(i) To find the probability that a randomly chosen packet weighs less than 1 kg, we standardize the variable:

\(P(X < 1) = P\left( Z < \frac{1 - 1.04}{0.017} \right) = P(Z < -2.353)\)

Using the standard normal distribution table, \(P(Z < -2.353) = 0.0093\).

(ii) The expected number of packets from 1000 kg is calculated by dividing the total weight by the mean weight per packet:

\(\frac{1000}{1.04} \approx 961 \text{ or } 962\)

(iii) Given that the probability of a packet weighing less than 1 kg is 0.0388, we find \(\mu\) by setting up the equation:

\(P(Z < -1.765) = 0.0388\)

Standardizing gives:

\(-1.765 = \frac{1 - \mu}{0.017}\)

Solving for \(\mu\):

\(1 - \mu = -1.765 \times 0.017\)

\(\mu = 1.03\)

(iv) With the new mean, the expected number of packets is:

\(\frac{1000}{1.03} \approx 971 \text{ or } 970\)

Let the mean be \(\mu\) and the standard deviation be \(\sigma\).

For the time less than 36 minutes, 23% corresponds to a \(z\)-score of approximately \(-0.739\). Thus,

\(z_1 = \frac{36 - \mu}{\sigma} = -0.739\)

For the time greater than 54 minutes, 10% corresponds to a \(z\)-score of approximately \(1.282\). Thus,

\(z_2 = \frac{54 - \mu}{\sigma} = 1.282\)

We have two equations:

1. \(\frac{36 - \mu}{\sigma} = -0.739\)
2. \(\frac{54 - \mu}{\sigma} = 1.282\)

Solving these equations simultaneously, we find:

\(\mu = 42.6\)

\(\sigma = 8.91\)

Given that X ~ N(20, 49), we have a normal distribution with mean bc = 20 and variance 3c2 = 49, so the standard deviation 3c = 7.

\(We need to find k such that P(X > k) = 0.25. This implies that P(X 3c k) = 0.75.\)

Using the standard normal distribution table, we find the z-score corresponding to 0.75 is approximately 0.674.

Standardizing, we have:

\(3cbr>z = 3cfrac{k - 20}{7}\)

Substituting the z-score, we get:

\(0.674 = 3cfrac{k - 20}{7}\)

Solving for k:

\(k - 20 = 0.674 3c 7\)

\(k = 0.674 3c 7 + 20\)

\(k = 4.718 + 20\)

\(k = 24.7\)

(i) We know that 15.5% of desks have a height greater than 70 cm. This corresponds to a z-score of approximately 1.015 (from standard normal distribution tables).

Using the standardization formula:

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

Substitute the known values:

\(1.015 = \frac{70 - 69}{\sigma}\)

Solve for \(\sigma\):

\(\sigma = \frac{1}{1.015} = 0.985\)

(ii) Jodu's knees are at 58 cm, and they need to be at least 9 cm below the desk top, so the desk height must be greater than 67 cm (58 + 9).

Standardize 67 cm:

\(P(>67) = P\left( z > \frac{67 - 69}{0.985} \right)\)

\(= P(z > -2.03)\)

From standard normal distribution tables, \(P(z > -2.03) = 0.9788\).

Estimate the number of comfortable desks:

\(300 \times 0.9788 = 293.64\)

Therefore, approximately 293 desks are comfortable for Jodu.

We are given that the time is normally distributed with mean \(\mu = 9.5\) and standard deviation \(\sigma = 1.3\). We need to find the value of \(t\) such that 90% of the time, Peter takes longer than \(t\) minutes. This means that 10% of the time, he takes less than \(t\) minutes.

Using the standard normal distribution table, we find the z-score corresponding to the 10th percentile, which is \(z = -1.282\).

We use the standardization formula:

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

Substituting the known values:

\(-1.282 = \frac{t - 9.5}{1.3}\)

Solving for \(t\):

\(t - 9.5 = -1.282 \times 1.3\)

\(t - 9.5 = -1.6666\)

\(t = 9.5 - 1.6666\)

\(t = 7.8334\)

Rounding to two decimal places, \(t = 7.83\).