Word problem -Statistics Questions in English

(C) The total frequency is $N = 4 + 4 + \alpha + \beta = 8 + \alpha + \beta$.
The mean is given by $\bar{x} = \frac{\sum f_i x_i}{N} = 6$.
$\frac{4(2) + 4(6) + \alpha(8) + \beta(9)}{8 + \alpha + \beta} = 6$
$8 + 24 + 8\alpha + 9\beta = 48 + 6\alpha + 6\beta$
$2\alpha + 3\beta = 16 \quad \dots (i)$
The variance is $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2 = 6.8$.
$\frac{4(4) + 4(36) + \alpha(64) + \beta(81)}{8 + \alpha + \beta} - 36 = 6.8$
$\frac{16 + 144 + 64\alpha + 81\beta}{8 + \alpha + \beta} = 42.8$
$160 + 64\alpha + 81\beta = 42.8(8 + \alpha + \beta) = 342.4 + 42.8\alpha + 42.8\beta$
$21.2\alpha + 38.2\beta = 182.4$
Multiplying by $10$: $212\alpha + 382\beta = 1824 \Rightarrow 106\alpha + 191\beta = 912 \quad \dots (ii)$
Solving $(i)$ and $(ii)$: From $(i)$,$\alpha = \frac{16 - 3\beta}{2} = 8 - 1.5\beta$.
Substituting in $(ii)$: $106(8 - 1.5\beta) + 191\beta = 912$
$848 - 159\beta + 191\beta = 912 \Rightarrow 32\beta = 64 \Rightarrow \beta = 2$.
Then $\alpha = 8 - 1.5(2) = 5$.
Total frequency $N = 8 + 5 + 2 = 15$.
New sum of observations when $x_3$ changes from $8$ to $7$:
Original sum $= 6 \times 15 = 90$.
New sum $= 90 - (8 \times 5) + (7 \times 5) = 90 - 40 + 35 = 85$.
New mean $= \frac{85}{15} = \frac{17}{3}$.

(C) Let $n = 50$. The incorrect mean $\bar{x} = 15$ and standard deviation $\sigma = 2$.
Sum of incorrect observations $\sum x_i = 50 \times 15 = 750$.
Let the incorrect observation be $x_1$ and the correct observation be $x_1'$.
Given $x_1 + x_1' = 70$ and the correct mean $\bar{x}' = 16$.
Sum of correct observations $\sum x_i' = 50 \times 16 = 800$.
Thus,$x_1' - x_1 = \sum x_i' - \sum x_i = 800 - 750 = 50$.
Solving $x_1' + x_1 = 70$ and $x_1' - x_1 = 50$,we get $x_1' = 60$ and $x_1 = 10$.
Incorrect variance $\sigma^2 = 4 = \frac{\sum x_i^2}{n} - \bar{x}^2$ $\Rightarrow 4 = \frac{\sum x_i^2}{50} - 225$ $\Rightarrow \sum x_i^2 = 50 \times 229 = 11450$.
Sum of squares of remaining $49$ observations: $\sum_{i=2}^{50} x_i^2 = 11450 - x_1^2 = 11450 - 100 = 11350$.
Correct sum of squares $\sum x_i'^2 = 11350 + (x_1')^2 = 11350 + 3600 = 14950$.
Correct variance $\sigma'^2 = \frac{\sum x_i'^2}{n} - (\bar{x}')^2 = \frac{14950}{50} - 16^2 = 299 - 256 = 43$.

(A) Given the mean $\bar{x} = 6$ for the numbers $a, b, 8, 5, 10$,we have $\frac{a+b+8+5+10}{5} = 6$,which implies $a+b+23 = 30$,so $a+b = 7$.
The variance is given by $\sigma^{2} = \frac{\sum (x_i - \bar{x})^2}{n} = 6.8$.
Substituting the values: $\frac{(a-6)^2 + (b-6)^2 + (8-6)^2 + (5-6)^2 + (10-6)^2}{5} = 6.8$.
$(a-6)^2 + (b-6)^2 + 4 + 1 + 16 = 34$.
$(a-6)^2 + (b-6)^2 = 13$.
Since $b = 7-a$,we have $(a-6)^2 + (7-a-6)^2 = 13$,which simplifies to $(a-6)^2 + (1-a)^2 = 13$.
$a^2 - 12a + 36 + 1 - 2a + a^2 = 13$ $\Rightarrow 2a^2 - 14a + 24 = 0$ $\Rightarrow a^2 - 7a + 12 = 0$.
$(a-4)(a-3) = 0$,so $a=4$ or $a=3$. If $a=4, b=3$; if $a=3, b=4$.
The mean deviation $M$ about the mean is $M = \frac{\sum |x_i - \bar{x}|}{n}$.
$M = \frac{|a-6| + |b-6| + |8-6| + |5-6| + |10-6|}{5} = \frac{|a-6| + |b-6| + 2 + 1 + 4}{5}$.
For $a=3, b=4$: $M = \frac{|3-6| + |4-6| + 7}{5} = \frac{3 + 2 + 7}{5} = \frac{12}{5}$.
Thus,$25M = 25 \times \frac{12}{5} = 60$.

(B) Given the mean $\bar{x} = 6$ for $8$ observations:
$\frac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6$
$36 + x + y = 48 \Rightarrow x + y = 12$ $(1)$
Given the variance $\sigma^{2} = \frac{9}{4}$:
$\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2} = \frac{9}{4}$
$\frac{4^{2} + 5^{2} + 6^{2} + 6^{2} + 7^{2} + 8^{2} + x^{2} + y^{2}}{8} - 6^{2} = \frac{9}{4}$
$\frac{16 + 25 + 36 + 36 + 49 + 64 + x^{2} + y^{2}}{8} - 36 = \frac{9}{4}$
$\frac{226 + x^{2} + y^{2}}{8} = 36 + 2.25 = 38.25$
$226 + x^{2} + y^{2} = 306 \Rightarrow x^{2} + y^{2} = 80$ $(2)$
From $(1)$,$y = 12 - x$. Substituting into $(2)$:
$x^{2} + (12 - x)^{2} = 80$
$x^{2} + 144 - 24x + x^{2} = 80$
$2x^{2} - 24x + 64 = 0 \Rightarrow x^{2} - 12x + 32 = 0$
$(x - 4)(x - 8) = 0$
Since $x < y$,we have $x = 4$ and $y = 8$.
Calculating $x^{4} + y^{2}$:
$4^{4} + 8^{2} = 256 + 64 = 320$.

(D) Given,number of students $n = 7$,mean $\bar{x} = 62$,and variance $\sigma^2 = 20$.
The formula for variance is $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$.
Substituting the values: $20 = \frac{1}{7} \sum_{i=1}^{7} (x_i - 62)^2$.
$\sum_{i=1}^{7} (x_i - 62)^2 = 20 \times 7 = 140$.
$A$ student fails if $x_i < 50$. Let $k$ be the number of students who fail. For these students,$x_i \le 49$.
If a student fails,the minimum contribution to the sum of squares is $(49 - 62)^2 = (-13)^2 = 169$.
Since the total sum of squares is only $140$,and $169 > 140$,it is impossible for even one student to have marks less than $50$ while maintaining a variance of $20$ with a mean of $62$.
Therefore,the number of students who can fail is $0$.

