Given that the probability \(P(X > 195) = 0.128\), we need to find the mean \(\mu\) of the normal distribution.

First, find the z-score corresponding to the probability 0.128. From standard normal distribution tables, \(P(Z > 1.136) = 0.128\), so \(z = 1.136\).

Using the standardization formula:

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

Substitute the known values:

\(1.136 = \frac{195 - \mu}{22}\)

Solving for \(\mu\):

\(1.136 \times 22 = 195 - \mu\)

\(24.992 = 195 - \mu\)

\(\mu = 195 - 24.992\)

\(\mu = 170.008\)

Rounding to the nearest whole number, \(\mu = 170\).

(a)(i) To find the probability that Zak takes less than 30 minutes, we standardize the variable:

\(z = \frac{30 - 35.2}{4.7} = -1.106\)

Using the standard normal distribution table, \(P(z < -1.106) = 0.1345\).

The expected number of days in a year is \(0.1345 \times 52 = 6.99\), which rounds to 7 days.

(a)(ii) We know \(\Phi(t) = 0.648\) and \(z = 0.380\).

Standardizing, \(\frac{t - 35.2}{4.7} = 0.380\).

Solving for \(t\), we get \(t = 37.0\).

(b) We have two equations from the given probabilities:

\(\frac{7 - \mu}{\sigma} = -0.8\) and \(\frac{10 - \mu}{\sigma} = 0.44\).

Solving these equations simultaneously, we find \(\mu = 8.94\) and \(\sigma = 2.42\).

Let the random variable \(X\) represent the weights of the bags of sugar. We know \(X\) is normally distributed with mean \(\mu = 1.04\) kg and standard deviation \(\sigma = 0.06\) kg.

We are given that \(P(X < w) = 0.81\). To find \(w\), we standardize \(X\) to a standard normal variable \(Z\):

\(P\left(Z < \frac{w - 1.04}{0.06}\right) = 0.81\)

Using the standard normal distribution table, we find that the z-score corresponding to 0.81 is approximately 0.878.

Thus, we have:

\(\frac{w - 1.04}{0.06} = 0.878\)

Solving for \(w\):

\(w - 1.04 = 0.878 \times 0.06\)

\(w - 1.04 = 0.05268\)

\(w = 1.04 + 0.05268\)

\(w = 1.09268\)

Therefore, \(1.09 \leq w \leq 1.093\).

Given that the probability \(P(3.2 < X < \mu) = 0.475\), we can find the probability \(P(X < 3.2) = 0.5 - 0.475 = 0.025\).

Using the standard normal distribution table, \(P(Z < z) = 0.025\) corresponds to \(z = -1.96\).

Standardizing, we have:

\(z = \frac{3.2 - \mu}{0.714} = -1.96\)

Solving for \(\mu\):

\(3.2 - \mu = -1.96 \times 0.714\)

\(\mu = 3.2 + 1.96 \times 0.714\)

\(\mu = 4.6\)

(i) To find the proportion of gem stones in each category, we standardize the weights using the formula:

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

For Small (\(X < 1.2\)): \(z = \frac{1.2 - 1.9}{0.55} = -1.2727\)

\(P(X < 1.2) = P(z < -1.2727) = 0.1014\)

For Large (\(X > 2.5\)): \(z = \frac{2.5 - 1.9}{0.55} = 1.0909\)

\(P(X > 2.5) = P(z > 1.0909) = 0.138\)

For Medium (\(1.2 < X < 2.5\)): \(P(1.2 < X < 2.5) = 1 - P(X < 1.2) - P(X > 2.5) = 1 - 0.101 - 0.138 = 0.761\)

(ii) We need \(P(k < X < 2.5) = 0.8\). We know \(P(X < 2.5) = 0.8623\), so \(P(X > k) = 0.9377\).

Find \(z\) such that \(P(z) = 0.9377\), which gives \(z = -1.536\).

Using the standardization formula:

\(-1.536 = \frac{k - 1.9}{0.55}\)

Solving for \(k\):

\(k = 1.9 + (-1.536 \times 0.55) = 1.06\)