ and $xy = 32$,the quadratic equation $t^2 - 12t + 32 = 0$ gives the values.
$(t - 8)(t - 4) = 0$.
Thus,the remaining observations are $4$ and $8$.

(A) Let the remaining two observations be $x$ and $y$.
The observations are $2, 4, 10, 12, 14, x, y$.
Mean,$\bar{x} = \frac{2+4+10+12+14+x+y}{7} = 8$.
$\Rightarrow 42 + x + y = 56$ $\Rightarrow x + y = 14$ (Equation $1$).
Variance,$\sigma^2 = 16 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.
$16 = \frac{2^2 + 4^2 + 10^2 + 12^2 + 14^2 + x^2 + y^2}{7} - 8^2$.
$16 + 64 = \frac{4 + 16 + 100 + 144 + 196 + x^2 + y^2}{7}$.
$80 \times 7 = 460 + x^2 + y^2$.
$560 = 460 + x^2 + y^2 \Rightarrow x^2 + y^2 = 100$ (Equation $2$).
From $(x+y)^2 = x^2 + y^2 + 2xy$,we have $14^2 = 100 + 2xy$.
$196 = 100 + 2xy$ $\Rightarrow 2xy = 96$ $\Rightarrow xy = 48$.
Since $x+y = 14$ and $xy = 48$,the quadratic equation $t^2 - 14t + 48 = 0$ gives the values.
$(t-6)(t-8) = 0 \Rightarrow t = 6, 8$.
Thus,the remaining observations are $6$ and $8$.

(A) Number of observations $(n) = 20$.
Incorrect mean $(\bar{x}) = 10$.
Incorrect standard deviation $(\sigma) = 2$.
$\sum x_i = n \times \bar{x} = 20 \times 10 = 200$.
Correct sum of observations $= 200 - 8 = 192$.
Correct mean $= \frac{192}{19} \approx 10.105$.
Incorrect $\sum x_i^2 = n(\sigma^2 + \bar{x}^2) = 20(2^2 + 10^2) = 20(4 + 100) = 2080$.
Correct $\sum x_i^2 = 2080 - 8^2 = 2080 - 64 = 2016$.
Correct standard deviation $= \sqrt{\frac{\sum x_i^2}{n} - (\text{mean})^2} = \sqrt{\frac{2016}{19} - (10.105)^2} = \sqrt{106.105 - 102.111} = \sqrt{3.994} \approx 2.00$.

(B) Given $n = 20$,Incorrect Mean $\bar{x} = 10$,Incorrect Standard Deviation $\sigma = 2$.
Incorrect sum of observations $\sum x_i = n \times \bar{x} = 20 \times 10 = 200$.
Correct sum of observations $= 200 - 8 + 12 = 204$.
Correct mean $= \frac{204}{20} = 10.2$.
Using the formula $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$,we have $2^2 = \frac{1}{20} \sum x_i^2 - 10^2$.
$4 = \frac{1}{20} \sum x_i^2 - 100 \Rightarrow \sum x_i^2 = 20 \times 104 = 2080$.
Correct $\sum x_i^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160$.
Correct standard deviation $= \sqrt{\frac{2160}{20} - (10.2)^2} = \sqrt{108 - 104.04} = \sqrt{3.96} \approx 1.9899$.

(B) Given $n = 100$,$\bar{x} = 20$,$\sigma = 3$.
Sum of observations $\sum x_i = 100 \times 20 = 2000$.
Sum of squares of observations $\sum x_i^2 = n(\sigma^2 + \bar{x}^2) = 100(3^2 + 20^2) = 100(9 + 400) = 40900$.
Incorrect observations are $21, 21, 18$. Sum $= 60$. Sum of squares $= 21^2 + 21^2 + 18^2 = 441 + 441 + 324 = 1206$.
New $n = 100 - 3 = 97$.
New sum $\sum x_i' = 2000 - 60 = 1940$.
New mean $\bar{x}' = \frac{1940}{97} = 20$.
New sum of squares $\sum x_i'^2 = 40900 - 1206 = 39694$.
New variance $\sigma'^2 = \frac{\sum x_i'^2}{n'} - (\bar{x}')^2 = \frac{39694}{97} - 20^2 = 409.216 - 400 = 9.216$.
New standard deviation $\sigma' = \sqrt{9.216} \approx 3.036$.

