Question

Let \({\rm{X}}\) denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < 0}}}\\{\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{4}}}}&{{\rm{0£ x < 2}}}\\{\rm{1}}&{{\rm{2\£ x}}}\end{array}} \right.\)

a. Calculate\({\rm{P(X£ 1)}}\).

b. Calculate\({\rm{P(}}{\rm{.5£ X£ 1)}}\).

c. Calculate\({\rm{P(X > 1}}{\rm{.5)}}\).

d. What is the median checkout duration \({\rm{\tilde \mu }}\) ? (solve\({\rm{5 = F(\tilde \mu ))}}\).

e. Obtain the density function\({\rm{f(x)}}\).

f. Calculate\({\rm{E(X)}}\).

g. Calculate \({\rm{V(X)}}\)and\({{\rm{\sigma }}_{\rm{X}}}\).

h. If the borrower is charged an amount \({\rm{h(X) = }}{{\rm{X}}^{\rm{2}}}\) when checkout duration is\({\rm{X}}\), compute the expected charge\({\rm{E(h(X))}}\).

Short answer

(a) The solution is \({\rm{0}}{\rm{.25}}\).

(b) The solution is \({\rm{0}}{\rm{.1875}}\).

(c) The solution is \({\rm{0}}{\rm{.4375}}\).

(d) The solution is \({\rm{1}}{\rm{.414}}\).

(e) The solution is \({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < 0}}}\\{\frac{{\rm{x}}}{{\rm{2}}}}&{{\rm{0\poundsx < 2}}}\\{\rm{1}}&{{\rm{2\poundsx}}}\end{array}} \right.\).

(f) The solution is \(\frac{{\rm{4}}}{{\rm{3}}}\).

(g) The solution is \({\rm{V(X) = 0}}{\rm{.2222;}}{{\rm{\sigma }}_{\rm{X}}}{\rm{ = 0}}{\rm{.4714}}\).

(h) The solution is \({\rm{2}}\).

Step-by-step solution

Definition

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.

Calculating \({\rm{P(X\pounds1)}}\)

The cdf \({\rm{F(x)}}\) is given as:

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < 0}}}\\{\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{4}}}}&{{\rm{0\poundsx < 2}}}\\{\rm{1}}&{{\rm{2\poundsx}}}\end{array}} \right.\)

(a)

Then using the given cdf, we can write:

\(\begin{array}{l}{\rm{P(X\pounds1) = F(1) = }}\frac{{{{\rm{1}}^{\rm{2}}}}}{{\rm{4}}}\\{\rm{P(X\pounds1) = 0}}{\rm{.25}}\end{array}\)

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

\({\rm{P(X\poundsa) = F(a)}}\)

Calculating \({\rm{P(}}{\rm{.5\poundsX\pounds1)}}\)

(b)

From the given cdf, we can write:

\(\begin{array}{c}{\rm{P(0}}{\rm{.5\poundsX\pounds1) = F(1) - F(0}}{\rm{.5)}}\\{\rm{ = }}\frac{{{{\rm{1}}^{\rm{2}}}}}{{\rm{4}}}{\rm{ - }}\frac{{{\rm{0}}{\rm{.}}{{\rm{5}}^{\rm{2}}}}}{{\rm{4}}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ - }}\frac{{{\rm{0}}{\rm{.25}}}}{{\rm{4}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.75}}}}{{\rm{4}}}\\{\rm{P(0}}{\rm{.5\poundsX\pounds1) = 0}}{\rm{.1875}}\end{array}\)

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

\({\rm{P(a\poundsX\poundsb) = F(b) - F(a)}}\)

Calculating \({\rm{P(X > 1}}{\rm{.5)}}\)

(c)

From the given cdf, we can write:

\(\begin{array}{c}{\rm{P(X > 1}}{\rm{.5) = 1 - F(1}}{\rm{.5)}}\\{\rm{ = 1 - }}\frac{{{\rm{1}}{\rm{.}}{{\rm{5}}^{\rm{2}}}}}{{\rm{4}}}\\{\rm{ = 1 - 0}}{\rm{.5625}}\\{\rm{P(X > 1}}{\rm{.5) = 0}}{\rm{.4375}}\end{array}\)

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

\({\rm{P(X > a) = 1 - F(a)}}\)

What is the median checkout duration \({\rm{\tilde \mu }}\)

(d)

According to the definition of the median\({\rm{(\tilde \mu )}}\), we can write:

\({\rm{F(\tilde \mu ) = 0}}{\rm{.5}}\)

For the given cdf, we write it as

\(\begin{array}{l}\frac{{{{{\rm{\tilde \mu }}}^{\rm{2}}}}}{{\rm{4}}}{\rm{ = 0}}{\rm{.5}}{{{\rm{\tilde \mu }}}^{\rm{2}}}\\{\rm{ = 2\tilde \mu }}\\{\rm{ = }}\sqrt {\rm{2}} \\{\rm{ = 1}}{\rm{.414}}\end{array}\)

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

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

For median the value of \({\rm{p}}\) is \({\rm{0}}{\rm{.5}}\)

Obtain the density function \({\rm{f(x)}}\)

(e)

For values of \({\rm{x}}\) for which \({{\rm{F}}^{\rm{\cent}}}{\rm{(x)}}\) exists, we define \({\rm{f(x)}}\) as: \({\rm{f(x) = }}{{\rm{F}}^{\rm{\cent}}}{\rm{(x)}}\)

Since \({\rm{F(x)}}\) is constant for interval \({\rm{x < 0}}\) and, hence \({\rm{f(x) = 0}}\) for these intervals.

