= Solution
The smallest positive normal <half-precision floating-point number> uses $E=1$ and a zero fraction, while the largest uses $E=30$ and all ten fraction <bits> equal to one. Hence
$$
\boxed{x_{\min,\mathrm{normal}}=1.0_2\,2^{1-15}=2^{-14},\qquad x_{\max,\mathrm{normal}}=(2-2^{-10})2^{15}=65504.}
$$
The corresponding words are `0 00001 0000000000` and `0 11110 1111111111`. The stored <exponent> $31$ is reserved and cannot extend the finite normal range.
For a <subnormal number>, $E=0$ and the implicit leading <bit> becomes zero, while the scale remains $2^{1-15}=2^{-14}$. The least positive nonzero fraction is $0.0000000001_2=2^{-10}$, giving
$$
\boxed{x_{\min,\mathrm{subnormal}}=2^{-10}2^{-14}=2^{-24},\qquad\text{word }0\;00000\;0000000001.}
$$
Using scale $2^{-15}$ here would be incorrect: the <subnormal number> convention uses the minimum normal scale, not the ordinary biased-<exponent> formula at $E=0$.
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