(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).