Let the random variable representing the number of people with blood group A+ be denoted by \(X\). Given that 35% of the population are group A+, we have \(p = 0.35\).

The sample size is \(n = 150\). The mean \(\mu\) and variance \(\sigma^2\) of \(X\) are given by:

\(\mu = n \times p = 150 \times 0.35 = 52.5\)

\(\sigma^2 = n \times p \times (1-p) = 150 \times 0.35 \times 0.65 = 34.125\)

We use the normal approximation to the binomial distribution. Applying continuity correction, we find:

\(P(X > 60) = P\left( Z > \frac{60.5 - 52.5}{\sqrt{34.125}} \right)\)

Calculating the standard score:

\(Z = \frac{60.5 - 52.5}{\sqrt{34.125}} = 1.369\)

Using standard normal distribution tables, we find:

\(P(Z > 1.369) = 1 - \Phi(1.369) = 0.0854 \text{ or } 0.0855\)