÷ÈÓ°Ö±²¥

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.

Given: Weibull distribution with

\(\begin{array}{l}{\rm{\alpha = 9}}\\{\rm{\beta = 180}}\end{array}\)

(a) Density function of the Weibull distribution:

\({\rm{f(x) = }}\frac{{\rm{\alpha }}}{{{{\rm{\beta }}^{\rm{\alpha }}}}}{{\rm{x}}^{{\rm{\alpha - 1}}}}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}\)

Replace \({\rm{\alpha }}\) by \({\rm{9}}\) and \({\rm{\beta }}\) by\({\rm{180}}\):

\({\rm{f(x) = }}\frac{{\rm{9}}}{{{\rm{18}}{{\rm{0}}^{\rm{9}}}}}{{\rm{x}}^{{\rm{9 - 1}}}}{{\rm{e}}^{{\rm{ - (x/180}}{{\rm{)}}^{\rm{9}}}}}{\rm{ = }}\frac{{\rm{9}}}{{{\rm{18}}{{\rm{0}}^{\rm{9}}}}}{{\rm{x}}^{\rm{8}}}{{\rm{e}}^{{\rm{ - (x/180}}{{\rm{)}}^{\rm{9}}}}}\)

(b) Cumulative distribution function of a Weibull distribution:

\(F(x;\alpha ,\beta ) = \left\{ {\begin{array}{*{20}{c}} 0&{x < 0} \\{1 - {e^{ - {{(x/\beta )}^\alpha }}}}&{x0}\end{array}} \right.\)

The cumulative distribution at \({\rm{X = x}}\) is the probability to the left of \({\rm{x}}\):

\(\begin{aligned}P(X > 175) &= 1 - P(X£175) \\&= 1 - F(175;9,180) = {e^{ - {{(175/180)}^9}}} \\\gg 0.4602 &= 46.02\% \\P(150 < X < 175) &= P(X£175) - P(X£150) \\&= F(175;9,180) - F(150;9,180) \\&= \left( {1 - {e^{ - {{(175/180)}^9}}}} \right) - \left( {1 - {e^{-{{(150/180)}^9}}}} \right) \\&={e^{-{{(150/180)}^9}}}-{e^{-{{(175/180)}^9}}}\\\gg& 0.3636 \\&= 36.36\% \\ \end{aligned} \)

(c) Multiplication rule for independent events:

\({\rm{P(A and B) = P(A) \times P(B)}}\)

Complement rule:

\({\rm{P(notA) = 1 - P(A)}}\)

Use the complement rule:

\(\begin{array}{l}{\rm{P( not 150 < X < 175) = 1 - P(150 < X < 175)}}\\{\rm{ = 1 - 0}}{\rm{.3636}}\\{\rm{ = 0}}{\rm{.6364}}\end{array}\)

If two specimens are chosen at random, we may safely assume that they are independent, and so the multiplication formula for independent events can be applied:

\(\begin{aligned}P({\text{ }}both{\text{ }}not{\text{ }}150 < X < 175) \\&= P({\text{ }}not{\text{ }}150 < X < 175) \times P({\text{ }}not{\text{ }}150 < X < 175) \\&= 0.6364 \times 0.6364 \\\gg& 0.4050 \\\end{aligned} \)

Use the complement rule again:

\(\begin{aligned}P({\text{ }}at{\text{ }}least{\text{ }}one{\text{ }}150 < X < 175) \\&= 1 - P({\text{ }}both{\text{ }}not{\text{ }}150 < X < 175) \\&= 1 - 0.4050 \\&= 0.5950 \\&= 59.50\% \\\end{aligned} \)

(d) The boundary for the weakest \({\rm{10\% }}\) has the property that the probability to its left is \({\rm{10\% }}\) or \({\rm{0}}{\rm{.10}}\) :

\(P(X£x) = 0.10\)

The cumulative distribution at \({\rm{X = x}}\) is the probability to the left of\({\rm{x}}\):

\({\rm{F(x) = 0}}{\rm{.10}}\)

Fill in the known expression for the cumulative distribution function:

\({\rm{1 - }}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}{\rm{ = 0}}{\rm{.10}}\)

Subtract 1 from each side of the equation:

\({\rm{ - }}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{a}}}}}{\rm{ = - 0}}{\rm{.90}}\)

Multiply each side of the equation by\({\rm{ - 1}}\):

\({{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{a}}}}}{\rm{ = 0}}{\rm{.90}}\)

Take the natural logarithm of each side:

\({\rm{ - }}{\left( {\frac{{\rm{x}}}{{\rm{\beta }}}} \right)^{\rm{\alpha }}}{\rm{ = ln}}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{a}}}}}{\rm{ = ln0}}{\rm{.90}}\)

Multiply each side by\({\rm{ - 1}}\):

\({\left( {\frac{{\rm{x}}}{{\rm{\beta }}}} \right)^{\rm{\alpha }}}{\rm{ = - ln0}}{\rm{.90}}\)

Take the \({\rm{1/\alpha }}\)th power of each side:

\(\frac{{\rm{x}}}{{\rm{\beta }}}{\rm{ = ( - ln0}}{\rm{.90}}{{\rm{)}}^{{\rm{1/\alpha }}}}\)

Multiply each side of the equation by\({\rm{\beta }}\):

\({\rm{x = \beta ( - ln0}}{\rm{.90}}{{\rm{)}}^{{\rm{1/\alpha }}}}\)

Replace \({\rm{\alpha }}\) by\({\rm{\;9}}\) and \({\rm{\beta }}\) by\({\rm{180}}\):

\({\rm{x = 180( - ln0}}{\rm{.90}}{{\rm{)}}^{{\rm{1/9}}}}{\rm{\gg 140}}{\rm{.1784}}\)

Thus, the strength value \({\rm{140}}{\rm{.1784MPa}}\)separates the weakest \({\rm{10\% }}\) of all specimens from the remaining\({\rm{90\% }}\).