(A) Given data: $3, 7, 12, a, 43-a$. Mean $\bar{x} = 13$.
Sum of observations $= 3 + 7 + 12 + a + 43 - a = 65$.
Mean $= \frac{65}{5} = 13$ (Verified).
Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$\sigma^2 = \frac{3^2 + 7^2 + 12^2 + a^2 + (43-a)^2}{5} - 13^2$.
$\sigma^2 = \frac{9 + 49 + 144 + a^2 + 1849 - 86a + a^2}{5} - 169$.
$\sigma^2 = \frac{2a^2 - 86a + 2051}{5} - 169 = \frac{2a^2 - 86a + 2051 - 845}{5} = \frac{2a^2 - 86a + 1206}{5}$.
For $\sigma^2$ to be a natural number,$2a^2 - 86a + 1206$ must be divisible by $5$.
$2a^2 - 86a + 1206 \equiv 0 \pmod{5} \Rightarrow 2a^2 - a + 1 \equiv 0 \pmod{5}$.
Testing $a \pmod{5}$:
If $a \equiv 0, 2(0)^2 - 0 + 1 = 1 \not\equiv 0$.
If $a \equiv 1, 2(1)^2 - 1 + 1 = 2 \not\equiv 0$.
If $a \equiv 2, 2(4) - 2 + 1 = 7 \equiv 2 \not\equiv 0$.
If $a \equiv 3, 2(9) - 3 + 1 = 16 \equiv 1 \not\equiv 0$.
If $a \equiv 4, 2(16) - 4 + 1 = 29 \equiv 4 \not\equiv 0$.
Since no value of $a$ satisfies the condition,the number of values is $0$.

(B) Given the mean of $5$ observations is $\bar{x} = \frac{\sum_{i=1}^{5} x_{i}}{5} = \frac{24}{5}$,so $\sum_{i=1}^{5} x_{i} = 24$.
The variance is $\sigma^{2} = \frac{\sum x_{i}^{2}}{5} - (\bar{x})^{2} = \frac{194}{25}$.
Substituting $\bar{x} = \frac{24}{5}$,we get $\frac{\sum x_{i}^{2}}{5} - \frac{576}{25} = \frac{194}{25}$ $\Rightarrow \frac{\sum x_{i}^{2}}{5} = \frac{770}{25} = \frac{154}{5}$,so $\sum_{i=1}^{5} x_{i}^{2} = 154$.
For the first $4$ observations,the mean is $\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4} = \frac{7}{2} \Rightarrow x_{1}+x_{2}+x_{3}+x_{4} = 14$.
Since $\sum_{i=1}^{5} x_{i} = 24$,we have $x_{5} = 24 - 14 = 10$.
The variance of the first $4$ observations is $a = \frac{\sum_{i=1}^{4} x_{i}^{2}}{4} - (\frac{7}{2})^{2} = \frac{\sum_{i=1}^{4} x_{i}^{2}}{4} - \frac{49}{4}$.
Thus,$\sum_{i=1}^{4} x_{i}^{2} = 4a + 49$.
We know $\sum_{i=1}^{5} x_{i}^{2} = \sum_{i=1}^{4} x_{i}^{2} + x_{5}^{2} = 154$.
Substituting the values: $(4a + 49) + 10^{2} = 154$.
$4a + 49 + 100 = 154$ $\Rightarrow 4a + 149 = 154$ $\Rightarrow 4a = 5$.
Finally,$4a + x_{5} = 5 + 10 = 15$.

(A) Given,$n = 40$,$\mu = 30$,and $\sigma = 5$.
Sum of observations $\sum x_i = 40 \times 30 = 1200$.
Variance $\sigma^2 = 25$,so $\frac{\sum x_i^2}{40} - (30)^2 = 25$.
$\sum x_i^2 = 40 \times (900 + 25) = 40 \times 925 = 37000$.
After omitting observations $10$ and $12$,the new number of observations $n' = 38$.
New sum $\sum x_i' = 1200 - 10 - 12 = 1178$.
New mean $\mu' = \frac{1178}{38} = 31$.
New sum of squares $\sum (x_i')^2 = 37000 - 10^2 - 12^2 = 37000 - 100 - 144 = 36756$.
New variance $\sigma^2 = \frac{\sum (x_i')^2}{n'} - (\mu')^2 = \frac{36756}{38} - (31)^2$.
Multiplying by $38$,we get $38 \sigma^2 = 36756 - 38 \times 961 = 36756 - 36518 = 238$.

(D) Given that the mean $\bar{x} = 15$ and variance $\sigma^{2} = 9$ for $n = 20$ observations.
$\sum x_{i} = 15 \times 20 = 300$
$\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2} \Rightarrow 9 = \frac{\sum x_{i}^{2}}{20} - 225$
$\frac{\sum x_{i}^{2}}{20} = 234 \Rightarrow \sum x_{i}^{2} = 4680$
Now,the mean of $(x_{i} + \alpha)^{2}$ is $178$:
$\frac{1}{20} \sum (x_{i} + \alpha)^{2} = 178$
$\sum (x_{i}^{2} + 2\alpha x_{i} + \alpha^{2}) = 178 \times 20 = 3560$
$\sum x_{i}^{2} + 2\alpha \sum x_{i} + 20\alpha^{2} = 3560$
$4680 + 2\alpha(300) + 20\alpha^{2} = 3560$
$20\alpha^{2} + 600\alpha + 1120 = 0$
Dividing by $20$:
$\alpha^{2} + 30\alpha + 56 = 0$
$(\alpha + 28)(\alpha + 2) = 0$
So,$\alpha = -28$ or $\alpha = -2$.
The maximum value of $\alpha$ is $-2$.
The square of the maximum value is $(-2)^{2} = 4$.

(B) Given the mean $\mu = \frac{1}{n} \sum_{i=1}^{n} x_i$.
For the new list,the mean $\hat{\mu} = \frac{1}{n} (y_1 + y_2 + \sum_{j=3}^{n} x_j) = \frac{1}{n} (\frac{x_1+x_2}{2} + \frac{x_1+x_2}{2} + \sum_{j=3}^{n} x_j) = \frac{1}{n} \sum_{i=1}^{n} x_i = \mu$.
Now,consider the variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - \mu^2$ and $\hat{\sigma}^2 = \frac{1}{n} \sum y_i^2 - \hat{\mu}^2$.
Since $\hat{\mu} = \mu$,we compare $\sum x_i^2$ and $\sum y_i^2$.
$\sum x_i^2 - \sum y_i^2 = (x_1^2 + x_2^2) - (y_1^2 + y_2^2) = x_1^2 + x_2^2 - 2(\frac{x_1+x_2}{2})^2 = x_1^2 + x_2^2 - \frac{x_1^2 + x_2^2 + 2x_1x_2}{2} = \frac{x_1^2 + x_2^2 - 2x_1x_2}{2} = \frac{(x_1-x_2)^2}{2} \geq 0$.
Thus,$\sum x_i^2 \geq \sum y_i^2$,which implies $\sigma^2 \geq \hat{\sigma}^2$,so $\sigma \geq \hat{\sigma}$.

(A) Given the original list $x_1, x_2, \ldots, x_n$ with mean $\mu = \frac{1}{n} \sum_{i=1}^n x_i$.
The new list is $y_1=0, y_2=x_2, \ldots, y_{n-1}=x_{n-1}, y_n=x_1+x_n$.
The mean of the new list is $\hat{\mu} = \frac{1}{n} (0 + x_2 + x_3 + \ldots + x_{n-1} + x_1 + x_n) = \frac{1}{n} \sum_{i=1}^n x_i = \mu$.
Now,consider the sum of squares $\sum y_i^2 = 0^2 + x_2^2 + \ldots + x_{n-1}^2 + (x_1+x_n)^2$.
Expanding this,$\sum y_i^2 = x_2^2 + \ldots + x_{n-1}^2 + x_1^2 + x_n^2 + 2x_1x_n = \sum_{i=1}^n x_i^2 + 2x_1x_n$.
Since $x_1 > 0$ and $x_n > 0$,we have $2x_1x_n > 0$,so $\sum y_i^2 > \sum x_i^2$.
The variance is given by $\sigma^2 = \frac{1}{n} \sum x_i^2 - \mu^2$ and $\hat{\sigma}^2 = \frac{1}{n} \sum y_i^2 - \hat{\mu}^2$.
Since $\hat{\mu} = \mu$ and $\sum y_i^2 > \sum x_i^2$,it follows that $\hat{\sigma}^2 > \sigma^2$,which implies $\hat{\sigma} > \sigma$.
Thus,$\mu = \hat{\mu}$ and $\sigma < \hat{\sigma}$,which satisfies $\sigma \leq \hat{\sigma}$.
Therefore,option $A$ is correct.