(N/A) Let the two sets of observations be $x_{i}$ $(i=1, 2, \ldots, n_{1})$ and $y_{j}$ $(j=1, 2, \ldots, n_{2})$.
The combined mean $\bar{x}$ is given by $\bar{x} = \frac{n_{1}\bar{x}_{1} + n_{2}\bar{x}_{2}}{n_{1} + n_{2}}$.
The variance of the combined set is $\sigma^{2} = \frac{1}{n_{1}+n_{2}} [\sum (x_{i}-\bar{x})^{2} + \sum (y_{j}-\bar{x})^{2}]$.
Using the identity $\sum (x_{i}-\bar{x})^{2} = \sum (x_{i}-\bar{x}_{1} + \bar{x}_{1}-\bar{x})^{2} = \sum (x_{i}-\bar{x}_{1})^{2} + n_{1}(\bar{x}_{1}-\bar{x})^{2} = n_{1}s_{1}^{2} + n_{1}d_{1}^{2}$,where $d_{1} = \bar{x}_{1}-\bar{x}$.
Similarly,$\sum (y_{j}-\bar{x})^{2} = n_{2}s_{2}^{2} + n_{2}d_{2}^{2}$,where $d_{2} = \bar{x}_{2}-\bar{x}$.
Substituting $d_{1} = \frac{n_{2}(\bar{x}_{1}-\bar{x}_{2})}{n_{1}+n_{2}}$ and $d_{2} = \frac{n_{1}(\bar{x}_{2}-\bar{x}_{1})}{n_{1}+n_{2}}$,we get:
$\sigma^{2} = \frac{n_{1}s_{1}^{2} + n_{2}s_{2}^{2}}{n_{1}+n_{2}} + \frac{n_{1}d_{1}^{2} + n_{2}d_{2}^{2}}{n_{1}+n_{2}}$.
Simplifying the term $\frac{n_{1}d_{1}^{2} + n_{2}d_{2}^{2}}{n_{1}+n_{2}}$ leads to $\frac{n_{1}n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}$.
Thus,$SD = \sqrt{\frac{n_{1}s_{1}^{2}+n_{2}s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}}$.

(D) Given,$n_{1}=20, \sigma_{1}=5, \bar{x}_{1}=17$ and $n_{2}=20, \sigma_{2}=5, \bar{x}_{2}=22$.
We know that the combined standard deviation $\sigma$ is given by:
$\sigma=\sqrt{\frac{n_{1} \sigma_{1}^{2}+n_{2} \sigma_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}}$
Substituting the values:
$\sigma=\sqrt{\frac{20 \times(5)^{2}+20 \times(5)^{2}}{20+20}+\frac{20 \times 20(17-22)^{2}}{(20+20)^{2}}}$
$\sigma=\sqrt{\frac{500+500}{40}+\frac{400 \times (-5)^{2}}{40^{2}}}$
$\sigma=\sqrt{\frac{1000}{40}+\frac{400 \times 25}{1600}}$
$\sigma=\sqrt{25+\frac{10000}{1600}}$
$\sigma=\sqrt{25+6.25} = \sqrt{31.25} \approx 5.59$

