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.

Given:

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{1}}}{{{\rm{25}}}}{\rm{y}}}&{{\rm{0£ y < 5}}}\\{\frac{{\rm{2}}}{{\rm{5}}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{25}}}}{\rm{y}}}&{{\rm{5£ y £ 10}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

(a) The cumulative distribution function is the integral of the pdf of all values up to\({\rm{y}}\). First assume \({\rm{0£ y < 5}}\)

\(\begin{array}{c}{\rm{F(y) = }}\int_{{\rm{ - \yen}}}^{\rm{y}} {\rm{f}} {\rm{(y)dy}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{y}} {\frac{{\rm{1}}}{{{\rm{25}}}}} {\rm{ydy}}\\{\rm{ = }}\left. {\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{25(2)}}}}} \right)} \right|_{\rm{0}}^{\rm{y}}\\{\rm{ = }}\left. {\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{50}}}}} \right)} \right|_{\rm{0}}^{\rm{y}}\\{\rm{ = }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{50}}}}\end{array}\)

Next assume \({\rm{5£ y£ 10}}\) :

\(\begin{array}{c}{\rm{F(y) = }}\int_{{\rm{ - \yen}}}^{\rm{y}} {\rm{f}} {\rm{(y)dy}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{5}} {\frac{{\rm{1}}}{{{\rm{25}}}}} {\rm{ydy + }}\int_{\rm{5}}^{\rm{y}} {\frac{{\rm{2}}}{{\rm{5}}}} {\rm{ - }}\frac{{\rm{1}}}{{{\rm{25}}}}{\rm{ydy}}\\{\rm{ = }}\left. {\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{25(2)}}}}} \right)} \right|_{\rm{0}}^{\rm{5}}{\rm{ + }}\left. {\left( {\frac{{{\rm{2y}}}}{{\rm{5}}}{\rm{ - }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{25(2)}}}}} \right)} \right|_{\rm{5}}^{\rm{y}}\\{\rm{ = }}\left. {\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{50}}}}} \right)} \right|_{\rm{0}}^{\rm{5}}{\rm{ + }}\left. {\left( {\frac{{{\rm{2y}}}}{{\rm{5}}}{\rm{ - }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{50}}}}} \right)} \right|_{\rm{5}}^{\rm{y}}\\{\rm{ = }}\frac{{{{\rm{5}}^{\rm{2}}}}}{{{\rm{50}}}}{\rm{ + }}\frac{{{\rm{2y}}}}{{\rm{5}}}{\rm{ - }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{50}}}}{\rm{ - }}\left( {\frac{{{\rm{2(5)}}}}{{\rm{5}}}{\rm{ - }}\frac{{{{{\rm{(5)}}}^{\rm{2}}}}}{{{\rm{50}}}}} \right)\\{\rm{ = - }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{50}}}}{\rm{ + }}\frac{{{\rm{2y}}}}{{\rm{5}}}{\rm{ - 1}}\end{array}\)

Since the cumulative distribution function is zero for all values smaller than the first nonzero value of the pdf and since the cumulative distribution function is one for all values larger than the last nonzero value of the pdf, we then obtain:

\({\rm{F(y) = }}\left\{ {\begin{array}{*{20}{c}}{\rm{0}}&{{\rm{y < 0}}}\\{\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{50}}}}}&{{\rm{0£ y < 5}}}\\{{\rm{ - }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{50}}}}{\rm{ + }}\frac{{{\rm{2y}}}}{{\rm{5}}}{\rm{ - 1}}}&{{\rm{5£ y£ 10}}}\\{\rm{1}}&{{\rm{y > 10}}}\end{array}} \right.\)

(b)

Since\(\int_{\rm{0}}^{\rm{5}} {\frac{{\rm{1}}}{{{\rm{25}}}}} {\rm{ydy = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.5}}\), we know that the if \({\rm{p}}\) is between 0 and\({\rm{0}}{\rm{.5}}\), then the percentile falls between 0 and\({\rm{5}}\). If \({\rm{p}}\) is between \({\rm{0}}{\rm{.5}}\) and\({\rm{1}}\), then the percentile falls between \({\rm{5}}\) and \({\rm{10}}\).

The \({\rm{100}}\) the percentile is the value of \({\rm{y}}\) for which the probability of values smaller than \({\rm{y}}\) is equal to\({\rm{p}}\).