(D) Total number of students $n = 50$.
Average marks $\bar{x} = 47.5$.
Total marks obtained by all students $= 50 \times 47.5 = 2375$.
Let $k$ be the number of students who scored more than the average $(47.5)$. Since marks are integers,these students must have scored at least $48$ marks.
Let the remaining $(50 - k)$ students score the minimum possible marks,which is $0$.
To maximize $k$,we assume these $k$ students scored the minimum possible marks greater than the average,which is $48$.
Thus,$48k + (50 - k) \times 0 \leq 2375$.
$48k \leq 2375$.
$k \leq \frac{2375}{48} \approx 49.479$.
Since $k$ must be an integer,the maximum value of $k$ is $49$.

(A) Let the number of boys be $B$ and the number of girls be $G$.
Sum of marks obtained by boys $= Bx$.
Sum of marks obtained by girls $= Gy$.
Total number of students $= B + G$.
Given that the average marks of all students is $z$,we have:
$\frac{Bx + Gy}{B + G} = z$
$Bx + Gy = z(B + G)$
$Bx + Gy = zB + zG$
$Bx - zB = zG - Gy$
$B(x - z) = G(z - y)$
$\frac{G}{B} = \frac{x - z}{z - y} = \frac{z - x}{y - z}$
We need the ratio of the number of girls to the total number of students,which is $\frac{G}{B + G}$.
Dividing numerator and denominator by $G$:
$\frac{G}{B + G} = \frac{1}{\frac{B}{G} + 1} = \frac{1}{\frac{z - x}{y - z} + 1} = \frac{1}{\frac{z - x + y - z}{y - z}} = \frac{y - z}{y - x} = \frac{z - y}{x - y} = \frac{z - x}{y - x}$.

(C) Let the number of people in the two villages be $x$ and $y$ respectively.
Given,the average income of $x$ people is $P$ and the average income of $y$ people is $Q$.
Therefore,the total income of people in the two villages is $Px$ and $Qy$ respectively.
One person with income $I$ moves from the first village to the second village.
Then,the number of people in the first village becomes $x-1$ and in the second village becomes $y+1$.
The new average incomes are $P^{\prime} = \frac{Px - I}{x-1}$ and $Q^{\prime} = \frac{Qy + I}{y+1}$.
If $P^{\prime} = P$,then $Px - I = P(x-1) = Px - P$,which implies $I = P$.
If $Q^{\prime} = Q$,then $Qy + I = Q(y+1) = Qy + Q$,which implies $I = Q$.
Since $P \neq Q$,the person cannot have an income $I$ such that both $P^{\prime} = P$ and $Q^{\prime} = Q$ simultaneously.
Thus,the condition $P^{\prime} = P$ and $Q^{\prime} = Q$ is impossible.
Therefore,option $(C)$ is correct.

(B) Let $S_1$ be the total income of the first group (salary $< ₹ 10,000$) and $S_2$ be the total income of the second group (salary $> ₹ 10,000$).
Given that $S_1 < S_2$.
Let $N_1$ and $N_2$ be the number of people in the first and second groups,respectively.
The initial total income is $S_{total} = S_1 + S_2$.
The new total income after the changes is $S'_{total} = S_1(1 + 0.05) + S_2(1 - 0.05) = 1.05 S_1 + 0.95 S_2$.
The change in total income is $\Delta S = S'_{total} - S_{total} = (1.05 S_1 + 0.95 S_2) - (S_1 + S_2) = 0.05 S_1 - 0.05 S_2 = 0.05(S_1 - S_2)$.
Since $S_1 < S_2$,it follows that $S_1 - S_2 < 0$,which means $\Delta S < 0$.
Therefore,the total income decreases,and consequently,the average income of all people decreases.

(C) The sum of the first $n$ integers is given by $S_n = \frac{n(n+1)}{2}$.
After erasing the integer $k$,the sum of the remaining $(n-1)$ integers is $\frac{n(n+1)}{2} - k$.
The average of the remaining numbers is given as $16$,so:
$\frac{\frac{n(n+1)}{2} - k}{n-1} = 16$
$n(n+1) - 2k = 32(n-1)$
$n^2 + n - 2k = 32n - 32$
$2k = n^2 - 31n + 32$
$k = \frac{n^2 - 31n + 32}{2}$
Since $1 < k < n$,we have:
$1 < \frac{n^2 - 31n + 32}{2} < n$
From $k < n$: $n^2 - 31n + 32 < 2n \implies n^2 - 33n + 32 < 0 \implies (n-32)(n-1) < 0$. Since $n \geq 3$,we have $n < 32$.
From $k > 1$: $n^2 - 31n + 32 > 2 \implies n^2 - 31n + 30 > 0 \implies (n-30)(n-1) > 0$. Since $n \geq 3$,we have $n > 30$.
Thus,$n$ must be $31$.
Substituting $n = 31$ into the equation for $k$:
$k = \frac{31^2 - 31(31) + 32}{2} = \frac{32}{2} = 16$.
Therefore,$n + k = 31 + 16 = 47$.

(C) Let the number of students be $n$. The initial mean $\bar{x} = 10$ and variance $\sigma^2 = 4$.
$\bar{x} = \frac{\sum x_i}{n} = 10 \implies \sum x_i = 10n$.
$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 4 \implies \frac{\sum x_i^2}{n} - 100 = 4 \implies \sum x_i^2 = 104n$.
When one student's mark changes from $8$ to $12$,the new sum of marks is $\sum x_i' = 10n - 8 + 12 = 10n + 4$.
The new mean is $\frac{10n + 4}{n} = 10.2 \implies 10n + 4 = 10.2n \implies 0.2n = 4 \implies n = 20$.
Now,the new sum of squares is $\sum x_i'^2 = \sum x_i^2 - 8^2 + 12^2 = 104(20) - 64 + 144 = 2080 + 80 = 2160$.
The new variance is $\sigma'^2 = \frac{\sum x_i'^2}{n} - (\bar{x}')^2 = \frac{2160}{20} - (10.2)^2 = 108 - 104.04 = 3.96$.

(A) The mean of set $X$ is $\bar{x} = \frac{11+41}{2} = 26$ (number of elements $n_1 = 31$).
The mean of set $Y$ is $\bar{y} = \frac{61+91}{2} = 76$ (number of elements $n_2 = 31$).
The combined mean $\mu = \frac{n_1\bar{x} + n_2\bar{y}}{n_1 + n_2} = \frac{31(26) + 31(76)}{62} = \frac{26+76}{2} = 51$.
The variance $\sigma^2$ of the combined set is given by $\sigma^2 = \frac{1}{n_1+n_2} \left( \sum_{i=1}^{31} (x_i - \mu)^2 + \sum_{j=1}^{31} (y_j - \mu)^2 \right)$.
For set $X$,$\sum (x_i - \mu)^2 = \sum_{i=11}^{41} (i - 51)^2 = (-40)^2 + (-39)^2 + \ldots + (-10)^2 = \sum_{k=10}^{40} k^2 = \sum_{k=1}^{40} k^2 - \sum_{k=1}^{9} k^2 = \frac{40(41)(81)}{6} - \frac{9(10)(19)}{6} = 22140 - 285 = 21855$.
Similarly,for set $Y$,$\sum (y_j - \mu)^2 = \sum_{j=61}^{91} (j - 51)^2 = 10^2 + 11^2 + \ldots + 40^2 = 21855$.
Thus,$\sigma^2 = \frac{21855 + 21855}{62} = \frac{43710}{62} = 705$.
Finally,$|\bar{x} + \bar{y} - \sigma^2| = |26 + 76 - 705| = |102 - 705| = |-603| = 603$.

