$($II$)$ On an audio compact disc $($CD$),$ digital bits of information are encoded sequentially along a spiral path. Each bit occupies about $0\mathrm{.}28m.$ A CD player's readout laser scans along the spiral's sequence of bits at a constant speed of about $1\mathrm{.}2\frac{m}{s}$ as the CD spins. $($a$)$ Determine the number $N$ of digital bits that a CD player reads every second $\frac{?}{(b)}$ The audio information is sent to each of the two loudspeakers $44,100$ times per second. Each of these samplings requires $16$ bits and so one would $($at first glance$)$ think the required bit rate for a $CD$ player is

$N_0=2(44,100\frac{\text{s}\text{a}\text{m}\text{p}\text{l}\text{i}\text{n}\text{g}\text{s}}{second})(16\frac{\text{b}\text{i}\text{t}\text{s}}{\text{s}\text{a}\text{m}\text{p}\text{l}\text{i}\text{n}\text{g}})=1\mathrm{.}4{10}^6\frac{\text{b}\text{i}\text{t}\text{s}}{second}$

where the $2$ is for the $2$ loudspeakers $($the $2$ stereo channels$).$ Note that $N_0$ is less than the number $N$ of bits actually read per second by a $CD$ player. The excess number of bits $)=(N-N_0$ is needed for encoding and error$-$correction. What percentage of the bits on a $CD$ are dedicated to encoding and error$-$correction?