(N/A) Sum of frequencies:
$(x-2) + x + x^2 + (x+1)^2 + 2x + (x+1) = 60$
$x^2 - 2 + x + x^2 + x^2 + 2x + 1 + 2x + x + 1 = 60$
$3x^2 + 7x - 60 = 0$
$(3x - 12)(x + 5) = 0$ (Wait,re-calculating: $x-2+x+x^2+x^2+2x+1+2x+x+1 = 2x^2+7x = 60 \implies 2x^2+7x-60=0$)
$(2x+15)(x-4)=0$
Since $x$ is a positive integer,$x=4$.
Frequency table for $x=4$:
$\begin{array}{|c|c|c|c|c|} \hline x_i & f_i & d_i=x_i-3 & f_i d_i & f_i d_i^2 \\ \hline 0 & 2 & -3 & -6 & 18 \\ \hline 1 & 4 & -2 & -8 & 16 \\ \hline 2 & 16 & -1 & -16 & 16 \\ \hline 3 & 25 & 0 & 0 & 0 \\ \hline 4 & 8 & 1 & 8 & 8 \\ \hline 5 & 5 & 2 & 10 & 20 \\ \hline \text{Total} & \Sigma f_i=60 & & \Sigma f_i d_i=-12 & \Sigma f_i d_i^2=78 \\ \hline \end{array}$
Mean $= A + \frac{\Sigma f_i d_i}{\Sigma f_i} = 3 + (\frac{-12}{60}) = 3 - 0.2 = 2.8$
Standard Deviation $\sigma = \sqrt{\frac{\Sigma f_i d_i^2}{\Sigma f_i} - (\frac{\Sigma f_i d_i}{\Sigma f_i})^2} = \sqrt{\frac{78}{60} - (-0.2)^2} = \sqrt{1.3 - 0.04} = \sqrt{1.26} \approx 1.12$

Given: $n_{1}=60, \bar{x}_{1}=650, s_{1}=8$ and $n_{2}=80, \bar{x}_{2}=660, s_{2}=7$.
The formula for the combined standard deviation $\sigma$ is:
$\sigma = \sqrt{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2}(\bar{x}_{1} - \bar{x}_{2})^{2}}{(n_{1} + n_{2})^{2}}}$
Substituting the values:
$\sigma = \sqrt{\frac{60(8)^{2} + 80(7)^{2}}{60 + 80} + \frac{60 \times 80(650 - 660)^{2}}{(60 + 80)^{2}}}$
$\sigma = \sqrt{\frac{60(64) + 80(49)}{140} + \frac{4800(-10)^{2}}{(140)^{2}}}$
$\sigma = \sqrt{\frac{3840 + 3920}{140} + \frac{480000}{19600}}$
$\sigma = \sqrt{\frac{7760}{140} + \frac{4800}{196}} = \sqrt{55.428 + 24.489} = \sqrt{79.917} \approx 8.94$ hours.

Given,$n=100, \bar{x}=40$ and $\sigma=10$.
$\Sigma x_{i} = n \times \bar{x} = 100 \times 40 = 4000$.
Corrected $\Sigma x_{i} = 4000 - 30 - 70 + 3 + 27 = 3930$.
Corrected mean $\bar{x}_{\text{new}} = \frac{3930}{100} = 39.3$.
We know $\sigma^{2} = \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2}$.
$100 = \frac{\Sigma x_{i}^{2}}{100} - (40)^{2} \Rightarrow \Sigma x_{i}^{2} = 100(100 + 1600) = 170000$.
Corrected $\Sigma x_{i}^{2} = 170000 - (30)^{2} - (70)^{2} + (3)^{2} + (27)^{2} = 170000 - 900 - 4900 + 9 + 729 = 164938$.
Corrected $\sigma_{\text{new}} = \sqrt{\frac{\Sigma x_{i}^{2}}{n} - (\bar{x}_{\text{new}})^{2}} = \sqrt{\frac{164938}{100} - (39.3)^{2}}$.
$= \sqrt{1649.38 - 1544.49} = \sqrt{104.89} \approx 10.24$.

(N/A) Given $n=10, \bar{x}=45$ and $\sigma^{2}=16$.
$\bar{x} = \frac{\Sigma x_{i}}{n} = 45 \Rightarrow \Sigma x_{i} = 450$.
Corrected $\Sigma x_{i} = 450 - 52 + 25 = 423$.
Corrected mean $\bar{x} = \frac{423}{10} = 42.3$.
Using $\sigma^{2} = \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2}$:
$16 = \frac{\Sigma x_{i}^{2}}{10} - (45)^{2} \Rightarrow \Sigma x_{i}^{2} = 10(16 + 2025) = 20410$.
Corrected $\Sigma x_{i}^{2} = 20410 - (52)^{2} + (25)^{2} = 20410 - 2704 + 625 = 18331$.
Corrected $\sigma^{2} = \frac{18331}{10} - (42.3)^{2} = 1833.1 - 1789.29 = 43.81$.