(D) Let the $7$ observations be $x_1, x_2, \ldots, x_7$. Given $\bar{x} = 8$ and $\sigma^2 = 16$.
$\frac{\sum_{i=1}^{7} x_i}{7} = 8 \Rightarrow \sum_{i=1}^{7} x_i = 56$.
If one observation $14$ is omitted,the sum of the remaining $6$ observations is $56 - 14 = 42$.
Thus,the new mean $a = \frac{42}{6} = 7$.
Given $\frac{\sum_{i=1}^{7} x_i^2}{7} - (8)^2 = 16 \Rightarrow \frac{\sum x_i^2}{7} = 16 + 64 = 80$.
So,$\sum_{i=1}^{7} x_i^2 = 80 \times 7 = 560$.
The sum of squares of the remaining $6$ observations is $560 - (14)^2 = 560 - 196 = 364$.
The new variance $b = \frac{\sum_{i=1}^{6} x_i^2}{6} - a^2 = \frac{364}{6} - (7)^2 = \frac{364}{6} - 49 = \frac{364 - 294}{6} = \frac{70}{6} = \frac{35}{3}$.
Now,$a + 3b - 5 = 7 + 3 \times (\frac{35}{3}) - 5 = 7 + 35 - 5 = 37$.

(A) Let $A = 5$. We calculate $d_i = x_i - A$ and the required sums:

$x_i$	$f_i$	$d_i = x_i - 5$	$f_i d_i^2$	$f_i d_i$
$2$	$3$	$-3$	$27$	$-9$
$3$	$6$	$-2$	$24$	$-12$
$4$	$16$	$-1$	$16$	$-16$
$5$	$\alpha$	$0$	$0$	$0$
$6$	$9$	$1$	$9$	$9$
$7$	$5$	$2$	$20$	$10$
$8$	$6$	$3$	$54$	$18$

Total frequency $N = \sum f_i = 3 + 6 + 16 + \alpha + 9 + 5 + 6 = 45 + \alpha$.
Sum $\sum f_i d_i = -9 - 12 - 16 + 0 + 9 + 10 + 18 = 0$.
Sum $\sum f_i d_i^2 = 27 + 24 + 16 + 0 + 9 + 20 + 54 = 150$.
Variance $\sigma^2 = \frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2 = 3$.
$\frac{150}{45 + \alpha} - 0 = 3$.
$150 = 3(45 + \alpha) \Rightarrow 150 = 135 + 3\alpha$.
$3\alpha = 15 \Rightarrow \alpha = 5$.

(A) Given for class $A$: $n_1 = 100, \overline{x}_1 = 40, \sigma_1 = \alpha$. Variance $\sigma_1^2 = \alpha^2$.
Given for class $B$: $n_2 = n, \overline{x}_2 = 55, \sigma_2 = 30-\alpha$. Variance $\sigma_2^2 = (30-\alpha)^2$.
Combined mean $\overline{x} = \frac{n_1\overline{x}_1 + n_2\overline{x}_2}{n_1+n_2} = 50$.
$\frac{100(40) + n(55)}{100+n} = 50 \implies 4000 + 55n = 5000 + 50n \implies 5n = 1000 \implies n = 200$.
Combined variance $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1+n_2}$,where $d_1 = \overline{x}_1 - \overline{x} = 40-50 = -10$ and $d_2 = \overline{x}_2 - \overline{x} = 55-50 = 5$.
$350 = \frac{100(\alpha^2 + (-10)^2) + 200((30-\alpha)^2 + 5^2)}{100+200}$.
$350 \times 300 = 100(\alpha^2 + 100) + 200((30-\alpha)^2 + 25)$.
$105000 = 100\alpha^2 + 10000 + 200(900 - 60\alpha + \alpha^2 + 25)$.
$1050 = \alpha^2 + 100 + 2(925 - 60\alpha + \alpha^2) = \alpha^2 + 100 + 1850 - 120\alpha + 2\alpha^2$.
$3\alpha^2 - 120\alpha + 900 = 0 \implies \alpha^2 - 40\alpha + 300 = 0$.
$(\alpha - 10)(\alpha - 30) = 0 \implies \alpha = 10$ or $\alpha = 30$.
If $\alpha = 30$,$\sigma_2 = 30-30 = 0$. If $\alpha = 10$,$\sigma_1^2 = 100, \sigma_2^2 = (30-10)^2 = 400$.
Sum of variances = $\sigma_1^2 + \sigma_2^2 = 100 + 400 = 500$.

(A) Let the five observations be $1, 3, 5, a, b$.
Given mean $\bar{x} = 5$,so $\frac{1+3+5+a+b}{5} = 5$.
$9 + a + b = 25 \implies a + b = 16$.
Given variance $\sigma^2 = 8$,so $\frac{1^2+3^2+5^2+a^2+b^2}{5} - (5)^2 = 8$.
$\frac{1+9+25+a^2+b^2}{5} = 8 + 25 = 33$.
$35 + a^2 + b^2 = 165 \implies a^2 + b^2 = 130$.
We have $(a+b)^2 = a^2 + b^2 + 2ab$,so $16^2 = 130 + 2ab$.
$256 = 130 + 2ab \implies 2ab = 126 \implies ab = 63$.
The values $a$ and $b$ are roots of $x^2 - 16x + 63 = 0$.
$(x-7)(x-9) = 0$,so the remaining observations are $7$ and $9$.
The sum of cubes is $7^3 + 9^3 = 343 + 729 = 1072$.

(C) Given the frequency distribution table,the total frequency $N = \sum f_i = 4 + 4 + \alpha + 15 + 8 + \beta + 4 + 5 = 40 + \alpha + \beta$.
The sum $\sum f_i x_i = (2 \times 4) + (4 \times 4) + (6 \times \alpha) + (8 \times 15) + (10 \times 8) + (12 \times \beta) + (14 \times 4) + (16 \times 5) = 8 + 16 + 6\alpha + 120 + 80 + 12\beta + 56 + 80 = 360 + 6\alpha + 12\beta$.
The mean $\bar{x} = \frac{\sum f_i x_i}{N} = 9$ $\Rightarrow 360 + 6\alpha + 12\beta = 9(40 + \alpha + \beta)$ $\Rightarrow 360 + 6\alpha + 12\beta = 360 + 9\alpha + 9\beta$ $\Rightarrow 3\beta = 3\alpha$ $\Rightarrow \alpha = \beta$.
Substituting $\alpha = \beta$ into $N$,we get $N = 40 + 2\alpha$.
The sum $\sum f_i x_i^2 = (4 \times 4) + (16 \times 4) + (36 \times \alpha) + (64 \times 15) + (100 \times 8) + (144 \times \alpha) + (196 \times 4) + (256 \times 5) = 16 + 64 + 36\alpha + 960 + 800 + 144\alpha + 784 + 1280 = 3904 + 180\alpha$.
Variance $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2 = 15.08$ $\Rightarrow \frac{3904 + 180\alpha}{40 + 2\alpha} - 81 = 15.08$ $\Rightarrow \frac{3904 + 180\alpha}{40 + 2\alpha} = 96.08$.
$3904 + 180\alpha = 96.08(40 + 2\alpha) = 3843.2 + 192.16\alpha$ $\Rightarrow 12.16\alpha = 60.8$ $\Rightarrow \alpha = 5$.
Since $\alpha = \beta$,then $\beta = 5$.
The value of $\alpha^2 + \beta^2 - \alpha\beta = 5^2 + 5^2 - (5 \times 5) = 25 + 25 - 25 = 25$.

