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) pdf of \({\rm{X}}\) is given to us as:

As we know that every pdf \({\rm{f(x)}}\)must satisfy the following condition:

\(\int_{{\rm{ - \yen}}}^{\rm{\yen}} {\rm{f}} {\rm{(x) \times dx = 1}}\)

Hence for the given pdf, we can write:

\(\int_{\rm{0}}^{\rm{\yen}} {\rm{k}} {\left( {{\rm{1 + }}\frac{{\rm{x}}}{{{\rm{2}}{\rm{.5}}}}} \right)^{{\rm{ - 7}}}}{\rm{ \times dx = 1}}\)

Let us assume

\({\rm{1 + }}\frac{{\rm{x}}}{{{\rm{2}}{\rm{.5}}}}{\rm{ = t}}\)

, then

\(\begin{array}{l}{\rm{x = 2}}{\rm{.5(t - 1)}}\\{\rm{dx = (2}}{\rm{.5) \times dt}}\end{array}\)

As \({\rm{x}}\) goes from \({\rm{0}}\) to \(\infty ,t\) goes from \({\rm{1}}\) to\(\infty \). Substituting these in the integration we get:

\(\begin{array}{l}\int_{\rm{1}}^{\rm{\yen}} {\rm{k}} {\rm{ \times }}{{\rm{t}}^{{\rm{ - 7}}}}{\rm{ \times (2}}{\rm{.5dt) = 1(2}}{\rm{.5)k}}\left( {\frac{{{{\rm{t}}^{{\rm{ - 6}}}}}}{{{\rm{ - 6}}}}} \right)_{\rm{1}}^{\rm{\yen}}\\{\rm{ = 1(2}}{\rm{.5)k}}\left( {{\rm{0 - }}\frac{{\rm{1}}}{{{\rm{ - 6}}}}} \right)\\{\rm{ = 1k}}\\{\rm{ = }}\frac{{\rm{6}}}{{{\rm{2}}{\rm{.5}}}}{\rm{k}}\\{\rm{ = 2}}{\rm{.4}}\end{array}\)

Conditions satisfied by a pdf: For a pdf to be a legitimate pdf, it must satisfy the following two conditions:

(i) for all \({\rm{x}}\)

(ii) \(\int_{{\rm{ - \yen}}^{\rm{\yen}} {\rm{f}} {\rm{(x) \times dx = 1}}\)

(b)

Using the value of \({\rm{k}}\) calculated in part(a), the pdf of \({\rm{X}}\) can be finally written as:

It's graph of \({\rm{f(x)}}\) is given as below:

(c) The expected value of \({\rm{f(x)}}\) is written as:

\(\begin{array}{l}{\rm{E(X) = }}\int_{{\rm{ - \yen}}}^{\rm{\yen}} {\rm{x}} {\rm{ \times f(x) \times dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{\yen}} {\rm{x}} {\rm{ \times (2}}{\rm{.4)}}{\left( {{\rm{1 + }}\frac{{\rm{x}}}{{{\rm{2}}{\rm{.5}}}}} \right)^{{\rm{ - 7}}}}{\rm{ \times dx}}\end{array}\)

Let us assume

\({\rm{1 + }}\frac{{\rm{x}}}{{{\rm{2}}{\rm{.5}}}}{\rm{ = t}}\)

, then

\(\begin{array}{l}{\rm{x = 2}}{\rm{.5(t - 1)}}\\{\rm{dx = (2}}{\rm{.5) \times dt}}\end{array}\)

As \({\rm{x}}\) goes from \({\rm{0}}\) to \(\infty ,t\) goes from \({\rm{1}}\) to\(\infty \). Substituting these in eq(\({\rm{1}}{\rm{.1}}\)) we get :

