Let $X$ be the annual rainfall: $X \sim N(\mu = 1200, \sigma = 200)$.
Convert $X = 1400$ to the standard normal variate $Z$:
$$Z = \frac{X - \mu}{\sigma} = \frac{1400 - 1200}{200} = \frac{200}{200} = 1.0$$We need $P(X \gt 1400) = P(Z \gt 1)$:
$$P(Z \gt 1) = P(Z \gt 0) - P(0 \lt Z \lt 1) = 0.50 - 0.34 = 0.16$$मानक प्रसामान्य चर $Z$ में रूपांतरण करने पर:
$$Z = \frac{X - \mu}{\sigma} = \frac{1400 - 1200}{200} = 1.0$$वर्षा 1400 मिमी से अधिक होने की प्रायिकता:
$$P(X \gt 1400) = P(Z \gt 1) = 0.50 - P(0 \lt Z \lt 1) = 0.50 - 0.34 = 0.16$$