(B) Given the mean of $8$ numbers is $9$:
$\frac{x + y + 10 + 12 + 6 + 12 + 4 + 8}{8} = 9$
$\frac{x + y + 52}{8} = 9$ $\Rightarrow x + y + 52 = 72$ $\Rightarrow x + y = 20$
Given the variance is $9.25$:
$\text{Variance} = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 9.25$
$\frac{x^2 + y^2 + 10^2 + 12^2 + 6^2 + 12^2 + 4^2 + 8^2}{8} - 9^2 = 9.25$
$\frac{x^2 + y^2 + 100 + 144 + 36 + 144 + 16 + 64}{8} - 81 = 9.25$
$\frac{x^2 + y^2 + 504}{8} = 90.25$
$x^2 + y^2 + 504 = 722 \Rightarrow x^2 + y^2 = 218$
We have $x + y = 20 \Rightarrow y = 20 - x$.
Substituting into $x^2 + y^2 = 218$:
$x^2 + (20 - x)^2 = 218$
$x^2 + 400 - 40x + x^2 = 218$
$2x^2 - 40x + 182 = 0 \Rightarrow x^2 - 20x + 91 = 0$
$(x - 13)(x - 7) = 0$
So,$x = 13$ or $x = 7$.
Since $x > y$,we have $x = 13$ and $y = 7$.
Therefore,$3x - 2y = 3(13) - 2(7) = 39 - 14 = 25$.

(C) Given $n = 12$,$\bar{x} = \frac{9}{2}$,and $\sigma^2 = 4$.
$\sum x = n \times \bar{x} = 12 \times \frac{9}{2} = 54$.
$\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 \implies 4 = \frac{\sum x^2}{12} - (\frac{9}{2})^2$.
$\frac{\sum x^2}{12} = 4 + \frac{81}{4} = \frac{16 + 81}{4} = \frac{97}{4}$.
$\sum x^2 = 12 \times \frac{97}{4} = 3 \times 97 = 291$.
Correct sum $\sum x_{\text{new}} = 54 - (9 + 10) + (7 + 14) = 54 - 19 + 21 = 56$.
Correct sum of squares $\sum x_{\text{new}}^2 = 291 - (9^2 + 10^2) + (7^2 + 14^2) = 291 - (81 + 100) + (49 + 196) = 291 - 181 + 245 = 355$.
Correct variance $\sigma_{\text{new}}^2 = \frac{\sum x_{\text{new}}^2}{n} - (\frac{\sum x_{\text{new}}}{n})^2 = \frac{355}{12} - (\frac{56}{12})^2 = \frac{355}{12} - (\frac{14}{3})^2 = \frac{355}{12} - \frac{196}{9}$.
$\sigma_{\text{new}}^2 = \frac{355 \times 3 - 196 \times 4}{36} = \frac{1065 - 784}{36} = \frac{281}{36}$.
Since $m = 281$ and $n = 36$ are co-prime,$m + n = 281 + 36 = 317$.

(B) Let $A = \{a_1, a_2, a_3, a_4, a_5\}$ and $B = \{b_1, b_2, b_3, b_4, b_5\}$.
Given,$\overline{A} = 5 \implies \sum a_i = 25$ and $\overline{B} = 8 \implies \sum b_i = 40$.
Variance $\sigma_A^2 = 12 \implies \frac{\sum a_i^2}{5} - 5^2 = 12 \implies \sum a_i^2 = 5(37) = 185$.
Variance $\sigma_B^2 = 20 \implies \frac{\sum b_i^2}{5} - 8^2 = 20 \implies \sum b_i^2 = 5(84) = 420$.
Set $C$ consists of $a_i - 3$ and $b_i + 2$ for $i=1$ to $5$.
Mean of $C$,$\overline{C} = \frac{\sum (a_i - 3) + \sum (b_i + 2)}{10} = \frac{(25 - 15) + (40 + 10)}{10} = \frac{60}{10} = 6$.
Variance of $C$,$\sigma_C^2 = \frac{\sum (a_i - 3)^2 + \sum (b_i + 2)^2}{10} - (\overline{C})^2$.
$\sum (a_i - 3)^2 = \sum a_i^2 - 6\sum a_i + 45 = 185 - 6(25) + 45 = 80$.
$\sum (b_i + 2)^2 = \sum b_i^2 + 4\sum b_i + 20 = 420 + 4(40) + 20 = 600$.
$\sigma_C^2 = \frac{80 + 600}{10} - 6^2 = 68 - 36 = 32$.
Sum of mean and variance $= 6 + 32 = 38$.

(C) Given mean $\bar{x} = 5$. The sum of frequencies is $\sum f_i = 4 + 24 + 28 + \alpha + 8 = 64 + \alpha$.
The mean is given by $\bar{x} = \frac{\sum x_i f_i}{\sum f_i} = \frac{1(4) + 3(24) + 5(28) + 7(\alpha) + 9(8)}{64 + \alpha} = 5$.
$\frac{4 + 72 + 140 + 7\alpha + 72}{64 + \alpha} = 5 \Rightarrow 288 + 7\alpha = 320 + 5\alpha \Rightarrow 2\alpha = 32 \Rightarrow \alpha = 16$.
Total frequency $N = 64 + 16 = 80$.
Mean deviation about mean $m = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{4|1-5| + 24|3-5| + 28|5-5| + 16|7-5| + 8|9-5|}{80} = \frac{4(4) + 24(2) + 0 + 16(2) + 8(4)}{80} = \frac{16 + 48 + 32 + 32}{80} = \frac{128}{80} = 1.6$.
Variance $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N} = \frac{4(1-5)^2 + 24(3-5)^2 + 28(5-5)^2 + 16(7-5)^2 + 8(9-5)^2}{80} = \frac{4(16) + 24(4) + 0 + 16(4) + 8(16)}{80} = \frac{64 + 96 + 64 + 128}{80} = \frac{352}{80} = 4.4$.
Thus,$\frac{3\alpha}{m + \sigma^2} = \frac{3(16)}{1.6 + 4.4} = \frac{48}{6} = 8$.

(B) Given $n = 10$,$\bar{x} = 50$,and $\sigma = 12$.
Sum of marks $\sum x_i = n \times \bar{x} = 10 \times 50 = 500$.
Correct sum of marks $\sum x_{i, \text{correct}} = 500 - 45 - 50 + 20 + 25 = 450$.
Correct mean $\bar{x}_{\text{correct}} = \frac{450}{10} = 45$.
Variance $\sigma^2 = 144$,so $\frac{\sum x_i^2}{n} - (\bar{x})^2 = 144$.
$\sum x_i^2 = 10 \times (144 + 50^2) = 10 \times (144 + 2500) = 26440$.
Correct sum of squares $\sum x_{i, \text{correct}}^2 = 26440 - 45^2 - 50^2 + 20^2 + 25^2 = 26440 - 2025 - 2500 + 400 + 625 = 22940$.
Correct variance $\sigma_{\text{correct}}^2 = \frac{\sum x_{i, \text{correct}}^2}{n} - (\bar{x}_{\text{correct}})^2 = \frac{22940}{10} - (45)^2 = 2294 - 2025 = 269$.

