(i) To find the probability that a pencil has a length greater than 10.9 cm, we standardize the normal distribution:

\(P(x > 10.9) = P\left(z > \frac{10.9 - 11}{0.095}\right)\)

\(= P(z > -1.0526)\)

Using the standard normal distribution table, \(P(z > -1.0526) = 0.8538\), which rounds to 0.854.

(ii) To find the probability that at least two pencils in a sample of 6 have lengths less than 10.9 cm, we first find the probability of a single pencil being less than 10.9 cm:

\(P(x < 10.9) = 1 - P(x > 10.9) = 1 - 0.8538 = 0.1462\)

We use the binomial distribution for 6 trials with probability \(p = 0.1462\):

\(P(\text{at least 2} < 10.9) = 1 - P(0, 1)\)

\(= 1 - \left(0.8538^6 + 6 \cdot 0.1462 \cdot 0.8538^5\right)\)

\(= 1 - (0.8538^6 + 6 \cdot 0.1462 \cdot 0.8538^5)\)

\(= 0.215\)

Let \(X\) be the daily minimum temperature in January in Ottawa. We have \(X \sim N(-15.1, 62.0)\).

We need to find \(P(X > 0)\).

Standardize the variable:

\(Z = \frac{X - (-15.1)}{\sqrt{62}}\)

For \(X = 0\),

\(Z = \frac{0 - (-15.1)}{\sqrt{62}} = \frac{15.1}{\sqrt{62}} \approx 1.918\)

Using the standard normal distribution table, find \(\Phi(1.918) \approx 0.9724\).

Thus, \(P(X > 0) = 1 - \Phi(1.918) = 1 - 0.9724 = 0.0276\).

First, standardize the values 1.82 and 1.92 using the formula for the z-score:

\(z_1 = \frac{1.92 - 1.9}{0.15} = 0.1333\)

\(z_2 = \frac{1.82 - 1.9}{0.15} = -0.5333\)

Next, find the probability that a single tyre has a pressure between 1.82 and 1.92 bars:

\(\text{area} = \Phi(0.1333) - \Phi(-0.5333)\)

Using standard normal distribution tables or a calculator:

\(\Phi(0.1333) = 0.5529\)

\(\Phi(-0.5333) = 1 - \Phi(0.5333) = 1 - 0.7029 = 0.2971\)

Thus, the probability for one tyre is:

\(0.5529 - 0.2971 = 0.256\)

Since the tyres are independent, the probability that all four tyres have pressures between 1.82 and 1.92 bars is:

\((0.256)^4 = 0.0043\)

(i) To find the proportion of melons classified as small, we calculate the z-score for 350 grams using the formula:

\(z = \frac{350 - 450}{120} = -0.833\)

The proportion of melons weighing less than 350 grams is given by the cumulative distribution function (CDF) for a standard normal distribution at \(z = -0.833\). From standard normal distribution tables, \(\Phi(-0.833) \approx 0.2025\). Therefore, the proportion of melons classified as small is 20.25%.

(ii) The remaining melons are divided equally between medium and large. The proportion of melons not classified as small is \(1 - 0.2025 = 0.7975\). Half of these are classified as large, so the proportion classified as large is \(0.7975 / 2 = 0.39875\).

We need to find the z-score corresponding to a cumulative probability of \(1 - 0.39875 = 0.60125\). From standard normal distribution tables, \(z \approx 0.257\).

Convert this z-score back to the original scale using:

\(x = 120 \times 0.257 + 450 = 481\)

Thus, the weight above which melons are classified as large is 481 grams.

(i) To find the proportion of people who spend between 7.8 days and 11.0 days in hospital, we use the standard normal distribution. First, calculate the z-scores:

\(z = \frac{11.0 - 7.8}{2.8} = 1.143\)

The probability that a person spends less than 11.0 days is given by:

\(P(T < 11.0) = \Phi(1.143) = 0.8735\)

Since 7.8 days is the mean, the probability of spending less than 7.8 days is 0.5. Therefore, the proportion of people who spend between 7.8 and 11.0 days is:

\(P(7.8 < T < 11.0) = 0.8735 - 0.5 = 0.3735\)

(ii) To find the probability that exactly 2 out of 3 people spend longer than 11.0 days, use the binomial distribution. The probability of one person spending more than 11.0 days is:

\(P(T > 11.0) = 1 - 0.8735 = 0.1265\)

The probability of exactly 2 out of 3 people spending more than 11.0 days is:

\((0.1265)^2 \times (0.8735) \times \binom{3}{2} = 0.0419\)

(iii) The box-and-whisker plot is not symmetric, indicating that the data is not normally distributed. Therefore, it does not agree with the model used in parts (i) and (ii).