Fermi distribution of 3s electrons of sodium

[N8]

Homework Statement

What fraction of the 3s-electrons of sodium is found within an energy k_b*T below the Fermi level? (Take room temperature at T= 300k) Fermi level for Na is 3.2 eV

Homework Equations

F(E) = 1 / {exp((E-E_f)/k_b*T) + 1}

The Attempt at a Solution

From what I can tell, do not need to know room temp, or even sodium's fermi level. Simply do (E=Ef-k_b*T) solve through and obtain 1/((e^-1)+1) which ends up being approximately 0.73.

However, book states the answer is 0.88. Why is this? I do not understand where I am going wrong. Any help would be much appreciated.

 

[TeethWhitener]

The question is very vaguely worded. I thought at first it meant to integrate the Fermi-Dirac distribution between 3.2-0.025 eV and 3.2+0.025 eV (fraction of electrons within $kT$ of the Fermi energy), but the answer is much too small. I happened to stumble on the fact that
$$\frac{1}{e^{-2}+1}=0.88$$
In other words, the answer they’re asking for is the average number of electrons with an energy of $E_F-2kT$. I’m not sure if there’s another way to get at the desired answer, or another way to interpret the question. I admit, I’m very confused by it.