Word problem -Statistics Questions in English

(D) Given that the mean of $n$ observations is $\bar{x}$.
Therefore,the sum of observations is $\sum_{i=1}^{n} x_{i} = n \bar{x}$.
When the observation $x_{q}$ is replaced by $x_{q}^{\prime}$,the new sum of observations becomes:
$\sum x_{new} = \sum x - x_{q} + x_{q}^{\prime} = n \bar{x} - x_{q} + x_{q}^{\prime}$.
The new mean $\bar{x}^{\prime}$ is given by:
$\bar{x}^{\prime} = \frac{\sum x_{new}}{n} = \frac{n \bar{x} - x_{q} + x_{q}^{\prime}}{n}$.

(D) Let the $10$ observations be $x_1, x_2, \ldots, x_9, \alpha$.
Given mean $\bar{x} = 10$,so $\frac{\sum_{i=1}^9 x_i + \alpha}{10} = 10 \Rightarrow \sum_{i=1}^9 x_i + \alpha = 100$.
Given variance $\sigma^2 = 2$,so $\frac{\sum_{i=1}^9 x_i^2 + \alpha^2}{10} - (10)^2 = 2 \Rightarrow \sum_{i=1}^9 x_i^2 + \alpha^2 = 1020$.
When $\alpha$ is replaced by $\beta$,the new mean is $10.1$,so $\frac{\sum_{i=1}^9 x_i + \beta}{10} = 10.1 \Rightarrow \sum_{i=1}^9 x_i + \beta = 101$.
Subtracting the first mean equation from this,$\beta - \alpha = 101 - 100 = 1 \Rightarrow \beta = \alpha + 1$.
The new variance is $1.99$,so $\frac{\sum_{i=1}^9 x_i^2 + \beta^2}{10} - (10.1)^2 = 1.99$.
$\sum_{i=1}^9 x_i^2 + \beta^2 = 10(1.99 + 102.01) = 10(104) = 1040$.
Subtracting the first variance equation from this,$\beta^2 - \alpha^2 = 1040 - 1020 = 20$.
Since $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha) = 20$ and $\beta - \alpha = 1$,we have $1 \times (\beta + \alpha) = 20$.
Therefore,$\alpha + \beta = 20$.

(D) The class marks $(x_i)$ are $6, 10, 14, 18$. The frequencies $(f_i)$ are $3, \lambda, 4, 7$. The sum of frequencies is $N = 14+\lambda$.
Mean $\mu = \frac{\sum f_i x_i}{N} = \frac{3(6) + 10\lambda + 4(14) + 7(18)}{14+\lambda} = \frac{18 + 10\lambda + 56 + 126}{14+\lambda} = \frac{200+10\lambda}{14+\lambda}$.
Variance $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2 = 19$.
Calculating $\sum f_i x_i^2 = 3(36) + 100\lambda + 4(196) + 7(324) = 108 + 100\lambda + 784 + 2268 = 3160 + 100\lambda$.
So,$\frac{3160+100\lambda}{14+\lambda} - \left(\frac{200+10\lambda}{14+\lambda}\right)^2 = 19$.
Solving for $\lambda$,we find $\lambda = 6$.
Substituting $\lambda = 6$ into the mean formula: $\mu = \frac{200+60}{14+6} = \frac{260}{20} = 13$.
Therefore,$\lambda + \mu = 6 + 13 = 19$.

(A) Given the mean of $8$ numbers is $\frac{7}{2}$.
Sum of numbers $= -10 - 7 - 1 + x + y + 9 + 2 + 16 = x + y + 9$.
$\frac{x + y + 9}{8} = \frac{7}{2}$ $\Rightarrow x + y + 9 = 28$ $\Rightarrow x + y = 19$ . . . $(1)$
Variance $= \frac{\sum x_i^2}{n} - (\text{mean})^2 = \frac{293}{4}$.
$\frac{(-10)^2 + (-7)^2 + (-1)^2 + x^2 + y^2 + 9^2 + 2^2 + 16^2}{8} - (\frac{7}{2})^2 = \frac{293}{4}$.
$\frac{100 + 49 + 1 + x^2 + y^2 + 81 + 4 + 256}{8} - \frac{49}{4} = \frac{293}{4}$.
$\frac{491 + x^2 + y^2}{8} = \frac{293 + 49}{4} = \frac{342}{4} = 85.5$.
$491 + x^2 + y^2 = 684 \Rightarrow x^2 + y^2 = 193$ . . . $(2)$
From $(1)$,$y = 19 - x$. Substituting in $(2)$:
$x^2 + (19 - x)^2 = 193 \Rightarrow x^2 + 361 - 38x + x^2 = 193$.
$2x^2 - 38x + 168 = 0 \Rightarrow x^2 - 19x + 84 = 0$.
$(x - 12)(x - 7) = 0$. So,$x = 12, y = 7$ or $x = 7, y = 12$.
The $4$ numbers are $x, y, x + y + 1, |x - y|$.
Substituting values: $12, 7, 12 + 7 + 1, |12 - 7| = 12, 7, 20, 5$.
Mean $= \frac{12 + 7 + 20 + 5}{4} = \frac{44}{4} = 11$.