(A) Let the two remaining observations be $a$ and $b$.
Given the mean $\bar{x} = 10$ for $8$ observations:
$\frac{5+7+10+12+14+15+a+b}{8} = 10$
$63 + a + b = 80 \Rightarrow a + b = 17 \quad (1)$
Given the variance $\sigma^2 = 13.5$:
$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$
$13.5 = \frac{5^2+7^2+10^2+12^2+14^2+15^2+a^2+b^2}{8} - 10^2$
$113.5 = \frac{25+49+100+144+196+225+a^2+b^2}{8}$
$908 = 739 + a^2 + b^2 \Rightarrow a^2 + b^2 = 169 \quad (2)$
From $(a+b)^2 = a^2 + b^2 + 2ab$,we have $17^2 = 169 + 2ab$ $\Rightarrow 289 = 169 + 2ab$ $\Rightarrow 2ab = 120$ $\Rightarrow ab = 60$.
Now,$(a-b)^2 = (a+b)^2 - 4ab = 17^2 - 4(60) = 289 - 240 = 49$.
Thus,$|a-b| = \sqrt{49} = 7$.

(B) Given mean $\bar{x} = 5$ for data $3, 5, 7, a, b$.
$\frac{3+5+7+a+b}{5} = 5$ $\Rightarrow 15+a+b = 25$ $\Rightarrow a+b = 10$.
Given standard deviation $\sigma = 2$.
$\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2} = 4$.
$\frac{3^{2}+5^{2}+7^{2}+a^{2}+b^{2}}{5} - 5^{2} = 4$.
$\frac{9+25+49+a^{2}+b^{2}}{5} = 29$.
$83+a^{2}+b^{2} = 145 \Rightarrow a^{2}+b^{2} = 62$.
We know $(a+b)^{2} = a^{2}+b^{2}+2ab$.
$10^{2} = 62+2ab$ $\Rightarrow 100-62 = 2ab$ $\Rightarrow 2ab = 38$ $\Rightarrow ab = 19$.
The quadratic equation with roots $a$ and $b$ is $x^{2} - (a+b)x + ab = 0$.
Substituting the values,we get $x^{2} - 10x + 19 = 0$.

(B) Let the sum of the ages of the $25$ teachers be $\sum x_i$.
Given,$\frac{\sum x_i}{25} = 40$,so $\sum x_i = 1000$.
Let the age of the newly appointed teacher be $N$.
After the retirement of the $60$-year-old teacher and the appointment of the new teacher,the new sum of ages is $\sum x_i - 60 + N$.
The new mean age is $39$ years for $25$ teachers.
$\frac{1000 - 60 + N}{25} = 39$
$940 + N = 39 \times 25$
$940 + N = 975$
$N = 975 - 940 = 35$.
Thus,the age of the newly appointed teacher is $35$ years.

(C) The formula for the combined variance $\sigma^{2}$ of two sets is given by:
$\sigma^{2} = \frac{n_{1}\sigma_{1}^{2} + n_{2}\sigma_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1}n_{2}}{(n_{1} + n_{2})^{2}}(\bar{x}_{1} - \bar{x}_{2})^{2}$
Given values:
$n_{1} = 10, n_{2} = n, \sigma_{1}^{2} = 2, \sigma_{2}^{2} = 1$
$\bar{x}_{1} = 2, \bar{x}_{2} = 3, \sigma^{2} = \frac{17}{9}$
Substituting these values into the formula:
$\frac{17}{9} = \frac{10(2) + n(1)}{10 + n} + \frac{10n}{(10 + n)^{2}}(2 - 3)^{2}$
$\frac{17}{9} = \frac{20 + n}{10 + n} + \frac{10n}{(10 + n)^{2}}$
Multiply by $9(10 + n)^{2}$ to clear denominators:
$17(10 + n)^{2} = 9[(20 + n)(10 + n) + 10n]$
$17(100 + 20n + n^{2}) = 9[200 + 30n + n^{2} + 10n]$
$1700 + 340n + 17n^{2} = 1800 + 360n + 9n^{2}$
$8n^{2} - 20n - 100 = 0$
$2n^{2} - 5n - 25 = 0$
$(2n + 5)(n - 5) = 0$
Since $n$ must be positive,$n = 5$.

