$b_{XY}=r\sigma_X/\sigma_Y = 0.5$; regression line $X-\bar X = b_{XY}(Y-\bar Y)$ gives $X = 0.5Y + 40$, so at $Y=100$, $X=90$.

Given: $\bar X = 85$, $\bar Y = 90$, $\sigma_X = 5$, $\sigma_Y = 6$, $r = 0.6$.

Step 1 — Regression coefficient of $X$ on $Y$.

$b_{XY} = r\cdot\frac{\sigma_X}{\sigma_Y} = 0.6\times \frac{5}{6} = 0.5.$

Step 2 — Equation of the line of $X$ on $Y$.

$X - \bar X = b_{XY}\,(Y - \bar Y)$

$X - 85 = 0.5\,(Y - 90)$

$X - 85 = 0.5Y - 45$

$X = 0.5Y + 40.$

Step 3 — Estimate production when demand $Y = 100$.

$X = 0.5\times 100 + 40 = 50 + 40 = 90.$

Fitting $Y$ on $X$ by least squares gives $b_{YX} = 2$ and intercept $3$, so $Y = 2X + 3$.

We need the regression line of $Y$ on $X$: $Y - \bar{y} = b_{YX}(X - \bar{x})$.

With $n = 5$ and the data $X: 1,2,3,4,5$ and $Y: 5,7,9,11,13$:

$\sum X = 15,\quad \sum Y = 45,\quad \sum XY = 5+14+27+44+65 = 155,\quad \sum X^2 = 55.$

Means: $\bar{x} = \dfrac{15}{5} = 3$, $\bar{y} = \dfrac{45}{5} = 9$.

Slope:

$b_{YX} = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2 - (\sum X)^2} = \frac{5(155) - 15(45)}{5(55) - 15^2} = \frac{775 - 675}{275 - 225} = \frac{100}{50} = 2.$

Intercept from $\bar{y} = a + b_{YX}\bar{x}$: $a = 9 - 2(3) = 3$.

The regression coefficient $b_{yx}=3.75$, so the line of $Y$ on $X$ is $y = 3.75x - 39$.

The regression coefficient of $Y$ on $X$ is

$b_{yx} = \dfrac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^{2}}$

Since $\sigma_x^{2}=\dfrac{\sum(x_i-\bar{x})^{2}}{n}$, we have $\sum(x_i-\bar{x})^{2}=n\sigma_x^{2}=8\times 2^{2}=32$.

Therefore

$b_{yx} = \dfrac{120}{32} = 3.75$

The line of regression of $Y$ on $X$ passes through $(\bar{x},\bar{y})=(20,36)$:

$y-\bar{y} = b_{yx}(x-\bar{x})$

$y-36 = 3.75(x-20)$

$y-36 = 3.75x - 75$

$y = 3.75x - 39$