Let the number of small bands be a random variable, denoted by \(X\). Given that 60% of the bands are small, \(X\) follows a binomial distribution \(B(n=150, p=0.6)\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 150 \times 0.6 = 90\)

\(\sigma^2 = np(1-p) = 150 \times 0.6 \times 0.4 = 36\)

Thus, the standard deviation \(\sigma\) is \(\sqrt{36} = 6\).

We use a normal approximation to find \(P(88 < X < 97)\). Applying continuity correction, this becomes \(P(87.5 < X < 97.5)\).

Standardizing, we calculate:

\(P(87.5 < X < 97.5) = \Phi\left(\frac{97.5 - 90}{6}\right) - \Phi\left(\frac{87.5 - 90}{6}\right)\)

\(= \Phi(1.25) - \Phi(-0.4166)\)

Using standard normal distribution tables, \(\Phi(1.25) = 0.8944\) and \(\Phi(-0.4166) = 1 - 0.6616 = 0.3384\).

Thus, the probability is:

\(0.8944 - 0.3384 = 0.556\)