= Solution
Both signs are positive, so perform <addition> of the aligned <significands>. The true <exponents> are $-1$ and $3$, a difference of four. Shift the smaller <significand> right four places, retaining its low <bits>:
$$
\frac58=0.000101_2\,2^3,\qquad15\frac12=1.111100_2\,2^3.
$$
Add at the common <exponent>:
$$
\begin{array}{r}
1.111100_2\\
+\;0.000101_2\\\hline
10.000001_2
\end{array}
\qquad\text{so}\qquad10.000001_2\,2^3=1.0000001_2\,2^4.
$$
Renormalization raises the true <exponent> to $4$ and the biased <exponent> to $E=19=10011_2$. There are only seven fraction places to retain, so no <floating-point rounding> changes the result:
$$
\boxed{\frac58+15\frac12=16\frac18,\qquad\text{stored fields }0\;10011\;0000001000.}
$$
The complete word is `0100110000001000`.
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