\({\rm{F(y) = P(Y£ y) = p}}\)

Let us first assume that\({\rm{0£ y < 5}}\):

\({\rm{p = F(y) = }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{50}}}}\)

Multiply each side of the equation by\({\rm{50}}\):

\({{\rm{y}}^{\rm{2}}}{\rm{ = 50p}}\)

Take the square root of each side:

\({\rm{y = \pm }}\sqrt {{\rm{50p}}} \)

Since \({\rm{0£ y < 5,y}}\) cannot be negative:

\({\rm{y = }}\sqrt {{\rm{50p}}} \)

Next assume that\({\rm{5£ y < 10}}\):

\({\rm{p = F(y) = - }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{{\rm{50}}}}{\rm{ + }}\frac{{{\rm{2y}}}}{{\rm{5}}}{\rm{ - 1}}\)

Multiply both sides by\({\rm{ - 50}}\):

\({{\rm{y}}^{\rm{2}}}{\rm{ - 20y + 50 = - 50p}}\)

Subtract \({\rm{1}}\) from each side:

\({{\rm{y}}^{\rm{2}}}{\rm{ - 20y = - 50p - 50}}\)

Add \({\rm{100}}\) to each side:

\({{\rm{y}}^{\rm{2}}}{\rm{ - 20y + 100 = - 50p + 50}}\)

Factorize the left side of the equation:

\({{\rm{(y - 10)}}^{\rm{2}}}{\rm{ = - 50p + 50}}\)

Take the square root of each side:

\({\rm{y - 10 = \pm }}\sqrt {{\rm{ - 50p + 50}}} \)

Add 10 to each side:

\({\rm{y = 10 \pm }}\sqrt {{\rm{ - 50p + 50}}} \)

Since\({\rm{5£ y£ 10}}\):

\({\rm{y = 10 - }}\sqrt {{\rm{ - 50p + 50}}} \)

Combining the results, we then obtain:

\({\rm{ Percentile = }}\left\{ {\begin{array}{*{20}{c}}{\sqrt {{\rm{50p}}} }&{{\rm{0 < p < 0}}{\rm{.5}}}\\{{\rm{10 - }}\sqrt {{\rm{ - 50p + 50}}} }&{{\rm{0}}{\rm{.5 < p < 1}}}\end{array}} \right.\)

(c)

The expected value (or mean) \({\rm{\mu }}\)is the sum (integral) of the product of each possibility \({\rm{x}}\) with its probability\({\rm{P(x)}}\):

\(\begin{aligned}V(Y) &= \int_{ - \yen}^{ + \yen} {{{(y - \mu )}^2}} f(y)dy \\ &= \int_0^5 {{{(y - 5)}^2}} \frac{1}{{25}}ydy + \int_5^{10} {{{(y - 5)}^2}} \left( {\frac{2}{5} -\frac{1}{{25}}y} \right)dy \\ &= \left. {\left( {\frac{{{y^4}}}{{100}} - \frac{{2{y^3}}}{{15}} + \frac{{{y^2}}}{2}} \right)} \right|_1^{ + \yen} + \left. {\left( {10y - \frac{{5{y^2}}}{2} + \frac{{4{y^3}}}{{15}} - \frac{{{y^4}}}{{100}}} \right)} \right|_5^{10} \\ &= \frac{{25}}{{12}} + \frac{{25}}{{12}} \\ &= \frac{{25}}{6} \\ \gg 4.1667 \\ \end{aligned} \)

Assume a uniform distribution on \({\rm{(0,5)}}\) with \({\rm{a = 0}}\) and \({\rm{b = 5}}\) (for a single bus \({\rm{X}}\)):

The mean of a uniform distribution is the average of the boundaries:

\({\rm{E(X) = \mu = }}\frac{{{\rm{\alpha + \beta }}}}{{\rm{2}}}{\rm{ = }}\frac{{{\rm{0 + 5}}}}{{\rm{2}}}{\rm{ = }}\frac{{\rm{5}}}{{\rm{2}}}{\rm{ = 2}}{\rm{.5}}\)

The standard deviation of a uniform distribution is the difference of the boundaries divided by the square root of12:

\({\rm{\sigma = }}\frac{{{\rm{\beta - \alpha }}}}{{\sqrt {{\rm{12}}} }}{\rm{ = }}\frac{{{\rm{5 - (0)}}}}{{\sqrt {{\rm{12}}} }}{\rm{ = }}\frac{{\rm{5}}}{{\sqrt {{\rm{12}}} }}\)

The variance is the square of the standard deviation:

\({\rm{V(X) = }}{{\rm{\sigma }}^{\rm{2}}}{\rm{ = }}\frac{{{{\rm{5}}^{\rm{2}}}}}{{{{{\rm{(}}\sqrt {{\rm{12}}} {\rm{)}}}^{\rm{2}}}}}{\rm{ = }}\frac{{{\rm{25}}}}{{{\rm{12}}}}\)

We then note that the expected waiting time for a single bus is half the expected waiting time for two buses and the variance for a single bus is half the variance of two buses.