91影视

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

(a) We manipulate the given pdf \({\rm{f(\nu )}}\) as follows:

\(\begin{aligned}f(\nu ) &= \frac{\nu }{{{\sigma ^2}}} \times \exp \left( {\frac{{ - {\nu ^2}}}{{2{\sigma ^2}}}} \right) \\&= \frac{{2\nu }}{{2{\sigma ^2}}} \times \exp \left( {\frac{{ - {\nu ^2}}}{{2{\sigma ^2}}}} \right) \\&= \frac{{2\nu }}{{{{(\sqrt 2 \sigma )}^2}}} \times \exp \left( {\frac{{ - {\nu ^2}}}{{{{(\sqrt 2 \sigma )}^2}}}} \right) \\&= \frac{2}{{{{(\sqrt 2 \sigma )}^2}}} \times {\nu ^{2 - 1}} \times \exp \left( {\frac{{ - {\nu ^2}}}{{{{(\sqrt 2 \sigma )}^2}}}} \right) \\\end{aligned} \)

\((for{\text{ }}\nu > 0)\)

Now observing the pdf, we can see that the pdf \({\rm{f(\nu )}}\) is a from Weibull distribution of family with parameters \({\rm{\alpha = 2}}\) and \({\rm{\beta = }}\sqrt {\rm{2}} {\rm{\sigma }}\)

(b) If \({\rm{\sigma = 20\;mm}}\) then the pdf \({\rm{f(\nu )}}\) can be rewritten as:

\({\rm{f(\nu ) = }}\frac{{\rm{\nu }}}{{{\rm{400}}}}{\rm{ \times exp}}\left( {\frac{{{\rm{ - }}{{\rm{\nu }}^{\rm{2}}}}}{{{\rm{800}}}}} \right)\)

\({\rm{(for \nu > 0)}}\)

Then the probability that a dart will land within \({\rm{25\;mm}}\)of the target can be written as \(P(\nu 拢25)\). Then using the definition of pdf, we can write:

\(P(\nu 拢25) = \int_{ - 楼}^{25} f (\nu ) \times d\nu = \int_0^{25} {\frac{\nu }{{400}}} \times \exp \left( {\frac{{ - {\nu ^2}}}{{800}}} \right) \times d\nu \)

Now we make the following substitution:

\(\begin{aligned}{}{\frac{{{{\rm{\nu }}^{\rm{2}}}}}{{{\rm{800}}}}{\rm{ = t}}}\\{\frac{{{\rm{2\nu \times d\nu }}}}{{{\rm{800}}}}{\rm{ = dt}}}\\{\frac{{{\rm{\nu \times d\nu }}}}{{{\rm{400}}}}{\rm{ = dt}}}\end{aligned}\)

As \({\rm{\nu }}\) goes from \({\rm{0}}\) to \({\rm{25,t}}\) goes from \({\rm{0}}\) to\({\rm{0}}{\rm{.78125}}\). Hence the limits of the new integral wrt \({\rm{t}}\) are \({\rm{0}}\) to \({\rm{0}}{\rm{.78125}}\)

Hence the integration becomes:

\(\begin{aligned}P(\nu 拢25)&= \int_0^{0.78125} {{e^{ - t}}} \times dt \\&= \left[ { - {e^{ - t}}} \right]_0^{0.78125} \\&= \left[ { - {e^{ - 0.78125}} - ( - 1)} \right]P(\nu 拢25) \\&= 0.5422 \\\end{aligned} \)