= Solution
For a normal <half-precision floating-point number>, write the stored fields as sign $s$, biased <exponent> $E$ and ten-<bit> fraction $f$. The value is $(-1)^s(1.f)_2\,2^{E-15}$ for $1\leq E\leq30$. The leading one is implicit: the ten stored fraction <bits> give eleven significant <binary digits> including that one. The normalized <binary representations> and stored fields are
|| Value
|| Normalized <binary representation>
|| Sign, biased <exponent>, fraction
| $1/2$
| $1.0_2\,2^{-1}$
| `0 01110 0000000000`
| $5/8$
| $1.01_2\,2^{-1}$
| `0 01110 0100000000`
| $-3/16$
| $-1.1_2\,2^{-3}$
| `1 01100 1000000000`
| $15\frac12$
| $1.1111_2\,2^3$
| `0 10010 1111000000`
Thus the four 16-<bit> words are
$$
\boxed{0011100000000000,\quad0011100100000000,\quad1011001000000000,\quad0100101111000000.}
$$
Every conversion is exact: all four numbers have terminating <binary representations> within the available precision.
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