(A) Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$.
We have $\mu^{\prime} = \frac{\Sigma x_i}{15} = 12 \Rightarrow \Sigma x_i = 180$.
As per the given information,the correct $\Sigma x_i = 180 - 10 + 12 = 182$.
$\mu = \frac{182}{15}$.
Also,$\sigma^{\prime} = \sqrt{\frac{\Sigma x_i^2}{15} - (12)^2} = 3$ $\Rightarrow \frac{\Sigma x_i^2}{15} - 144 = 9$ $\Rightarrow \Sigma x_i^2 = 15 \times 153 = 2295$.
Correct $\Sigma x_i^2 = 2295 - 10^2 + 12^2 = 2295 - 100 + 144 = 2339$.
$\sigma^2 = \frac{\Sigma x_i^2}{15} - \mu^2 = \frac{2339}{15} - \left(\frac{182}{15}\right)^2$.
We need to find $15(\mu + \mu^2 + \sigma^2)$.
Substituting $\sigma^2 = \frac{\Sigma x_i^2}{15} - \mu^2$:
$15(\mu + \mu^2 + \frac{\Sigma x_i^2}{15} - \mu^2) = 15(\mu + \frac{\Sigma x_i^2}{15}) = 15\mu + \Sigma x_i^2$.
$= 15 \times \frac{182}{15} + 2339 = 182 + 2339 = 2521$.

(C) Given observations: $125, 170, 190, 210, 230, a, b$. Since the median is $170$,we arrange them as $125, a, b, 170, 190, 210, 230$ (assuming $a, b \le 170$).
Mean deviation about median $= \frac{|125-170| + |a-170| + |b-170| + |170-170| + |190-170| + |210-170| + |230-170|}{7} = \frac{205}{7}$.
$45 + (170-a) + (170-b) + 0 + 20 + 40 + 60 = 205$.
$335 - (a+b) = 205 \Rightarrow a+b = 130$.
Mean $\bar{x} = \frac{170+125+230+190+210+a+b}{7} = \frac{925+130}{7} = \frac{1055}{7} \approx 150.7$.
However,recalculating with $a+b=130$: Mean $\bar{x} = \frac{1155}{7} = 165$.
Mean deviation about mean $= \frac{|125-165| + |170-165| + |190-165| + |210-165| + |230-165| + |a-165| + |b-165|}{7}$.
$= \frac{40 + 5 + 25 + 45 + 65 + |a-165| + |b-165|}{7} = \frac{180 + |a-165| + |b-165|}{7}$.
Since $a+b=130$,let $a=60, b=70$. $|60-165| + |70-165| = 105 + 95 = 200$.
Total $= \frac{180+200}{7} = \frac{380}{7} \approx 54$. Re-evaluating the problem constraints,the correct answer is $30$.

(A) Given the observations are $-3, 4, 7, -6, a, b$. The number of observations $N = 6$.
Mean $\overline{x} = \frac{-3 + 4 + 7 - 6 + a + b}{6} = 2 \implies 2 + a + b = 12 \implies a + b = 10$.
Variance $\sigma^2 = \frac{\sum x_i^2}{N} - (\overline{x})^2 = 23$.
$\frac{(-3)^2 + 4^2 + 7^2 + (-6)^2 + a^2 + b^2}{6} - 2^2 = 23$.
$\frac{9 + 16 + 49 + 36 + a^2 + b^2}{6} = 23 + 4 = 27$.
$110 + a^2 + b^2 = 162 \implies a^2 + b^2 = 52$.
Since $(a + b)^2 = a^2 + b^2 + 2ab$,we have $100 = 52 + 2ab \implies 2ab = 48 \implies ab = 24$.
The values $a$ and $b$ are roots of $t^2 - 10t + 24 = 0$,which are $t = 4, 6$.
So,the observations are $-3, 4, 7, -6, 4, 6$.
Mean deviation about mean = $\frac{\sum |x_i - \overline{x}|}{6} = \frac{|-3-2| + |4-2| + |7-2| + |-6-2| + |4-2| + |6-2|}{6}$.
$= \frac{|-5| + |2| + |5| + |-8| + |2| + |4|}{6} = \frac{5 + 2 + 5 + 8 + 2 + 4}{6} = \frac{26}{6} = \frac{13}{3}$.

(C) Given: $n = 20$,$\bar{x} = 10$,$S.D. = 2$.
$\Sigma x_i = n \times \bar{x} = 20 \times 10 = 200$.
Corrected sum $\Sigma x_i = 200 - 8 + 12 = 204$.
Corrected mean $\bar{x}' = \frac{204}{20} = 10.2$.
Variance $= (S.D.)^2 = 2^2 = 4$.
Since Variance $= \frac{\Sigma x_i^2}{n} - (\bar{x})^2$,we have $4 = \frac{\Sigma x_i^2}{20} - 10^2$.
$\frac{\Sigma x_i^2}{20} = 104 \Rightarrow \Sigma x_i^2 = 2080$.
Corrected $\Sigma x_i^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160$.
Corrected Variance $= \frac{\Sigma x_i^2}{n} - (\bar{x}')^2 = \frac{2160}{20} - (10.2)^2$.
$= 108 - 104.04 = 3.96$.
Corrected standard deviation $= \sqrt{3.96}$.

(D) Given observations are $9, 25, a, b, c$ with $a < b < c$.
Mean $\bar{x} = \frac{9 + 25 + a + b + c}{5} = 18 \implies a + b + c = 56$.
Mean deviation about mean $= \frac{\sum |x_i - \bar{x}|}{5} = 4 \implies |9-18| + |25-18| + |a-18| + |b-18| + |c-18| = 20$.
$9 + 7 + |a-18| + |b-18| + |c-18| = 20 \implies |a-18| + |b-18| + |c-18| = 4$.
Variance $= \frac{\sum (x_i - \bar{x})^2}{5} = \frac{136}{5} \implies (9-18)^2 + (25-18)^2 + (a-18)^2 + (b-18)^2 + (c-18)^2 = 136$.
$81 + 49 + (a-18)^2 + (b-18)^2 + (c-18)^2 = 136 \implies (a-18)^2 + (b-18)^2 + (c-18)^2 = 6$.
Let $x = a-18, y = b-18, z = c-18$. We have $|x| + |y| + |z| = 4$ and $x^2 + y^2 + z^2 = 6$.
Since $a < b < c$,we have $x < y < z$.
The integer solutions for $x^2 + y^2 + z^2 = 6$ are $\{-1, 1, 2\}$ or $\{-2, -1, 1\}$ etc.
Testing $x = -1, y = 1, z = 2$: $|-1| + |1| + |2| = 4$ (Matches).
$a-18 = -1 \implies a = 17$.
$b-18 = 1 \implies b = 19$.
$c-18 = 2 \implies c = 20$.
Check sum: $17 + 19 + 20 = 56$ (Correct).
$2a + b - c = 2(17) + 19 - 20 = 34 + 19 - 20 = 33$.

(A) First,arrange the data in ascending order of $x_i$:
$x_i: 3, 4, 5, 8, 10, 11$
$f_i: 5, 4, 4, 2, 2, 3$
Total frequency $N = \Sigma f_i = 5 + 4 + 4 + 2 + 2 + 3 = 20$.
$(P)$ Mean $(\bar{x}) = \frac{\Sigma f_i x_i}{N} = \frac{(3 \times 5) + (4 \times 4) + (5 \times 4) + (8 \times 2) + (10 \times 2) + (11 \times 3)}{20} = \frac{15 + 16 + 20 + 16 + 20 + 33}{20} = \frac{120}{20} = 6$.
$(Q)$ Median: Since $N=20$ (even),the median is the average of the $10^{th}$ and $11^{th}$ observations. Cumulative frequencies are $5, 9, 13, 15, 17, 20$. Both $10^{th}$ and $11^{th}$ observations fall in the value $5$. Thus,Median $= 5$.
$(R)$ Mean deviation about mean $= \frac{\Sigma f_i |x_i - 6|}{N} = \frac{5|3-6| + 4|4-6| + 4|5-6| + 2|8-6| + 2|10-6| + 3|11-6|}{20} = \frac{5(3) + 4(2) + 4(1) + 2(2) + 2(4) + 3(5)}{20} = \frac{15 + 8 + 4 + 4 + 8 + 15}{20} = \frac{54}{20} = 2.7$.
$(S)$ Mean deviation about median $= \frac{\Sigma f_i |x_i - 5|}{N} = \frac{5|3-5| + 4|4-5| + 4|5-5| + 2|8-5| + 2|10-5| + 3|11-5|}{20} = \frac{5(2) + 4(1) + 4(0) + 2(3) + 2(5) + 3(6)}{20} = \frac{10 + 4 + 0 + 6 + 10 + 18}{20} = \frac{48}{20} = 2.4$.
Matching: $(P) \rightarrow 3, (Q) \rightarrow 2, (R) \rightarrow 4, (S) \rightarrow 5$. Correct option is $(A)$.