(B) Let the numbers be $x_1, x_2, \ldots, x_{2n}$ and $y_1, y_2, \ldots, y_n$.
The mean of the first $2n$ numbers is $\bar{x} = 6$,so $\sum x_i = 12n$.
The mean of the remaining $n$ numbers is $\bar{y} = 3$,so $\sum y_i = 3n$.
The overall mean $\bar{X} = \frac{12n + 3n}{3n} = 5$.
The variance is given by $\sigma^2 = \frac{\sum x_i^2 + \sum y_i^2}{3n} - (\bar{X})^2 = 4$.
$4 = \frac{\sum x_i^2 + \sum y_i^2}{3n} - 25 \implies \sum x_i^2 + \sum y_i^2 = 87n$.
In the new set,the numbers are $(x_i + 1)$ and $(y_i - 1)$.
The new mean $\bar{X}' = \frac{\sum (x_i + 1) + \sum (y_i - 1)}{3n} = \frac{12n + 2n + 3n - n}{3n} = \frac{16n}{3n} = \frac{16}{3}$.
The new variance $k = \frac{\sum (x_i + 1)^2 + \sum (y_i - 1)^2}{3n} - (\bar{X}')^2$.
$k = \frac{\sum x_i^2 + 2\sum x_i + 2n + \sum y_i^2 - 2\sum y_i + n}{3n} - (\frac{16}{3})^2$.
$k = \frac{87n + 2(12n) + 2n - 2(3n) + n}{3n} - \frac{256}{9} = \frac{108n}{3n} - \frac{256}{9} = 36 - \frac{256}{9} = \frac{324 - 256}{9} = \frac{68}{9}$.
Thus,$9k = 68$.

(D) Given: $n = 20$,$\bar{x} = 10$,$\sigma = 2.5$.
Incorrect sum of observations: $\Sigma x_i = n \times \bar{x} = 20 \times 10 = 200$.
Correct sum of observations: $\Sigma x_i = 200 - 25 + 35 = 210$.
Correct mean $\alpha = \frac{210}{20} = 10.5$.
Incorrect sum of squares: $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2 \implies (2.5)^2 = \frac{\Sigma x_i^2}{20} - 10^2$.
$6.25 = \frac{\Sigma x_i^2}{20} - 100 \implies \Sigma x_i^2 = 20 \times 106.25 = 2125$.
Correct sum of squares: $\Sigma x_i^2 = 2125 - 25^2 + 35^2 = 2125 - 625 + 1225 = 2725$.
Correct variance $\beta = \frac{\Sigma x_i^2}{n} - (\alpha)^2 = \frac{2725}{20} - (10.5)^2 = 136.25 - 110.25 = 26$.
Thus,$(\alpha, \beta) = (10.5, 26)$.

(C) Given the mean of $3, 7, x, y$ is $5$:
$\frac{3+7+x+y}{4} = 5$ $\Rightarrow 10+x+y = 20$ $\Rightarrow x+y = 10$
Given the variance is $10$:
$\frac{3^2+7^2+x^2+y^2}{4} - (5)^2 = 10$
$\frac{9+49+x^2+y^2}{4} = 35$ $\Rightarrow 58+x^2+y^2 = 140$ $\Rightarrow x^2+y^2 = 82$
We have $x+y=10$ and $x^2+y^2=82$.
Since $(x+y)^2 = x^2+y^2+2xy$,we get $100 = 82+2xy$ $\Rightarrow 2xy = 18$ $\Rightarrow xy = 9$.
Solving $x+y=10$ and $xy=9$,we get $x=9$ and $y=1$ (since $x>y$).
The four numbers are $3+2(9)=21$,$7+2(1)=9$,$9+1=10$,and $9-1=8$.
The mean of these four numbers is $\frac{21+9+10+8}{4} = \frac{48}{4} = 12$.