(D) Let the $n$ observations be $x_1, x_2, \dots, x_n$.
Given: Mean $\bar{x} = 8$ and Variance $\sigma^2 = 16$.
For $n$ observations: $\sum_{i=1}^n x_i = 8n$ and $\frac{\sum x_i^2}{n} - (8)^2 = 16 \implies \sum x_i^2 = 80n$.
Let the sum of the first $(n-1)$ observations be $S_{n-1} = 48$ and the sum of squares be $Q_{n-1} = 496$.
The $n$-th observation is $x_n = \sum_{i=1}^n x_i - \sum_{i=1}^{n-1} x_i = 8n - 48$.
Also, $x_n^2 = \sum_{i=1}^n x_i^2 - \sum_{i=1}^{n-1} x_i^2 = 80n - 496$.
Substituting $x_n$: $(8n - 48)^2 = 80n - 496$.
$64n^2 - 768n + 2304 = 80n - 496$.
$64n^2 - 848n + 2800 = 0$.
Dividing by $16$: $4n^2 - 53n + 175 = 0$.
Factoring: $(4n - 25)(n - 7) = 0$.
Since $n$ must be an integer, $n = 7$.

(C) The mean of the data is given by $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
Mid-values $(x_i)$: $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$.
Total frequency $\sum f_i = 2 + k + 28 + 54 + (k + 1) + 5 = 90 + 2k$.
Sum of $f_i x_i = (2 \times 7.5) + (k \times 12.5) + (28 \times 17.5) + (54 \times 22.5) + ((k + 1) \times 27.5) + (5 \times 32.5) = 15 + 12.5k + 490 + 1215 + 27.5k + 27.5 + 162.5 = 1910 + 40k$.
Given $\bar{x} = 21$,so $\frac{1910 + 40k}{90 + 2k} = 21$.
$1910 + 40k = 21(90 + 2k) \Rightarrow 1910 + 40k = 1890 + 42k$.
$2k = 20 \Rightarrow k = 10$.
Testing $k = 10$ in the given equations:
$A) 2(10)^2 - 23(10) - 10 = 200 - 230 - 10 = -40 \neq 0$.
$B) 4(10)^2 - 35(10) + 24 = 400 - 350 + 24 = 74 \neq 0$.
$C) 2(10)^2 - 19(10) - 10 = 200 - 190 - 10 = 0$.
$D) 2(10)^2 - 35(10) + 98 = 200 - 350 + 98 = -52 \neq 0$.
Therefore,$k = 10$ is a root of the equation $2x^2 - 19x - 10 = 0$.

(B) The sum of the $9$ numbers is $21+8+17+a+51+103+b+13+67 = 280+a+b$.
Given the mean is $40$,we have $(280+a+b)/9 = 40$,which implies $280+a+b = 360$,so $a+b = 80$.
Arranging the numbers in ascending order: $8, 13, 17, 21, b, a, 51, 67, 103$ (since $a > b$ and the median is $21$,$b$ must be $21$ or less,but for the median to be $21$ in a set of $9$ numbers,the $5^{th}$ term must be $21$).
Thus,$b = 21$. Substituting this into $a+b = 80$,we get $a = 80 - 21 = 59$.
Checking the mean deviation about the median $(21)$: $\frac{1}{9} (|8-21| + |13-21| + |17-21| + |21-21| + |21-21| + |59-21| + |51-21| + |67-21| + |103-21|) = \frac{1}{9} (13+8+4+0+0+38+30+46+82) = \frac{221}{9} \approx 24.55$.
Given the mean deviation is $26$,there might be a slight discrepancy in the problem statement's provided value,but based on the constraints $a+b=80$ and median $=21$,$a=59$ is the unique solution.
Therefore,$2a = 2 \times 59 = 118$. The closest option is $117$.

(C) For the first set of observations: $n_1 = 4$,$\bar{x}_1 = 1$,and $\sigma_1^2 = 13$.
Using the formula $\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2$,we have $\frac{\sum x_1^2}{4} - (1)^2 = 13$,which gives $\sum x_1^2 = 4(14) = 56$.
For the second set of observations: $n_2 = 6$,$\bar{x}_2 = 2$,and $\sigma_2^2 = 1$.
Similarly,$\frac{\sum x_2^2}{6} - (2)^2 = 1$,which gives $\sum x_2^2 = 6(5) = 30$.
Now,the combined mean $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{4(1) + 6(2)}{4 + 6} = \frac{16}{10} = 1.6$.
The combined variance $\sigma^2 = \frac{\sum (x_1^2 + x_2^2)}{n_1 + n_2} - (\bar{x})^2 = \frac{56 + 30}{10} - (1.6)^2 = 8.6 - 2.56 = 6.04$.

(B) Given observations: $2, 4, \alpha, 8, \beta, 12, 14$. Number of observations $n = 7$.
Mean $\bar{x} = \frac{2+4+\alpha+8+\beta+12+14}{7} = 8$.
$40 + \alpha + \beta = 56 \Rightarrow \alpha + \beta = 16$ (Equation $1$).
Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 16$.
$\frac{4 + 16 + \alpha^2 + 64 + \beta^2 + 144 + 196}{7} - 8^2 = 16$.
$\frac{424 + \alpha^2 + \beta^2}{7} = 16 + 64 = 80$.
$424 + \alpha^2 + \beta^2 = 560 \Rightarrow \alpha^2 + \beta^2 = 136$ (Equation $2$).
Using $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$,we get $16^2 = 136 + 2\alpha\beta$.
$256 - 136 = 2\alpha\beta \Rightarrow 2\alpha\beta = 120 \Rightarrow \alpha\beta = 60$.
Since $\alpha + \beta = 16$ and $\alpha\beta = 60$,the values are $\alpha = 6$ and $\beta = 10$ (as $\alpha < \beta$).
The roots of the required quadratic equation are $3\alpha + 2 = 3(6) + 2 = 20$ and $2\beta + 1 = 2(10) + 1 = 21$.
Sum of roots $= 20 + 21 = 41$.
Product of roots $= 20 \times 21 = 420$.
The quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$,which is $x^2 - 41x + 420 = 0$.