(i) To find the value of t, we use the standard normal distribution. We know that 90% of calls take longer than t, so 10% take less time. This corresponds to a z-score of -1.282 (since the left tail is 0.10).

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

\(-1.282 = \frac{t - 6.5}{1.76}\)

Solving for t gives:

\(t = 6.5 + (-1.282 \times 1.76) = 4.24\) minutes

(ii) We need to find the probability that more than 7 out of 9 calls are within 1 standard deviation of the mean. The probability of a call being within 1 standard deviation is given by:

\(P(\text{within 1 sd of mean}) = 2\Phi(1) - 1 = 0.6826\)

We use the binomial distribution with \(n = 9\) and \(p = 0.6826\). We need \(P(X > 7) = P(X = 8) + P(X = 9)\).

\(P(X = 8) = \binom{9}{8} (0.6826)^8 (0.3174)^1\)

\(P(X = 9) = \binom{9}{9} (0.6826)^9\)

Calculating these gives:

\(P(X = 8) = 9 \times 0.6826^8 \times 0.3174\)

\(P(X = 9) = 0.6826^9\)

Adding these probabilities gives \(P(X > 7) = 0.167\).