Cut-off frequency in Fourier-Spectrum of a noise message
I think you consider that pulse in the spectrum as the noise. Then you should remove it by a lowpass filter of cutoff frequency 
$$ f_c = 5 ~\text{kHz.}$$

Slightly less than this should be used for guarding against the transition width of the lowpass filter. So you better select something like $4.8$ kHz for the lowpass filter cutoff frequency.

Note that your sampling frequency $F_s$, probably, is $44.1$ kHz and not $44.8$ kHz. Then your digital frequency for cutoff into the Matlab functions will be: 
$$f_{cd} = \frac{ 2 f_c }{F_s} $$

with the given values of $F_s$ and $F_c$ your digital cutoff will be: 

$$f_{cd} = \frac{ 2 \times 5  \times 10^3 }{44.1 10^3} = 0.2268 $$

or better 

$$f_{cd} = \frac{ 2 \times  4.8 \times 10^3 }{44.1 10^3} = 0.2177 $$