Given that 88% of shoppers spend more than t minutes, this implies that 12% spend less than t minutes. We need to find the z-score corresponding to the 12th percentile of a standard normal distribution.

The z-score for the 12th percentile is approximately -1.175.

Using the standardization formula:

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

Substitute \(z = -1.175\), \(\mu = 96\), and \(\sigma = 18\):

\(-1.175 = \frac{t - 96}{18}\)

Solving for \(t\):

\(t - 96 = -1.175 \times 18\)

\(t - 96 = -21.15\)

\(t = 96 - 21.15\)

\(t = 74.85\)

Therefore, the value of \(t\) is approximately between 74.85 and 74.9.

Given that the times are normally distributed with a mean of 62 seconds and a standard deviation of 5 seconds, we need to find the time t such that 13% of the members take more than t minutes to swim 100 metres.

\(First, convert t minutes to seconds: t minutes = 60t seconds.\)

We know that 13% corresponds to a z-score of approximately 1.127 (from standard normal distribution tables).

Using the z-score formula:

\(\frac{60t - 62}{5} = 1.127\)

Solving for t:

\(60t - 62 = 5 \times 1.127 = 5.635\)

\(60t = 5.635 + 62 = 67.635\)

\(t = \frac{67.635}{60} \approx 1.13\)

We are given that 75% of the time, Pia takes longer than t minutes. This implies that 25% of the time, she takes less than t minutes. We need to find the z-score corresponding to the 25th percentile of a standard normal distribution.

The z-score for the 25th percentile is approximately (-0.674).

Using the standardization formula:

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

Substitute the known values:

\(-0.674 = \frac{t - 10.1}{1.3}\)

Solve for \(t\):

\(t - 10.1 = -0.674 \times 1.3\)

\(t - 10.1 = -0.8762\)

\(t = 10.1 - 0.8762\)

\(t = 9.2238\)

Rounding to two decimal places, \(t = 9.22\).

(a)(i) To find the probability that a randomly chosen member has a height less than 170 cm, we use the standard normal distribution formula:

\(P(X < 170) = P\left(Z < \frac{170 - 166}{10}\right) = P(Z < 0.4)\)

Using standard normal distribution tables, \(P(Z < 0.4) = 0.655\).

(a)(ii) We know that 40% of the members have heights greater than \(h\) cm, so:

\(P\left(Z > \frac{h - 166}{10}\right) = 0.4\)

This implies:

\(\frac{h - 166}{10} = 0.253\)

Solving for \(h\):

\(h = 166 + 10 \times 0.253 = 168.53\)

(b) Given \(\sigma = \frac{2}{3}\mu\), we need to find \(P(X > 0)\):

\(P(X > 0) = P\left(Z > \frac{0 - \mu}{\sigma}\right) = P\left(Z > \frac{0 - \mu}{\frac{2}{3}\mu}\right)\)

\(= P\left(Z > -\frac{3}{2}\right)\)

Using standard normal distribution tables, \(P(Z > -1.5) = 0.933\).

We are given that the time is normally distributed with mean \(\mu = 3.5\) and standard deviation \(\sigma = 0.9\).

We need to find the value of \(t\) such that 90% of the time is greater than \(t\). This implies that 10% of the time is less than \(t\).

Using the standard normal distribution, we find the critical value for 10%, which corresponds to \(z = -1.282\).

We use the standardization formula:

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

Substitute the known values:

\(-1.282 = \frac{t - 3.5}{0.9}\)

Solving for \(t\):

\(t - 3.5 = -1.282 \times 0.9\)

\(t - 3.5 = -1.1538\)

\(t = 3.5 - 1.1538\)

\(t = 2.3462\)

Rounding to two decimal places, \(t = 2.35\).