For \({\rm{0\poundsx < 2,f(x)}}\) can be written as:

\({\rm{f(x) = }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{4}}}} \right){\rm{ = }}\frac{{\rm{x}}}{{\rm{2}}}\)

Finally, we can write \({\rm{f(x)}}\) as:

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < 0}}}\\{\frac{{\rm{x}}}{{\rm{2}}}}&{{\rm{0\poundsx < 2}}}\\{\rm{1}}&{{\rm{2\poundsx}}}\end{array}} \right.\)

Proposition: If \({\rm{X}}\) is a continuous rv with \({\rm{pdf(x)}}\) and cdf\({\rm{F(x)}}\), then at every \({\rm{x}}\) at which the derivative \({{\rm{F}}^{\rm{\cent}}}{\rm{(x)}}\) exists,

\({\rm{f(x) = }}{{\rm{F}}^{\rm{\cent}}}{\rm{(x)}}\)

Calculating \({\rm{E(X)}}\)

(f) For the pdf \({\rm{f(x)}}\) obtained in the last part:

\(\begin{array}{c}{\rm{E(X) = }}\int_{{\rm{ - \currency}}}^{\rm{\currency}} {\rm{x}} {\rm{ \times f(x) \times dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{2}} {\rm{x}} {\rm{ \times }}\frac{{\rm{x}}}{{\rm{2}}}{\rm{ \times dx}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{0}}^{\rm{2}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times dx}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{\rm{0}}^{\rm{2}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {\frac{{{{\rm{2}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ - 0}}} \right)\\{\rm{E(X) = }}\frac{{\rm{4}}}{{\rm{3}}}\end{array}\)

Definition: The expected or mean value of a continuous rv \({\rm{X}}\) with pdf \({\rm{f(x)}}\)is

\({\rm{\mu = E(X) = }}\int_{{\rm{ - \currency}}}^{\rm{\currency}} {\rm{x}} {\rm{ \times f(x) \times dx}}\)

Calculating \({\rm{V(X)}}\)and \({{\rm{\sigma }}_{\rm{X}}}\)

(g)

For the derived pdf\({\rm{f(x)}}\), we can write:

\(\begin{array}{c}{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = }}\int_{{\rm{ - \currency}}}^{\rm{\currency}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times f(x) \times dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{2}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times }}\frac{{\rm{x}}}{{\rm{2}}}{\rm{ \times dx}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{0}}^{\rm{2}} {{{\rm{x}}^{\rm{3}}}} {\rm{ \times dx}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {\frac{{{{\rm{x}}^{\rm{4}}}}}{{\rm{4}}}} \right)_{\rm{0}}^{\rm{2}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {\frac{{{{\rm{2}}^{\rm{4}}}}}{{\rm{4}}}{\rm{ - 0}}} \right)\\{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 2}}\end{array}\)

As we have already calculated \({\rm{E(X)}}\) in the last part: \({\rm{E(X) = }}\frac{{\rm{4}}}{{\rm{3}}}\)

Proposition: \({\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - E(X}}{{\rm{)}}^{\rm{2}}}\)

substituting values of \({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\) and\({\rm{E(X)}}\), we can write

\({\rm{V(X) = 2 - }}{\left( {\frac{{\rm{4}}}{{\rm{3}}}} \right)^{\rm{2}}}{\rm{ = 0}}{\rm{.2222}}\)

Now, standard deviation \(\left( {{{\rm{\sigma }}_{\rm{X}}}} \right)\) can be written as:

\(\begin{array}{c}{{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} {\rm{ = }}\sqrt {{\rm{0}}{\rm{.2222}}} \\{\rm{ = 0}}{\rm{.4714}}\end{array}\)

Definition: If \({\rm{X}}\) is a continuous rv with pdf \({\rm{f(x)}}\) and \({\rm{h(X)}}\)is any function of\({\rm{X}}\), then

\({\rm{E(h(x)) = }}\int_{{\rm{ - \currency}}}^{\rm{\currency}} {\rm{h}} {\rm{(x) \times f(x) \times dx}}\)

If the borrower is charged an amount \({\rm{h(X) = }}{{\rm{X}}^{\rm{2}}}\) when checkout duration is \({\rm{X}}\), compute the expected charge \({\rm{E(h(X))}}\)

(h)

It is given that\({\rm{h(X) = }}{{\rm{X}}^{\rm{2}}}\), then

\({\rm{E(h(X)) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\)

As we have already calculated \({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\) in the last part, hence

\({\rm{E(h(X)) = 2}}\)