(i) To find the probability that a randomly chosen person sleeps for less than 8 hours, we standardize the variable:

\(P(X < 8) = P\left( Z < \frac{8 - 7.15}{0.88} \right)\)

\(= P(Z < 0.9659)\)

Using the standard normal distribution table, \(\Phi(0.9659) = 0.833\).

Thus, the probability is 0.833.

(ii) To find the value of \(q\) such that \(P(X < q) = 0.75\), we find the z-value corresponding to 0.75 in the standard normal distribution table, which is approximately 0.674.

Using the standardization formula:

\(\frac{q - 7.15}{0.88} = 0.674\)

Solving for \(q\):

\(q = 0.674 \times 0.88 + 7.15\)

\(q = 7.74\)