Let the random variable \(X\) represent the number of students who play exactly one musical instrument in a sample of 90 students. Given that 52% of students play exactly one instrument, \(X\) follows a binomial distribution \(B(n=90, p=0.52)\).

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

\(\mu = np = 90 \times 0.52 = 46.8\)

\(\sigma^2 = np(1-p) = 90 \times 0.52 \times 0.48 = 22.464\)

We use a continuity correction to find \(P(X < 40)\), which is approximated by \(P(X < 39.5)\) in the normal distribution.

Standardizing, we find the z-score:

\(z = \frac{39.5 - 46.8}{\sqrt{22.464}} = \frac{-7.3}{4.74} \approx -1.540\)

Using standard normal distribution tables, we find:

\(P(Z < -1.540) = 1 - 0.9382 = 0.0618\)

Thus, the probability that fewer than 40 students play exactly one musical instrument is approximately 0.0618.