= Solution
For the fully gas-filled reservoir,
$$
Z_{p2}\simeq1.44\times10^6,\quad Z_{s2}\simeq0.81\times10^6\quad\mathrm{kg\,m^{-2}\,s^{-1}},
$$
so
$$
\boxed{R_p\simeq\frac{1.4375-2.25}{1.4375+2.25}\simeq-0.22,\qquad R_s\simeq\frac{0.8125-0.900}{0.8125+0.900}\simeq-0.05.}
$$
Both impedances are now smaller than those of the cap rock. Both reflections reverse <seismic reflection polarity>, with a substantially larger amplitude for the <P wave> reflection. Reading the right-hand $V_p$ endpoint instead as $1.20\ \mathrm{km\,s^{-1}}$ would give $R_p\simeq-0.20$, preserving the interpretation.
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