Statistics Part 2

Concept:For a linear transformation $Y = aX + b$, the mean of $Y$ is $a\mu_X + b$ and the variance of $Y$ is $a^2\sigma_X^2$. Here we use the given data to set up equations in $a$ and $b$.Formula:Mean: $\bar{Y} = \frac{1}{n}\sum_{i=1}^{n} (a x_i + b) = a\bar{x} + b$Variance: $\sigma_Y^2 = \frac{1}{n}\sum_{i=1}^{n} (a x_i + b)^2 - (\bar{Y})^2 = a^2\sigma_X^2$Sum of first $n$ natural numbers: $S = \frac{n(n+1)}{2}$Sum of squares: $S_2 = \frac{n(n+1)(2n+1)}{6}$Solution:$X = \{1,2,\ldots,19\}$, so $n=19$.Sum of $X$: $\sum x = \frac{19\cdot20}{2} = 190$Sum of squares: $\sum x^2 = \frac{19\cdot20\cdot39}{6} = 2470$Given mean of $Y$ is $30$:$\frac{1}{19}\sum (a x + b) = 30$ $\Rightarrow a(190) + 19b = 570$ $\Rightarrow 10a + b = 30 \quad \Rightarrow a = \frac{30-b}{10}$ …(1)Given variance of $Y$ is $750$:$\frac{1}{19}\sum (a x + b)^2 - (30)^2 = 750$ $\Rightarrow \frac{1}{19}\left(a^2\sum x^2 + 19b^2 + 2ab\sum x\right) = 750 + 900 = 1650$ $\Rightarrow a^2\cdot2470 + 19b^2 + 2ab\cdot190 = 19\cdot1650$Divide by $19$: $130a^2 + b^2 + 20ab = 1650$ …(2)Substitute $a$ from (1) into (2):$130\left(\frac{30-b}{10}\right)^2 + b^2 + 20b\left(\frac{30-b}{10}\right) = 1650$$\Rightarrow 130\cdot\frac{(30-b)^2}{100} + b^2 + 2b(30-b) = 1650$$\Rightarrow 1.3(900 - 60b + b^2) + b^2 + 60b - 2b^2 = 1650$$\Rightarrow 1170 - 78b + 1.3b^2 + b^2 + 60b - 2b^2 = 1650$$\Rightarrow (1170) + (-18b) + (0.3b^2) = 1650$Multiply by $10$: $11700 - 180b + 3b^2 = 16500$$\Rightarrow 3b^2 - 180b - 4800 = 0$$\Rightarrow b^2 - 60b - 1600 = 0$The sum of possible values of $b$ equals the sum of roots of this quadratic:$b_1 + b_2 = -\frac{\text{coefficient of }b}{\text{coefficient of }b^2} = -\frac{(-60)}{1} = 60$Answer:$60$, which corresponds to option D.