(a) To find the probability that the SBP of a randomly chosen adult is less than 132, we use the standard normal distribution. The z-score is calculated as:

\(z = \frac{132 - 125.4}{18.6} = 0.3548\)

The probability \(P(Z < 0.3548)\) is approximately 0.639.

(b) For the SBP of 12-year-old children, we know that 88% have SBP more than 108. This corresponds to a z-score of -1.175 (since 12% have SBP less than 108). Using the formula:

\(\frac{108 - 117}{\sigma} = -1.175\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{108 - 117}{-1.175} = 7.66\)

(c) To find the probability that each of these three adults has SBP within 1.5 standard deviations of the mean, we calculate:

\(P(-1.5 < Z < 1.5) = \Phi(1.5) - \Phi(-1.5) = 2\Phi(1.5) - 1\)

Using \(\Phi(1.5) \approx 0.9332\), we have:

\(2 \times 0.9332 - 1 = 0.8664\)

The probability that each of the three adults has SBP within this range is:

\(0.8664^3 = 0.650\)