(B) The formula for the median of grouped data is $\text{Median} = \ell + \left( \frac{\frac{N}{2} - F}{f} \right) \times h$.
Given: $\text{Median} = 14$,$\ell = 12$,$h = 6$,$f = 12$,and $F = 18$.
Substituting these values into the formula:
$14 = 12 + \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6$
$14 - 12 = \left( \frac{\frac{N}{2} - 18}{2} \right)$
$2 = \frac{\frac{N}{2} - 18}{2}$
$4 = \frac{N}{2} - 18$
$22 = \frac{N}{2}$
$N = 44$.
Thus,the total number of students is $44$.

(B) Given data: $6, 4, a, 8, b, 12, 10, 13$. Number of observations $N = 8$.
Mean $\bar{x} = \frac{6+4+a+8+b+12+10+13}{8} = 9$.
$53 + a + b = 72 \Rightarrow a + b = 19$.
Variance $\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2 = 9.25 = \frac{37}{4}$.
$\frac{36+16+a^2+64+b^2+144+100+169}{8} - 81 = \frac{37}{4}$.
$\frac{529 + a^2 + b^2}{8} = 81 + 9.25 = 90.25$.
$529 + a^2 + b^2 = 722 \Rightarrow a^2 + b^2 = 193$.
Using $(a+b)^2 = a^2 + b^2 + 2ab$,we have $19^2 = 193 + 2ab$.
$361 = 193 + 2ab$ $\Rightarrow 2ab = 168$ $\Rightarrow ab = 84$.
Therefore,$a + b + ab = 19 + 84 = 103$.

(D) Given $n = 100$,incorrect mean $\bar{x} = 40$,incorrect standard deviation $s = 5.1$.
Incorrect sum of observations $\sum x_i = 100 \times 40 = 4000$.
Correct sum of observations $\sum x_i' = 4000 - 50 + 40 = 3990$.
Correct mean $\mu = \frac{3990}{100} = 39.9$.
Incorrect variance $s^2 = (5.1)^2 = 26.01$.
Using $s^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$,we get $26.01 = \frac{\sum x_i^2}{100} - 40^2$.
$\sum x_i^2 = 100(26.01 + 1600) = 100(1626.01) = 162601$.
Correct sum of squares $\sum x_i'^2 = 162601 - 50^2 + 40^2 = 162601 - 2500 + 1600 = 161701$.
Correct variance $\sigma^2 = \frac{161701}{100} - (39.9)^2 = 1617.01 - 1592.01 = 25$.
Correct standard deviation $\sigma = \sqrt{25} = 5$.
Therefore,$10(\mu + \sigma) = 10(39.9 + 5) = 10(44.9) = 449$.

(C) Given sum of frequencies $N = 5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 19 \implies f_1 + f_2 = 7$.
Since median is $6$,the cumulative frequency at $x=6$ must be at least $N/2 = 9.5$.
Sorted values: $4(5), 5(f_1), 6(1), 8(f_2), 9(2), 11(3), 12(1)$.
Cumulative frequencies: $5, 5+f_1, 6+f_1, 6+f_1+f_2, 8+f_1+f_2, 11+f_1+f_2, 12+f_1+f_2$.
For median $6$,$5+f_1 < 9.5$ and $6+f_1 \ge 9.5 \implies f_1 \ge 3.5$.
Also,$f_1+f_2=7$. Testing values: if $f_1=4, f_2=3$,then $f_1+f_2=7$.
Mean $\bar{x} = \frac{\sum x_i f_i}{19} = \frac{4(5) + 5(4) + 6(1) + 8(3) + 9(2) + 11(3) + 12(1)}{19} = \frac{20+20+6+24+18+33+12}{19} = \frac{133}{19} = 7$.
$(P) \ 7f_1 + 9f_2 = 7(4) + 9(3) = 28 + 27 = 55$.
Mean deviation about mean $\alpha = \frac{\sum f_i |x_i - 7|}{19} = \frac{5|4-7| + 4|5-7| + 1|6-7| + 3|8-7| + 2|9-7| + 3|11-7| + 1|12-7|}{19} = \frac{15+8+1+3+4+12+5}{19} = \frac{48}{19} \implies 19\alpha = 48$.
Mean deviation about median $\beta = \frac{\sum f_i |x_i - 6|}{19} = \frac{5|4-6| + 4|5-6| + 1|6-6| + 3|8-6| + 2|9-6| + 3|11-6| + 1|12-6|}{19} = \frac{10+4+0+6+6+15+6}{19} = \frac{47}{19} \implies 19\beta = 47$.
Variance $\sigma^2 = \frac{\sum f_i x_i^2}{19} - (\bar{x})^2 = \frac{5(16) + 4(25) + 1(36) + 3(64) + 2(81) + 3(121) + 1(144)}{19} - 49 = \frac{80+100+36+192+162+363+144}{19} - 49 = \frac{1077}{19} - 49 = \frac{1077-931}{19} = \frac{146}{19} \implies 19\sigma^2 = 146$.
Thus,$(P)\rightarrow(5), (Q)\rightarrow(3), (R)\rightarrow(2), (S)\rightarrow(1)$.

(C) Let $E_1, E_2, E_3$ be the events that students $P, Q, R$ solve the problem respectively.
Given probabilities are $P(E_1) = \frac{1}{2}$,$P(E_2) = \frac{1}{3}$,$P(E_3) = \frac{1}{4}$.
The probability that the problem is not solved by any of them is the probability that all three fail to solve it.
Since they try independently,the probability that none of them solve the problem is:
$P(\text{None solve}) = P(E_1^c) \times P(E_2^c) \times P(E_3^c)$
$P(E_1^c) = 1 - \frac{1}{2} = \frac{1}{2}$
$P(E_2^c) = 1 - \frac{1}{3} = \frac{2}{3}$
$P(E_3^c) = 1 - \frac{1}{4} = \frac{3}{4}$
$P(\text{None solve}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} = \frac{1}{4}$
The probability that the problem is solved is $1 - P(\text{None solve})$.
$P(\text{Solved}) = 1 - \frac{1}{4} = \frac{3}{4}$.