(B) Let $n_1 = 20$ (boys) and $n_2 = 30$ (girls). Total candidates $N = 50$.
Given: $\bar{x}_b = 12$,$\sigma_b^2 = 2$,$\sigma_g^2 = 2$,and combined mean $\bar{x} = 15$.
First,find the mean marks of girls $(\mu = \bar{x}_g)$:
$N \bar{x} = n_1 \bar{x}_b + n_2 \bar{x}_g$
$50 \times 15 = 20 \times 12 + 30 \times \bar{x}_g$
$750 = 240 + 30 \bar{x}_g$
$30 \bar{x}_g = 510 \Rightarrow \bar{x}_g = 17 = \mu$.
Now,calculate the combined variance $\sigma^2$:
$\sigma^2 = \frac{n_1 \sigma_b^2 + n_2 \sigma_g^2}{n_1 + n_2} + \frac{n_1 n_2}{(n_1 + n_2)^2} (\bar{x}_b - \bar{x}_g)^2$
$\sigma^2 = \frac{20 \times 2 + 30 \times 2}{50} + \frac{20 \times 30}{50^2} (12 - 17)^2$
$\sigma^2 = \frac{100}{50} + \frac{600}{2500} (-5)^2$
$\sigma^2 = 2 + \frac{6}{25} \times 25 = 2 + 6 = 8$.
Finally,$\mu + \sigma^2 = 17 + 8 = 25$.

(C) Let the $7$ observations be $x_1, x_2, x_3, x_4, x_5, 6, 8$.
The mean is given by $\frac{\sum_{i=1}^{5} x_i + 6 + 8}{7} = 8$.
$\sum_{i=1}^{5} x_i + 14 = 56 \Rightarrow \sum_{i=1}^{5} x_i = 42$.
The variance is given by $\frac{\sum_{i=1}^{5} x_i^2 + 6^2 + 8^2}{7} - (8)^2 = 16$.
$\frac{\sum_{i=1}^{5} x_i^2 + 36 + 64}{7} = 16 + 64 = 80$.
$\sum_{i=1}^{5} x_i^2 + 100 = 560 \Rightarrow \sum_{i=1}^{5} x_i^2 = 460$.
Now,the variance of the remaining $5$ observations is $\frac{\sum_{i=1}^{5} x_i^2}{5} - \left(\frac{\sum_{i=1}^{5} x_i}{5}\right)^2$.
$= \frac{460}{5} - \left(\frac{42}{5}\right)^2 = 92 - \frac{1764}{25} = \frac{2300 - 1764}{25} = \frac{536}{25}$.

(A) Let the $6$ observations be $x_1, x_2, x_3, x_4, x_5, x_6$. Given $x_1=2, x_2=4, x_3=5, x_4=7$. Let $x_5=a$ and $x_6=b$.
The mean $\bar{x} = \frac{2+4+5+7+a+b}{6} = 6.5$.
$18+a+b = 39$ $\Rightarrow a+b = 21$ $\Rightarrow b = 21-a$.
The variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 10.25$.
$\frac{2^2+4^2+5^2+7^2+a^2+b^2}{6} - (6.5)^2 = 10.25$.
$\frac{4+16+25+49+a^2+b^2}{6} = 10.25 + 42.25 = 52.5$.
$94 + a^2 + b^2 = 315 \Rightarrow a^2 + b^2 = 221$.
Substitute $b = 21-a$: $a^2 + (21-a)^2 = 221$.
$a^2 + 441 - 42a + a^2 = 221$.
$2a^2 - 42a + 220 = 0 \Rightarrow a^2 - 21a + 110 = 0$.
$(a-10)(a-11) = 0$.
Thus,the observations are $10$ and $11$.

