Let the random variable \(X\) represent the number of students who do not like any sports. \(X\) follows a binomial distribution with parameters \(n = 90\) and \(p = 0.3\).

To approximate, we use a normal distribution with mean \(\mu = np = 0.3 \times 90 = 27\) and variance \(\sigma^2 = np(1-p) = 0.3 \times 90 \times 0.7 = 18.9\).

Thus, \(X \sim N(27, 18.9)\).

We want \(P(X < 32)\). Using continuity correction, this is approximated by \(P(X < 31.5)\).

Standardize: \(z = \frac{31.5 - 27}{\sqrt{18.9}} = \frac{4.5}{4.347} \approx 1.035\).

Using the standard normal distribution table, \(P(Z < 1.035) = \Phi(1.035) \approx 0.850\).