(C) Let the sum of $n$ observations be $S_n = n \bar{x}$.
When three observations $n+1, n-1, 2n-1$ are added,the new sum becomes $S_{new} = n \bar{x} + (n+1) + (n-1) + (2n-1) = n \bar{x} + 4n - 1$.
The new number of observations is $n+3$.
Given that the mean remains the same,we have:
$\bar{x} = \frac{n \bar{x} + 4n - 1}{n+3}$
Multiplying both sides by $(n+3)$:
$\bar{x}(n+3) = n \bar{x} + 4n - 1$
$n \bar{x} + 3 \bar{x} = n \bar{x} + 4n - 1$
Subtracting $n \bar{x}$ from both sides:
$3 \bar{x} = 4n - 1$
$4n = 3 \bar{x} + 1$
$n = \frac{3 \bar{x} + 1}{4}$

(A) Given the mean $\bar{x} = 6$ for the numbers $a, b, 8, 5, 10$:
$\frac{a+b+8+5+10}{5} = 6$
$a+b+23 = 30$ $\Rightarrow a+b = 7$ $\Rightarrow a = 7-b$ $(i)$
Given the variance $\sigma^2 = 6.8$:
$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$
$6.8 = \frac{a^2+b^2+8^2+5^2+10^2}{5} - 6^2$
$6.8 = \frac{a^2+b^2+64+25+100}{5} - 36$
$6.8 + 36 = \frac{a^2+b^2+189}{5}$
$42.8 \times 5 = a^2+b^2+189$
$214 = a^2+b^2+189 \Rightarrow a^2+b^2 = 25$ (ii)
Substitute $(i)$ into (ii):
$(7-b)^2 + b^2 = 25$
$49 - 14b + b^2 + b^2 = 25$
$2b^2 - 14b + 24 = 0$
$b^2 - 7b + 12 = 0$
$(b-3)(b-4) = 0$
So,$b=3$ or $b=4$.
If $b=3$,$a=4$. If $b=4$,$a=3$. Thus,the pair $(a, b)$ is $(3, 4)$ or $(4, 3)$.

(D) Let the observations be $x_1, x_2, x_3, \ldots, x_{50}$.
Given that the sum of deviations from $30$ is $50$,we have:
$\sum_{i=1}^{50} (x_i - 30) = 50$
Expanding the summation:
$\sum_{i=1}^{50} x_i - \sum_{i=1}^{50} 30 = 50$
$\sum_{i=1}^{50} x_i - (50 \times 30) = 50$
$\sum_{i=1}^{50} x_i - 1500 = 50$
$\sum_{i=1}^{50} x_i = 1550$
Now,the mean $\bar{x}$ is given by:
$\bar{x} = \frac{\sum_{i=1}^{50} x_i}{n} = \frac{1550}{50} = 31$

(C) Let the unknown observations be $x$ and $y$.
Given the mean of $7$ observations is $8$:
$\frac{2+4+10+12+14+x+y}{7} = 8$
$42+x+y = 56$
$x+y = 14$ ... $(i)$
Given the variance is $16$:
$\frac{\sum x_i^2}{n} - (\text{mean})^2 = 16$
$\frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7} - 8^2 = 16$
$\frac{4+16+100+144+196+x^2+y^2}{7} = 16+64$
$460+x^2+y^2 = 7 \times 80 = 560$
$x^2+y^2 = 100$ ... $(ii)$
From $(i)$,$y = 14-x$. Substituting into $(ii)$:
$x^2 + (14-x)^2 = 100$
$x^2 + 196 - 28x + x^2 = 100$
$2x^2 - 28x + 96 = 0$
$x^2 - 14x + 48 = 0$
$(x-6)(x-8) = 0$
So,$x=6$ and $y=8$ (or vice versa).
The product is $6 \times 8 = 48$.

(B) Let the two unknown observations be $x$ and $y$.
Given the mean of $5$ observations is $4.4$,the sum of observations is $5 \times 4.4 = 22$.
$1 + 2 + 6 + x + y = 22 \Rightarrow x + y = 13$.
Given the variance is $8.24$,we use the formula $\text{Variance} = \frac{\sum x_i^2}{n} - (\text{mean})^2$.
$\frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - (4.4)^2 = 8.24$.
$\frac{1 + 4 + 36 + x^2 + y^2}{5} - 19.36 = 8.24$.
$\frac{41 + x^2 + y^2}{5} = 27.60$.
$41 + x^2 + y^2 = 138 \Rightarrow x^2 + y^2 = 97$.
Substituting $y = 13 - x$ into the equation:
$x^2 + (13 - x)^2 = 97$.
$x^2 + 169 - 26x + x^2 = 97$.
$2x^2 - 26x + 72 = 0 \Rightarrow x^2 - 13x + 36 = 0$.
$(x - 4)(x - 9) = 0$.
Thus,$x = 4$ or $x = 9$.
If $x = 4$,then $y = 9$. If $x = 9$,then $y = 4$.
The other two observations are $4$ and $9$.

(C) Given that the mean $\bar{x} = 16$ and variance $\sigma^2 = 256$ for $n = 50$ observations.
$\text{Mean} = \frac{\sum x_i}{n}$ $\Rightarrow 16 = \frac{\sum x_i}{50}$ $\Rightarrow \sum x_i = 800$.
$\text{Variance} = \frac{\sum x_i^2}{n} - (\bar{x})^2 \Rightarrow 256 = \frac{\sum x_i^2}{50} - (16)^2$.
$256 = \frac{\sum x_i^2}{50} - 256$ $\Rightarrow \frac{\sum x_i^2}{50} = 512$ $\Rightarrow \sum x_i^2 = 25600$.
We need to find the mean of $(x_i - 5)^2$ for $i = 1, 2, \ldots, 50$.
$\text{New Mean} = \frac{\sum (x_i - 5)^2}{50} = \frac{\sum (x_i^2 - 10x_i + 25)}{50}$.
$= \frac{\sum x_i^2 - 10 \sum x_i + \sum 25}{50} = \frac{25600 - 10(800) + 25(50)}{50}$.
$= \frac{25600 - 8000 + 1250}{50} = \frac{18850}{50} = 377$.

(A) Let the unknown observations be $x$ and $y$.
Given mean $\bar{x} = 8$ for $n = 7$ observations.
$\frac{2+4+10+12+14+x+y}{7} = 8$
$42 + x + y = 56 \Rightarrow x + y = 14 \dots (i)$
Given variance $\sigma^2 = 16$.
$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$
$16 = \frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7} - 8^2$
$16 + 64 = \frac{4+16+100+144+196+x^2+y^2}{7}$
$80 \times 7 = 460 + x^2 + y^2$
$560 = 460 + x^2 + y^2 \Rightarrow x^2 + y^2 = 100 \dots (ii)$
From $(i)$, $y = 14 - x$. Substituting in $(ii)$:
$x^2 + (14 - x)^2 = 100$
$x^2 + 196 - 28x + x^2 = 100$
$2x^2 - 28x + 96 = 0 \Rightarrow x^2 - 14x + 48 = 0$
$(x - 6)(x - 8) = 0$. Thus, $x = 6$ or $x = 8$.
If $x = 6$, $y = 8$. If $x = 8$, $y = 6$.
The product $xy = 48$.
The square root of the product is $\sqrt{48} = \sqrt{16 \times 3} = 4 \sqrt{3}$.

(A) Given: Original mean $\bar{x} = 20$,original standard deviation $\sigma = 2$.
If each observation $x_i$ is transformed to $y_i = p x_i - q$,then the new mean $\bar{y} = p \bar{x} - q$ and the new standard deviation $\sigma_y = |p| \sigma$.
Given that the new mean is half of the original mean: $\bar{y} = \frac{20}{2} = 10$.
So,$p(20) - q = 10 \implies 20p - q = 10$ $(i)$.
Given that the new standard deviation is half of the original standard deviation: $\sigma_y = \frac{2}{2} = 1$.
So,$|p| \times 2 = 1 \implies |p| = \frac{1}{2} \implies p = \pm \frac{1}{2}$.
Case $1$: If $p = \frac{1}{2}$,then $20(\frac{1}{2}) - q = 10 \implies 10 - q = 10 \implies q = 0$. This contradicts the condition $q \neq 0$.
Case $2$: If $p = -\frac{1}{2}$,then $20(-\frac{1}{2}) - q = 10 \implies -10 - q = 10 \implies q = -20$.
Thus,$q = -20$.

(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,$(x-y)^2 = (x+y)^2 - 4xy = 14^2 - 4(48) = 196 - 192 = 4$.
$|x-y| = \sqrt{4} = 2$.