\(\begin{aligned}&= \int_{\rm{1}}^{\rm{\yen}} {\rm{2}} {\rm{.5(t - 1) \times (2}}{\rm{.4) \times }}{{\rm{t}}^{{\rm{ - 7}}}}{\rm{ \times (2}}{\rm{.5dt)}}\\&= 15\int_{\rm{1}}^{\rm{\yen}} {{\rm{(t - 1)}}} {\rm{ \times }}{{\rm{t}}^{{\rm{ - 7}}}}{\rm{ \times dx}}\\ &= 15\int_{\rm{1}}^{\rm{\yen}} {\left( {{{\rm{t}}^{{\rm{ - 6}}}}{\rm{ - }}{{\rm{t}}^{{\rm{ - 7}}}}} \right)} {\rm{ \times dx}}\\&= 15\left( {\frac{{{{\rm{t}}^{{\rm{ - 5}}}}}}{{{\rm{ - 5}}}}{\rm{ - }}\frac{{{{\rm{t}}^{{\rm{ - 6}}}}}}{{{\rm{ - 6}}}}} \right)_{\rm{1}}^{\rm{\yen}}\\&= 15\left( {{\rm{(0) - }}\left( {\frac{{{\rm{ - 1}}}}{{\rm{5}}}{\rm{ - }}\frac{{{\rm{ - 1}}}}{{\rm{6}}}} \right)} \right)\\&= 15\left( {\frac{{\rm{1}}}{{\rm{5}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{6}}}} \right)\\&= 15 \times \frac{{\rm{1}}}{{{\rm{30}}}}\\{\rm{E(X) = 0}}{\rm{.5}}\end{aligned}\)

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

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

Now we recall the following proposition:

Proposition: Variance \({\rm{V(X)}}\) and standard deviation \({{\rm{\sigma }}_{\rm{x}}}\) of a rv \({\rm{X}}\) with a given pdf can be written as:

\({\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}{\rm{ }}{{\rm{\sigma }}_{\rm{x}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} \)

First, we calculate \({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\)

\(\begin{array}{c}{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = }}\int_{{\rm{ - \yen}}}^{\rm{\yen}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times f(x) \times dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{\yen}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times (2}}{\rm{.4)}}{\left( {{\rm{1 + }}\frac{{\rm{x}}}{{{\rm{2}}{\rm{.5}}}}} \right)^{{\rm{ - 7}}}}{\rm{ \times dx}}\end{array}\)

Let us assume

\({\rm{1 + }}\frac{{\rm{x}}}{{{\rm{2}}{\rm{.5}}}}{\rm{ = t}}\)

then

\(\begin{array}{l}{\rm{x = 2}}{\rm{.5(t - 1)}}\\{\rm{dx = (2}}{\rm{.5) \times dt}}\end{array}\)

As\({\rm{x}}\) goes from \({\rm{0 to }}\infty ,{\rm{t}}\) goes from \({\rm{1}}\) to\(\infty \). Substituting these in eq(\({\rm{1}}{\rm{.2}}\)) we get :

\(\begin{array}{l}{\rm{ = }}\int_{\rm{1}}^{\rm{\yen}} {\rm{2}} {\rm{.}}{{\rm{5}}^{\rm{2}}}{{\rm{(t - 1)}}^{\rm{2}}}{\rm{ \times (2}}{\rm{.4) \times }}{{\rm{t}}^{{\rm{ - 7}}}}{\rm{ \times (2}}{\rm{.5dt)}}\\{\rm{ = 37}}{\rm{.5}}\int_{\rm{1}}^{\rm{\yen}} {\left( {{{\rm{t}}^{\rm{2}}}{\rm{ - 2t + 1}}} \right)} {\rm{ \times }}{{\rm{t}}^{{\rm{ - 7}}}}{\rm{ \times dx}}\\{\rm{ = 37}}{\rm{.5}}\int_{\rm{1}}^{\rm{\yen}} {\left( {{{\rm{t}}^{{\rm{ - 5}}}}{\rm{ - 2}}{{\rm{t}}^{{\rm{ - 6}}}}{\rm{ + }}{{\rm{t}}^{{\rm{ - 7}}}}} \right)} {\rm{ \times dx}}\\{\rm{ = 37}}{\rm{.5}}\left( {\frac{{{{\rm{t}}^{{\rm{ - 4}}}}}}{{{\rm{ - 4}}}}{\rm{ - 2}}\frac{{{{\rm{t}}^{{\rm{ - 5}}}}}}{{{\rm{ - 5}}}}{\rm{ + }}\frac{{{{\rm{t}}^{{\rm{ - 6}}}}}}{{{\rm{ - 6}}}}} \right)_{\rm{1}}^{\rm{\yen}}\\{\rm{ = 37}}{\rm{.5}}\left( {{\rm{(0) - }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ + }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{6}}}} \right)} \right)\\{\rm{ = 37}}{\rm{.5}}\left( {\frac{{{\rm{10}}}}{{{\rm{24}}}}{\rm{ - }}\frac{{\rm{2}}}{{\rm{5}}}} \right)\\{\rm{ = 37}}{\rm{.5 \times }}\frac{{\rm{1}}}{{{\rm{60}}}}{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\\{\rm{ = 0}}{\rm{.625}}\end{array}\)