(D) Given the mean $\bar{x} = 10$ for $6$ observations:
$\frac{7+10+11+15+a+b}{6} = 10$
$43+a+b = 60 \Rightarrow a+b = 17 \quad (i)$
Given the variance $\sigma^2 = \frac{20}{3}$:
$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$
$\frac{20}{3} = \frac{7^2+10^2+11^2+15^2+a^2+b^2}{6} - 10^2$
$\frac{20}{3} = \frac{49+100+121+225+a^2+b^2}{6} - 100$
$\frac{20}{3} + 100 = \frac{495+a^2+b^2}{6}$
$\frac{320}{3} = \frac{495+a^2+b^2}{6}$
$640 = 495 + a^2 + b^2 \Rightarrow a^2 + b^2 = 145 \quad (ii)$
Using $(a+b)^2 = a^2 + b^2 + 2ab$:
$17^2 = 145 + 2ab$ $\Rightarrow 289 = 145 + 2ab$ $\Rightarrow 2ab = 144$ $\Rightarrow ab = 72$
Now,$(a-b)^2 = (a+b)^2 - 4ab = 17^2 - 4(72) = 289 - 288 = 1$
$|a-b| = \sqrt{1} = 1$

(B) Given the total frequency $N = \sum f = 584$,we have $\alpha + 110 + 54 + 30 + \beta = 584$,which simplifies to $\alpha + \beta + 194 = 584$,so $\alpha + \beta = 390$.
The median is given as $45$,which lies in the class interval $40-50$. Thus,the lower limit $\ell = 40$,class size $h = 10$,frequency of the median class $f = 30$,and cumulative frequency of the class preceding the median class $c = \alpha + 164$.
The median formula is $Median = \ell + \left[\frac{\frac{N}{2} - c}{f}\right] \times h$.
Substituting the values: $45 = 40 + \left[\frac{292 - (\alpha + 164)}{30}\right] \times 10$.
$5 = \frac{128 - \alpha}{3}$.
$15 = 128 - \alpha$,which gives $\alpha = 113$.
Since $\alpha + \beta = 390$,we have $\beta = 390 - 113 = 277$.
Therefore,$|\alpha - \beta| = |113 - 277| = |-164| = 164$.

(C) Given $n_{1} = 100$,$\bar{x}_{1} = 15$,$\sigma_{1} = 3$.
Total items $n = 250$,combined mean $\bar{x} = 15.6$,combined variance $\sigma^{2} = 13.44$.
Since $n = n_{1} + n_{2}$,we have $n_{2} = 250 - 100 = 150$.
Using the combined mean formula $\bar{x} = \frac{n_{1}\bar{x}_{1} + n_{2}\bar{x}_{2}}{n_{1} + n_{2}}$:
$15.6 = \frac{100(15) + 150(\bar{x}_{2})}{250}$ $\Rightarrow 3900 = 1500 + 150\bar{x}_{2}$ $\Rightarrow 150\bar{x}_{2} = 2400$ $\Rightarrow \bar{x}_{2} = 16$.
Using 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 $d_{1} = \bar{x}_{1} - \bar{x} = 15 - 15.6 = -0.6$ and $d_{2} = \bar{x}_{2} - \bar{x} = 16 - 15.6 = 0.4$:
$13.44 = \frac{100(3^{2} + (-0.6)^{2}) + 150(\sigma_{2}^{2} + (0.4)^{2})}{250}$.
$13.44 \times 250 = 100(9 + 0.36) + 150(\sigma_{2}^{2} + 0.16)$.
$3360 = 936 + 150\sigma_{2}^{2} + 24$.
$3360 = 960 + 150\sigma_{2}^{2}$ $\Rightarrow 2400 = 150\sigma_{2}^{2}$ $\Rightarrow \sigma_{2}^{2} = 16$.
Thus,$\sigma_{2} = 4$.