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.

Let the time (in hours) to failure of solid insulating specimens subjected to \({\rm{AC}}\) voltage be denoted by \({\rm{X}}\). Then it is given that \({\rm{X}}\) has a Weibull distribution with following parameters:

\(\begin{array}{l}{\rm{\alpha = 2}}{\rm{.5}}\\{\rm{\beta = 200}}\end{array}\)

(a)

The cdf of given Weibull rv having parameters \({\rm{a = 2}}\) and \({\rm{b = 3}}\) is:

\({\text{F(x;a,b) = }}\left\{ {\begin{array}{*{20}{l}}{\text{0}}&{{\text{x < 0}}} \\ {{\text{1 - }}{{\text{e}}^{{\text{ - (x/200}}{{\text{)}}^{{\text{25}}}}}}}&{{\text{x0}}}\end{array}} \right.\)

The probability that a specimen's lifetime is at most \({\rm{250}}\) can be denoted by\({\rm{P(X£250)}}\). Then using the cdf, we can write this probability as:

\(\begin{aligned}P(X£ 250) &= F(250)\\ &= 1 - {{\rm{e}}^{{\rm{ - (250/200}}{{\rm{)}}^{{\rm{2}}{\rm{.5}}}}}}\\ &= 1 - {{\rm{e}}^{{\rm{ - 1}}{\rm{.747}}}} P(X£250)\\ &= 0{\rm{.8257}}\end{aligned}\)

The probability that a specimen's lifetime is less than \({\rm{250}}\) can be denoted by\({\rm{P(X < 250)}}\). Since Weibull distribution and continuous, hence probability of the rv being an exact value is zero. Hence

\({\rm{P(X < 250) = P(X£250) = 0}}{\rm{.8257}}\)

The probability that a specimen's lifetime is more than \({\rm{300}}\) is denoted by\({\rm{P(X > 300)}}\). Then using the cdf, we can write this probability as:

\(\begin{aligned}P(X > 300) &= 1 - F(300) \\ &= 1 - \left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - (300/200}}{{\rm{)}}^{{\rm{2}}{\rm{.5}}}}}}} \right)\\ &= {{\rm{e}}^{{\rm{ - 2}}{\rm{.756}}}}{\rm{P(X > 300)}}\\ &= 0 {\rm{.0636}}\end{aligned}\)

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

\({\rm{P(X£ a) = F(a)}}\)

(b)

The probability that a specimen's lifetime is between \({\rm{100}}\) and \({\rm{250}}\) is denoted as\({\rm{P(100£ X£ 250)}}\). Using the cdf, we can write \({\rm{P(100£ X£ 250)}}\) as:

\(\begin{aligned}P(100£X£250) &= F(250) - F(100) \\ &= 0{\text{.8257 - }}\left[ {{\text{1 - }}{{\text{e}}^{{\text{ - (100/200}}{{\text{)}}^{{\text{25}}}}}}} \right] \\&= 0{\text{.8257 - }}\left[ {{\text{1 - }}{{\text{e}}^{{\text{ - 0}}{\text{.1768}}}}} \right] \\ & = 0{\text{.8257 - 0}}{\text{.1620}} \\P(100£X£250) &= 0{\text{.6637}} \\ \end{aligned} \)

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

\({\rm{P(a£ X£ b) = F(b) - F(a)}}\)

(c)

Let \({\rm{x}}\) be the value such that exactly \({\rm{50\% }}\) of all specimens have lifetimes exceeding that value. Hence, we can write

\(\begin{aligned}P(X£x) &= 0{\rm{.51 - }}{{\rm{e}}^{{\rm{ - (x/200}}{{\rm{)}}^{{\rm{2}}{\rm{.5}}}}}}\\&= 0{\rm{.5}}{{\rm{e}}^{{\rm{ - (x/200}}{{\rm{)}}^{{\rm{2}}{\rm{.5}}}}}}\\&= 0 {\rm{.5}}{\left( {\frac{{\rm{x}}}{{{\rm{200}}}}} \right)^{{\rm{2}}{\rm{.5}}}}\\ &= - ln(0{\rm{.5)x}}\\&= 200 \times ( - ln(0{\rm{.5)}}{{\rm{)}}^{{\rm{1/2}}{\rm{.5}}}}{\rm{x}}\\&= 172{\rm{.727}}\end{aligned}\)