91Ó°ÊÓ

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

It is given that \({\rm{X}}\) is a random variable which denote the rainfall duration, and \({\rm{X}}\) has an exponential distribution with mean value \({\rm{2}}{\rm{.725}}\) hours. Since we know that for an exponential distribution:

\(\begin{array}{l}{\rm{\lambda = }}\frac{{\rm{1}}}{{\rm{\mu }}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{2}}{\rm{.725}}}}\\{\rm{\lambda = 0}}{\rm{.367}}\end{array}\)

Then pdf of exponential distribution is given as:

\({\text{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\text{(0}}{\text{.367)x}}{{\text{e}}^{{\text{ - (0}}{\text{.367)x}}}}}&{{\text{x0}}} \\ {\text{0}}&{{\text{ otherwise }}}\end{array}} \right.\)

And cof of exponential distribution is given as

\({\text{F(x) = }}\left\{ {\begin{array}{*{20}{l}}{\text{0}}&{{\text{x < 0}}} \\ {{\text{1 - }}{{\text{e}}^{{\text{ - (0}}{\text{.367)x}}}}}&{{\text{x0}}} \end{array}} \right.\)

(a) The probability that the duration of a particular rainfall event at this location is at least \({\rm{2}}\) hours is denoted as

\(\begin{array}{l}P(X \ge 2) = 1 - F(2)\\ = 1 - \left( {1 - {e^{ - (0.367) \cdot 2}}} \right)\\ = 1 - 0.52P(X \ge 2)\\ = 0.48\end{array}\)

The probability that the distance is at most \({\rm{3}}\) hours is denoted as \({\rm{P(X£3)}}\)

\(\begin{aligned}P(X£3) &= F(3) = 1 - {{\rm{e}}^{{\rm{ - (0}}{\rm{.367) \times 3}}}}{\rm{P(X£3)}}\\&= 0{\rm{.6674}}\end{aligned}\)

The probability that the distance is in between \({\rm{2}}\) and \({\rm{3}}\) hours is denoted as \({\rm{P(2£X£3)}}\)

\(\begin{aligned} P(2£X£3) &= F(3) - F(2) \\ &= 0 {\rm{.6674 - 0}}{\rm{.52P(2£X£3)}}\\ &= 0 {\rm{.1474}}\end{aligned}\)

Proposition: Let \({\rm{X}}\)be a continuous rv with pdf \({\rm{f(x)}}\)and cdf\({\rm{F(x)}}\). Then for any number a,

\({\rm{P(X£a) = F(a)}}\)

and for any two numbers a and \({\rm{b}}\) with\({\rm{a < b}}\),

\({\rm{P(a£X£b) = F(b) - F(a)}}\)

(b)

Since \({\rm{X}}\) has exponential distribution hence it's mean and standard deviations are equal. Hence

\({\rm{\mu = \sigma = 2}}{\rm{.725}}\)

The probability that rainfall duration exceeds the mean value by more than \({\rm{2}}\) standard deviations is denoted as \({\rm{P(X > \mu + }}\)\({\rm{2\sigma }}\)) Using cdf \({\rm{F(X)}}\) and values of \({\rm{\mu }}\)and\({\rm{\sigma }}\), we can write :

\(\begin{aligned}P(X > \mu + 2\sigma ) &= P(X > 8{\rm{.175)}}\\ &= 1 - F(8{\rm{.175)}}\\ &= 1 - \left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - (0}}{\rm{.367) \times (8}}{\rm{.175)}}}}} \right)\\ &= 0{\rm{.0498}}\end{aligned}\)

Since the rainfall duration always will be a positive value, which means it can never be less than mean by more than one standard deviation. Hence:

\({\rm{P(X < \mu - \sigma ) = 0}}\)