Solution:
$
n=\sum n_x=1000\\
$
$
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline \text { Gray lenel } & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline n_k & 0 & 100 & 400 & 50 & 200 & 50 & 200 & 0 \\
\hline \hline P_f=\frac{n_K}{n} & 0 & 0.1 & 0.4 & 0.05 & 0.2 & 0.05 & 0.2 & 0 \\
\hline \text { CDF }\left(S_k\right) & 0 & 0.1 & 0.5 & 0.55 & 0.75 & 0.8 & 1 & 1 \\
\hline(L-1) S_K & 0 & 0.7 & 3.5 & 3.85 & 5.25 & 5.6 & 7 & 7 \\
\hline \text { Rounding } & 0 & 1 & 4 & 4 & 5 & 6 & 7 & 7 \\
\hline & & & & & & & & \\
\hline
\end{array}
$
$
\begin{array}{|c|c|c|}
\hline \text { old gray level } & \text { No. of pixel } & \text { New gray level. } \\
\hline 0 & 0 & 0 \\
\hline 1 & 100 & 1 \\
\hline 2 & 400 & 4 \\
\hline 3 & 50 & 4 \\
\hline 4 & 200 & 5 \\
\hline 5 & 50 & 6 \\
\hline 6 & 200 & 7 \\
\hline 7 & 0 & 7 \\
\hline
\end{array}
$
$
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline \text { New graylerel } & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline \text { eq. No. of pixels } & 0 & 100 & 0 & 0 & 450 & 200 & 50 & 200 \\
\hline
\end{array}
$
In original histogram the pixels are evenly distributed so contrast is less.
Whereas in equalized image histogram the maximum pixels are on the higher side.