Comparing the given pdf with the standard normal density form $f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$:
Mean $\mu = 1$
Variance $\sigma^2 = 1 \implies \sigma = 1$

The normal distribution curve is perfectly symmetrical about its mean $\mu = 1$. Therefore, the total area under the probability density curve to the right of the mean is exactly half of the total area (which is 1):

Let $X$ be the annual rainfall: $X \sim N(\mu = 1200, \sigma = 200)$.

Convert $X = 1400$ to the standard normal variate $Z$:

We need $P(X \gt 1400) = P(Z \gt 1)$:

Analysis of dispersion relationships in normal distribution:
For a normal distribution $N(m, v^2)$, where the standard deviation is $\sigma = v$:

• Correct: Mean Deviation about the mean is given by:
  $$\text{MD} = \sqrt{\frac{2}{\pi}} \cdot \sigma \approx 0.7979 v \approx \frac{4}{5} v$$

Mathematical derivation of original score from Z-score:
Given $Z = 2.5$, $\mu = 45$, and $\sigma = 3$.

• Incorrect: 42 is obtained from $45 - 3$, which is $\mu - \sigma$.
• Correct: Using the definition of standard normal variate:
  $$Z = \frac{X - \mu}{\sigma}$$
  $$2.5 = \frac{X - 45}{3}$$
  $$X - 45 = 2.5 \times 3 = 7.5$$
  $$X = 45 + 7.5 = 52.5$$
• Incorrect: 47 is obtained if $Z = 2/3$.
• Incorrect: 3 is the standard deviation.