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)

It is given that\({\rm{X}}\)is the temperature in\(^{\rm{^\circ }}{\rm{C and}}\,{\rm{Y}}\) is the temperature in\(^{\rm{^\circ }}{\rm{F}}\), and\({\rm{Y = 1}}{\rm{.8X + 32}}\)

If we recall the definition of median, then we can write \({\rm{P(X\pounds\tilde \mu ) = 0}}{\rm{.5}}\)

Since \({\rm{\tilde \mu }}\) is the median of \({\rm{X}}\)

Now let us calculate \({\rm{P(Y\pounds1}}{\rm{.8\tilde \mu + 32)}}\)

\(\begin{array}{l}{\rm{P(Y\pounds1}}{\rm{.8\tilde \mu + 32)}}\\{\rm{ = P(1}}{\rm{.8X + 32\pounds1}}{\rm{.8\tilde \mu + 32)}}\\{\rm{ = P(1}}{\rm{.8X\pounds1}}{\rm{.8\tilde \mu )}}\\{\rm{ = P(X\pounds\tilde \mu )}}\\{\rm{ = 0}}{\rm{.5}}\end{array}\)

Hence \({\rm{1}}{\rm{.8\tilde \mu + 32}}\) is the median of\({\rm{Y}}\).

Definition: Let \({\rm{p}}\) be a number between\({\rm{0 and 1}}\). The \({{\rm{(100p)}}^{{\rm{th }}}}\) percentile of the distribution of a continuous rv\({\rm{X}}\), denoted by \({{\rm{\eta }}_{\rm{p}}}\) ), is defined by

\({\rm{p = F}}\left( {{{\rm{\eta }}_{\rm{p}}}} \right){\rm{ = P}}\left( {{\rm{x\pounds}}{{\rm{\eta }}_{\rm{p}}}} \right){\rm{ = }}\int_{{\rm{ - \currency}}}^{{{\rm{\eta }}_{\rm{p}}}} {\rm{f}} {\rm{(y)dy}}\)

For median: \({\rm{p = 0}}{\rm{.5}}\)

(b)

If we recall the definition of proportion, then we can write

\({\rm{P}}\left( {{\rm{X\pounds}}{{\rm{\eta }}_{{\rm{(0}}{\rm{.9)}}}}} \right){\rm{ = 0}}{\rm{.9}}\)

where \({{\rm{\eta }}_{{\rm{(0}}{\rm{.9)}}}}\) is the 90th percentile of \({\rm{X}}\)-distribution.

Now let us calculate \({\rm{P}}\left( {{\rm{Y\pounds1}}{\rm{.8}}{{\rm{\eta }}_{{\rm{(0}}{\rm{.9)}}}}{\rm{ + 32}}} \right)\)

\(\begin{array}{c}{\rm{P}}\left( {{\rm{Y\pounds1}}{\rm{.8}}{{\rm{\eta }}_{{\rm{(0}}{\rm{.9)}}}}{\rm{ + 32}}} \right)\\{\rm{ = P}}\left( {{\rm{1}}{\rm{.8X + 32\pounds1}}{\rm{.8}}{{\rm{\eta }}_{{\rm{(0}}{\rm{.9)}}}}{\rm{ + 32}}} \right)\\{\rm{ = P}}\left( {{\rm{1}}{\rm{.8X\pounds1}}{\rm{.8}}{{\rm{\eta }}_{{\rm{(0}}{\rm{.9)}}}}} \right)\\{\rm{ = P}}\left( {{\rm{X\pounds}}{{\rm{\eta }}_{{\rm{(0}}{\rm{.9)}}}}} \right)\\{\rm{ = 0}}{\rm{.9}}\end{array}\)

Hence \({\rm{1}}{\rm{.8}}{{\rm{\eta }}_{{\rm{(0}}{\rm{.9)}}}}{\rm{ + 32}}\) is the 9 0th percentile of \({\rm{Y}}\)-distribution.

(c)

If\({\rm{Y = aX + b}}\), and let \({{\rm{\eta }}_{\rm{p}}}\) denote the \({\rm{100}}{{\rm{p}}^{{\rm{th}}}}\) percentile of X-distribution.

Then corresponding \({\rm{100}}{{\rm{p}}^{{\rm{th }}}}\) percentile of the \({\rm{Y}}\)-distribution related to the corresponding percentile of the \({\rm{X}}\) distribution will be \({\rm{a}}{{\rm{\eta }}_{\rm{p}}}{\rm{ + b}}\)