Since we have already calculated\({\rm{E(X) = 0}}{\rm{.5}}\), hence \({\rm{V(X)}}\) is written as:

\(\begin{array}{c}{\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = 0}}{\rm{.625 - (0}}{\rm{.5}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = 0}}{\rm{.625 - 0}}{\rm{.25}}\\{\rm{V(X) = 0}}{\rm{.375}}\end{array}\)

Then standard deviation can be calculated as:

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

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

(d)

Let \({{\rm{X}}_{\rm{2}}}\) be total medical expanses when individual pays \({\rm{2500}}\) (including deductibles of \({\rm{500}}\) and \({\rm{20\% }}\) of the expanses exceeding\({\rm{500}}\)), Which means that at this point this \({\rm{20\% }}\) paid by the individual will be equal to\({\rm{2000}}\). Also keep in mind that the pdf \({\rm{f(x)}}\) is given for \({\rm{X}}\) which is in \({\rm{1000}}\)s of dollars.

\({\rm{0}}{\rm{.2}}\left( {{\rm{1000 \times }}{{\rm{X}}_{\rm{2}}}{\rm{ - 500}}} \right){\rm{ = 2000}}{{\rm{X}}_{\rm{2}}}{\rm{ = 10}}{\rm{.5}}\)

Finally, we can conclude that0:

\({\rm{y = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{X < 0}}{\rm{.5}}}\\{{\rm{0}}{\rm{.8(1000 \times X - 500)}}}&{{\rm{0}}{\rm{.5拢 X拢 10}}{\rm{.5}}}\\{{\rm{1000 \times X - 2500}}}&{{\rm{X > 10}}{\rm{.5}}}\end{array}} \right.\)

Now we can calculate expected value of y (let us denote it by\({{\rm{E}}_{\rm{y}}}\):

\(\begin{array}{l}{{\rm{E}}_{\rm{y}}}{\rm{ = }}\int_{{\rm{ - \yen}}}^{\rm{\yen}} {\rm{y}} {\rm{ \times f(x) \times dx}}\\{\rm{ = }}\int_{{\rm{0}}{\rm{.5}}}^{{\rm{10}}{\rm{.5}}} {\rm{0}} {\rm{.8(1000 \times x - 500) \times (2}}{\rm{.4)}}{\left( {{\rm{1 + }}\frac{{\rm{x}}}{{{\rm{2}}{\rm{.5}}}}} \right)^{{\rm{ - 7}}}}{\rm{ \times dx + }}\int_{{\rm{10}}{\rm{.5}}}^{\rm{\yen}} {{\rm{(1000x - 2500)}}} {\rm{ \times (2}}{\rm{.4)}}{\left( {{\rm{1 + }}\frac{{\rm{x}}}{{{\rm{2}}{\rm{.5}}}}} \right)^{{\rm{ - 7}}}}{\rm{ \times dx}}\end{array}\)

Let us assume

\({\rm{1 + }}\frac{{\rm{x}}}{{{\rm{2}}{\rm{.5}}}}{\rm{ = t}}\)

then

\(\begin{array}{l}{\rm{x = 2}}{\rm{.5(t - 1)}}\\{\rm{dx = (2}}{\rm{.5) \times dt}}\end{array}\)

As \({\rm{x}}\) goes from \({\rm{0}}{\rm{.5}}\) to \({\rm{10}}{\rm{.5,t}}\) goes from \({\rm{1}}{\rm{.2}}\) to\({\rm{5}}{\rm{.2}}\). And as \({\rm{x}}\) goes from \({\rm{10}}{\rm{.5}}\) to\(\infty \), \({\rm{t}}\) goes from \({\rm{5}}{\rm{.2}}\) to\(\infty \). Substituting these in che integral we get:

\(\begin{array}{l}{\rm{ = }}\int_{{\rm{1}}{\rm{.2}}}^{{\rm{5}}{\rm{.2}}} {\rm{0}} {\rm{.8(2500 \times (t - 1) - 500) \times (2}}{\rm{.4)}}{{\rm{t}}^{{\rm{ - 7}}}}{\rm{ \times (2}}{\rm{.5dt) + }}\int_{{\rm{5}}{\rm{.2}}}^{\rm{\yen}} {{\rm{(2500(}}} {\rm{t - 1) - 2500) \times (2}}{\rm{.4)}}{{\rm{t}}^{{\rm{ - 7}}}}{\rm{ \times (2}}{\rm{.5dt)}}\\{\rm{ = 4}}{\rm{.8}}\int_{{\rm{1}}{\rm{.2}}}^{{\rm{5}}{\rm{.2}}} {{\rm{(2500t - 3000)}}} {\rm{ \times }}{{\rm{t}}^{{\rm{ - 7}}}}{\rm{ \times dt + 6}}\int_{{\rm{5}}{\rm{.2}}}^{\rm{\yen}} {{\rm{(2500t - 5000)}}} {\rm{ \times }}{{\rm{t}}^{{\rm{ - 7}}}}{\rm{ \times dt}}\\{\rm{ = 4}}{\rm{.8}}\int_{{\rm{1}}{\rm{.2}}}^{{\rm{5}}{\rm{.2}}} {\left( {{\rm{2500 \times }}{{\rm{t}}^{{\rm{ - 6}}}}{\rm{ - 3000 \times }}{{\rm{t}}^{{\rm{ - 7}}}}} \right)} {\rm{ \times dt + 6}}\int_{{\rm{5}}{\rm{.2}}}^{\rm{\yen}} {\left( {{\rm{2500 \times }}{{\rm{t}}^{{\rm{ - 6}}}}{\rm{ - 5000 \times }}{{\rm{t}}^{{\rm{ - 7}}}}} \right)} {\rm{ \times dt}}\\{\rm{ = 4}}{\rm{.8}}\left( {{\rm{2500 \times }}\frac{{{{\rm{t}}^{{\rm{ - 5}}}}}}{{{\rm{ - 5}}}}{\rm{ - 3000 \times }}\frac{{{{\rm{t}}^{{\rm{ - 6}}}}}}{{{\rm{ - 6}}}}} \right)_{{\rm{1}}{\rm{.2}}}^{{\rm{5}}{\rm{.2}}}{\rm{ + 6}}\left( {{\rm{2500 \times }}\frac{{{{\rm{t}}^{{\rm{ - 5}}}}}}{{{\rm{ - 5}}}}{\rm{ - 5000 \times }}\frac{{{{\rm{t}}^{{\rm{ - 6}}}}}}{{{\rm{ - 6}}}}} \right)_{{\rm{5}}{\rm{.2}}}^{\rm{\yen}}\\{\rm{ = 4}}{\rm{.8}}\left( {{\rm{ - 500 \times }}\left( {{\rm{5}}{\rm{.}}{{\rm{2}}^{{\rm{ - 5}}}}{\rm{ - 1}}{\rm{.}}{{\rm{2}}^{{\rm{ - 5}}}}} \right){\rm{ + 500 \times }}\left( {{\rm{5}}{\rm{.}}{{\rm{2}}^{{\rm{ - 6}}}}{\rm{ - 1}}{\rm{.}}{{\rm{2}}^{{\rm{ - 6}}}}} \right)} \right){\rm{ + 6}}\left( {{\rm{ - 500 \times }}\left( {{\rm{0 - 5}}{\rm{.}}{{\rm{2}}^{{\rm{ - 5}}}}} \right){\rm{ + (833}}{\rm{.33) \times }}\left( {{\rm{0 - 5}}{\rm{.}}{{\rm{2}}^{{\rm{ - 6}}}}} \right)} \right)\\{\rm{ = 160}}{\rm{.24 + 0}}{\rm{.54}}\\{{\rm{E}}_{\rm{y}}}{\rm{ = 160}}{\rm{.78}}\end{array}\)