Question

The probability density function of evaporation E on any day during a year in a watershed is given by \(f\left( E \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{5}\;0 \le E \le 5\;mm/day}\\ {0\;otherwise} \end{array}} \right.\)

The probability that E lies in between 2 and 4 mm/day in a day in the watershed is (in decimal up to one decimal places) _____

Answer 1

Concept:

Let f(x) be any probability density function where ‘x’ is random variable.

The probability that x lies between x₁ and x­₂ is given by,

\(P\left( {{x_1} \le x \le {x_2}} \right) = \mathop \smallint \limits_{{x_1}}^{{x_2}} f\left( x \right)dx\)

In this case, x is evaporation and probability density f (x).

\(f\left( E \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{2}\;0 \le E \le 5\;mm/day}\\ {0\;;\;otherwise} \end{array}} \right.\)

Calculation:

The probability E lies between 2 and 4 mm/day is P(2 ≤ E ≤ 4)

\(P\left( {2 \le E \le 4} \right) = \mathop \smallint \limits_2^4 f\left( E \right)dE = \;\mathop \smallint \limits_2^4 \left( {\frac{1}{5}} \right)dE = \left. {\frac{1}{5}} \right|_2^4 = \frac{2}{5} = 0.4\)