= Solution
For $B=5$ T, the same <Spin-only magnetic moment>[spin-only] two-state probabilities apply, but the low-temperature argument is no longer small. At 300 K, $x=0.0111952$; at 2 K, $x=1.67928$. Hence
$$
\boxed{\begin{array}{c|cc}
T&\text{aligned population }p_+&\text{net polarization }M/M_s\\\hline
300\ \mathrm K&0.50559738&0.0111948\\
2\ \mathrm K&0.96638432&0.932769
\end{array}}
$$
The 2 K sample is about $93.3\%$ of its <Spin-one-half paramagnetic magnetization>[saturation magnetization], with about $96.6\%$ of the moments in the aligned state. The two percentages describe different observables. A linear extrapolation would predict $M/M_s\simeq x=1.679$, exceeding saturation and therefore contradicting the bounded two-state population.
For completeness, at this 2 K, 5 T bias the dimensionless secant <Magnetic susceptibility>[susceptibility] is approximately $\chi_0\tanh x/x=0.01199$, while the <Differential magnetic susceptibility>[differential susceptibility] is $\chi_0\operatorname{sech}^2x=0.002805$ under the same prescribed-field approximation. Neither equals the zero-field <Curie's law>[Curie slope] $0.02159$. Thus saturation reduces the incremental response even while the <magnetization> itself is large.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2016-spin-half-alignment.png]
{title=Parallel-state population and net spin polarization}
{description=Original independent-spin calculation. At zero field the parallel-state population is one half and net polarization is zero; the four specified field and temperature combinations are marked.}
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