91Ó°ÊÓ

Probability is the likelihood that an event will occur and is calculated by dividing the number of favourable outcomes by the total number of possible outcomes. The simplest example is a coin flip. When you flip a coin there are only two possible outcomes, the result is either heads or tails.

(a)

We know that mean for Weibull distribution is given as:

\(E(X) = b \cdot \Gamma \left( {1 + \frac{1}{a}} \right)\)

\( = 3 \cdot \Gamma \left( {1 + \frac{1}{2}} \right)\)

\( = 3 \cdot \Gamma \left( {\frac{3}{2}} \right)\)

\( = 3 \cdot \frac{1}{2} \cdot \sqrt \pi \)

\({\rm{E(X) \& = 2}}{\rm{.66}}\)

And the variance of the Weibull distribution is:

\(V(X){\rm{ }}{b^2} \cdot \left( {\Gamma \left( {1 + \frac{2}{a}} \right) - {{\left( {\Gamma \left( {1 + \frac{1}{a}} \right)} \right)}^2}} \right)\)

\( = {3^2} \cdot \left( {\Gamma \left( {1 + \frac{2}{2}} \right) - {{\left( {\Gamma \left( {1 + \frac{1}{2}} \right)} \right)}^2}} \right)\)

\( = 9 \cdot \left( {\Gamma (2) - {{\left( {\Gamma \left( {\frac{3}{2}} \right)} \right)}^2}} \right)\)

\( = 9 \cdot \left( {1 - {{\left( {\frac{{\sqrt \pi }}{2}} \right)}^2}} \right)\)

\( = 9 \cdot \left( {1 - \frac{\pi }{4}} \right)V(X) = 1.93\)

Therefore, the solution of \(E(X)\,\,\,and\,\,\,\,V(X)\)is \({\rm{E}}({\rm{X}}) = 2.66\,\,\,\,and\,\,\,\,{\rm{V}}\{ {\rm{X}}\} = 1.93\).

(b)

The parameterized cdf of a particular Weibull rv

\(a = 2\) and \(b = 3\) is :

\(F(x;a,b) = \left\{ {\begin{array}{*{20}{l}}0&{x < 0}\\{1 - {e^{ - {{(x/3)}^2}}}}&{x \ge 0}\end{array}} \right.\)

Then using the cdf, we can write \(P(X \le 6)\) as :

\(P(X \le 6) = F(6)\)

\( = 1 - {e^{ - {{(6/3)}^2}}}\)

\( = 1 - {e^{ - 4}}\)

\(P(X \le 6) = 0.9817\)

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

\(P(X \le a) = F(a)\)

Therefore, the solution of \(P(X \le 6)\)is \(0.9817\).

(c)

We may write \(P(X \le 6)\)as follows using the cdf:

\(P(1.5 \le X \le 6) = F(6) - F(1.5)\)

\( = 0.9817 - \left( {1 - {e^{ - {{(1.5/3)}^2}}}} \right)\)

\( = 0.9817 - \left( {1 - {e^{ - 0.25}}} \right)\)

\({\rm{ = 0}}{\rm{.9817 - 0}}{\rm{.2212}}\)

\(P(1.5 \le X \le 6) = 0.7605\)

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

\(P(a \le X \le b) = F(b) - F(a)\)

Therefore, the solution of \(P(1.5 \le X \le 6)\) is \(0.7605\).