91Ó°ÊÓ

Probability: Probability means possibility. It is a branch of mathematics which deals with the occurrence of a random event.

(a)

Considering the given information:

The information is based on the distance X that an animal moved from its birth site to territorial vacancy. Here, X follows the exponential distribution with parameter \({\rm{\lambda = 0}}{\rm{.01386}}\)

The variable X denotes the distance moved by the animal from its birth site to territorial vacancy.

Exponential distribution:

The cumulative distribution function is given by,

\(F(x;\lambda ) = \left\{ {\begin{array}{*{20}{l}}{0\quad x < 0}\\{1 - {e^{ - \lambda x}}\quad x \ge 0}\end{array}} \right.\)

The value of \(P(X \le 100)\)is obtained as shown below:

Putting \({\rm{\lambda = 0}}{\rm{.01386}}\)

\(\begin{array}{c}P(X \le 100) = 1 - {e^{ - 0.01386(100)}}\\ = 1 - {e^{ - 1.386}}\\ = 1 - 0.2501\\ = 0.7499\end{array}\)

As a result, the value of \(P(X \le 100)\)is \({\rm{0}}{\rm{.7499}}\).

The value of \(P(X \le 200)\) is obtained as shown below:

Putting \({\rm{\lambda = 0}}{\rm{.01386}}\) in the cumulative distribution function,

\(\begin{array}{c}P(X \le 200) = 1 - {e^{ - 0.01386(200)}}\\ = 1 - {e^{ - 2.772}}\\ = 1 - 0.0625\\ = 0.9375\end{array}\)

As a result, the value of \(P(X \le 200)\) is\({\rm{0}}{\rm{.9375}}\).

The \(P(100 \le X \le 200)\)is obtained as shown below:

\(\begin{array}{c}P(100 \le X \le 200) = P(X \le 200) - P(X \le 100)\\ = 0.9375 - 0.7499\\ = 0.1876\end{array}\)

As a result, the value of \(P(100 \le X \le 200)\) is \({\rm{0}}{\rm{.1876}}\).

(b)

Considering the given information:

Because X has an exponential distribution, the mean and standard deviations are calculated as follows:

\(\begin{array}{c}{\rm{\mu = }}\frac{{\rm{1}}}{{\rm{\lambda }}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{0}}{\rm{.01386}}}}\\{\rm{ = 72}}{\rm{.15}}\\{\rm{\sigma = }}\frac{{\rm{1}}}{{\rm{\lambda }}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{0}}{\rm{.01386}}}}\\{\rm{ = 72}}{\rm{.15}}\end{array}\)

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

\(\begin{array}{c}{\rm{P(X > \mu + 2\sigma ) = P(X > 216}}{\rm{.45)}}\\{\rm{ = 1 - F(216}}{\rm{.45)}}\\{\rm{ = 1 - }}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - (0}}{\rm{.01386) \times (216}}{\rm{.45)}}}}} \right)\\{\rm{ = 0}}{\rm{.0498}}\end{array}\)

Therefore, the required value is \({\rm{0}}{\rm{.0498}}\).

(c)

Considering the given information:

The value of the median distance is obtained as given below:

Putting \({\rm{x = \mu }}\)in the cumulative distribution function,

Here,\({\rm{\tilde \mu }}\)means the median of the Weibull distribution.

\(\begin{array}{c}F(x = \tilde \mu ) = 1 - {e^{ - \lambda \tilde \mu }}\\0.5 = 1 - {e^{ - \lambda \tilde \mu }}\end{array}\)

Since\({\rm{F(\tilde \mu ) = 0}}{\rm{.5}}\),

Putting \({\rm{\lambda = 0}}{\rm{.01386}}\)in the cumulative distribution function,

\(\begin{array}{c}{\rm{0}}{\rm{.5 = 1 - }}{{\rm{e}}^{{\rm{ - 0}}{\rm{.01386\tilde \mu }}}}\\{\rm{0}}{\rm{.5 = }}{{\rm{e}}^{{\rm{ - 0}}{\rm{.01386\tilde \mu }}}}\quad \\\frac{{{\rm{ - ln(0}}{\rm{.5)}}}}{{{\rm{0}}{\rm{.01386}}}}{\rm{ = \tilde \mu }}\\{\rm{\tilde \mu = 50}}{\rm{.01}}\end{array}\)

Therefore, the value of the median distance is \({\rm{50}}{\rm{.01\;m}}{\rm{.}}\)