(i) Any sensible set of data, such as heights, weights, or times, can be modeled by a normal distribution.

(ii) We know \(P(X > 10.0) = 0.7389\). This corresponds to a standard normal variable \(z\) such that \(P(Z > z) = 0.7389\), which gives \(z = 0.64\). Using the standardization formula:

\(z = \frac{\mu - 10}{\sqrt{21}}\)

Substitute \(z = 0.64\):

\(0.64 = \frac{\mu - 10}{\sqrt{21}}\)

Solving for \(\mu\):

\(\mu - 10 = 0.64 \times \sqrt{21}\)

\(\mu = 10 + 0.64 \times \sqrt{21}\)

\(\mu = 12.9\)

(iii) To find how many observations are greater than 22.0, we first find the \(z\)-score for 22:

\(z = \frac{22 - 12.9}{\sqrt{21}} = 1.986\)

Using the standard normal distribution, \(P(X > 22) = 1 - \Phi(1.986) \approx 0.0235\).

For 300 observations, the expected number greater than 22 is:

\(300 \times 0.0235 = 7.05\)

Rounding to the nearest integer, we estimate 7 observations.