Let X = x N : 1 x 19 and Y = ax+b : x X, a,bR for some a, b R.

Question

Let $X = \{x \in \mathbb{N} : 1 \le x \le 19\}$ and $Y = \{ax+b : x \in X, a,b\in\mathbb{R}\}$ for some $a, b \in \mathbb{R}$. If the mean and variance of the elements of $Y$ are 30 and 750, respectively, then the sum of all possible values of $b$ is

Accepted Answer

Mean of $Y$: $\bar{Y} = \dfrac{\sum_{i=1}^{19}(ax_i+b)}{19} = 30$

$$a(1+2+\cdots+19)+19b = 19\times30 \implies a\cdot\frac{19\times20}{2}+19b = 570 \implies 10a+b = 30 \implies a = \frac{30-b}{10}$$

Variance of $Y$: $\sigma^2 = \dfrac{\sum(ax_i+b)^2}{19} - (30)^2 = 750$

$$a^2\sum x_i^2 + 19b^2 + 2ab\sum x_i = (750+900)\times19 = 1650\times19$$

$$a^2\cdot\frac{19\times20\times39}{6} + 19b^2 + 2ab\cdot\frac{19\times20}{2} = 1650\times19$$

$130a^2+b^2+20ab = 1650$

Substituting $a = \dfrac{30-b}{10}$:

$$130\left(\frac{30-b}{10}\right)^2+b^2+20b\left(\frac{30-b}{10}\right)=1650$$
$13(900+b^2-60b)/10+b^2+60b-2b^2 = 1650$
$1170+1.3b^2-78b+10b^2+600b-20b^2 = 16500$
$$3b^2-180b-4800=0 \implies b^2-60b-1600=0$$

Sum of roots $b_1+b_2 = 60$ (by Vieta's formula: $-\frac{\text{coeff of }b}{\text{coeff of }b^2} = 60$).

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