Recover a missing score and compare model shapes

$\bar x=(2+2.4+3+3.8+4.8+7.2+8.2+9.4)/8=5.1$. The regression line passes through $(\bar x,\bar y)$, so $\bar y=1.9263+9.6223(5.1)=51.00003$. Also $\bar y=(378+p)/8$. Hence $p=8(51.00003)-378=30.00024$, giving the integer score $p=30$. The tiny excess is due to the printed rounded coefficients.

Use $p=30$ and plot all eight pairs. The scores rise with study time but the pattern levels off. The raw coefficient is $r_{x,y}=0.9442304387\approx0.944$.

Regress $y$ on $\ln x$: $y=-21.33408676+48.66979691\ln x$. Thus $a=-21.3$, $b=48.7$ (3 s.f.). The transformed correlation is $r_{\ln x,y}=0.9862411668\approx0.986$.

The logarithmic model is preferred: its correlation $0.986$ is closer to 1 than the raw $0.944$, and its decreasing positive gradient matches the scatter pattern. The relevant transformed calculation is in part (iii), despite the printed reference to (i) and (ii).

Using the unrounded logarithmic line at $x=5$, $\hat y=-21.33408676+48.66979691\ln5=56.99692957\approx57.0$. Since 5 lies within the observed study-time range $[2,9.4]$ and the chosen model has $r\approx0.986$, this is a reasonable interpolated estimate for the source exercise, not a guaranteed individual score.