The number of votes Anil received follows a binomial distribution with parameters \(n = 120\) and \(p = 0.4\).

We approximate this binomial distribution with a normal distribution with mean \(\mu = np = 120 \times 0.4 = 48\) and variance \(\sigma^2 = np(1-p) = 120 \times 0.4 \times 0.6 = 28.8\).

Using the normal approximation, we apply a continuity correction. We want to find \(P(36 \leq X \leq 54)\), which becomes \(P(35.5 < X < 54.5)\).

Standardizing, we calculate:

\(P\left( \frac{35.5 - 48}{\sqrt{28.8}} < Z < \frac{54.5 - 48}{\sqrt{28.8}} \right)\)

\(P\left( \frac{-12.5}{\sqrt{28.8}} < Z < \frac{6.5}{\sqrt{28.8}} \right)\)

\(P(-2.3292 < Z < 1.211)\)

Using standard normal distribution tables, we find:

\(P(Z < 1.211) = 0.8871\) and \(P(Z < -2.3292) = 0.0099\).

Thus, \(P(-2.3292 < Z < 1.211) = 0.8871 - 0.0099 = 0.8772\).

Therefore, the probability is approximately \(0.877\).