Sketch possible correlations and model monthly earnings

One authored construction uses $x=1,2,3,4,5,6,7,8$ and $y=1,2,3,4,4,3,2,1$. The symmetric cross-products cancel, giving zero linear correlation. The arch still has a nonlinear pattern, so this does not assert no relationship or statistical independence. Other possible eight-point sketches are valid; these are not the source’s unprovided measured values.

The second panel is an independently constructed eight-point example whose coefficient was checked to be exactly −0.8. It has visible spread about a falling trend, rather than a perfect straight line. These illustrative coordinates are not claimed to be printed data; the source supplies only the target coefficient and point count.

Keep age on the horizontal axis and monthly earnings in thousand dollars vertically. The third panel retains every pair and adds the full-precision regression line requested in (iii). The two illustrative eight-point panels in (a) are separate datasets and must not be substituted for these seven observations.

Using all seven age/earnings pairs gives the independently checked coefficient $0.9872953317\ldots$, matching the printed $0.987$. It summarises the strong positive linear pattern, not a universal age–income law.

The full fit is $y=0.10693414634146342x+0.5615170731707311$. Rounding the gradient and intercept to the requested 4 decimal places gives the printed equation. Keep full coefficients for the later prediction; the response remains in thousands of dollars.

At age 40, the full fit gives $y=4.8388829268\ldots$ thousand dollars, or 4838.8829 dollars. This rounds to the printed 4,840 dollars. Age 40 is within 18–55 and the scatter is approximately linear with high positive correlation. These support the model for a comparable worker, but do not guarantee an individual salary; the source asks for expected reliability, not certainty.

Adding $N$ thousand dollars to every earnings value translates the entire scatter vertically. Relative vertical deviations and the fitted gradient are unchanged; each fitted response rises by $N$, so only the intercept changes. No new regression calculation is needed. Keep $N$ in the same thousand-dollar units as $y$.