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.

(a) Let the reaction time (in seconds) to be denoted as a continuous random variable \({\rm{X}}\) with pdf \({\rm{f(x)}}\) given to us as:

\(f(x) = \left\{ {\begin{array}{*{20}{l}}{\frac{3}{{2 \times {x^2}}}}&{1£x£3} \\0&{{\text{ }}otherwise{\text{ }}} \end{array}} \right.\)

We recall the definition of cdf of a continuous variable.

Definition: The cumulative distribution function \({\rm{F(x)}}\) for a continuous rv \({\rm{X}}\) is defined for every number \({\rm{x}}\) by

\(F(x) = P(X£x) = \int_{ - \yen}^x f (y) \times dy\)

Since \({\rm{f(x)}}\) is non-zero only for \({\rm{x}}\) between\({\rm{1 and 3}}\). Hence For any number \({\rm{x}}\) between \({\rm{1 and 3}}\)

\(\begin{aligned}F(x)&=\int_{\rm{1}}^{\rm{x}} {\frac{{\rm{3}}}{{{\rm{2 \times }}{{\rm{y}}^{\rm{2}}}}}}{\rm{ \times dy}}\\&=\left( {\frac{{{\rm{ - 3}}}}{{{\rm{2 \times y}}}}}\right)_{\rm{1}}^{\rm{x}}\\&=\left( {\left( {\frac{{{\rm{ - 3}}}}{{{\rm{2 \times x}}}}} \right){\rm{ -}}\left( {\frac{{{\rm{ - 3}}}}{{\rm{2}}}} \right)} \right)\\F(x)&=\frac{{\rm{3}}}{{\rm{2}}}\left({{\rm{1 - }}\frac{{\rm{1}}}{{\rm{x}}}} \right)\end{aligned}\)

Thus \({\rm{F(x)}}\) can be given as:

\(F(x) = \left\{ {\begin{array}{*{20}{l}}0&{x < 1} \\ {\frac{3}{2}\left( {1 - \frac{1}{x}} \right)}&{1£x£3} \\ 1&{x > 3} \end{array}} \right.\)

(b)

Probability that the reaction time is at most \({\rm{2}}{\rm{.5sec}}\) is denoted by\({\text{P(X£ 2}}{\text{.5)}}\). Then using the given cdf from part(a), we can write:

\(\begin{aligned}P(X£2.5) &= F(2.5) \\ &= \frac{3}{2}\left( {1 - \frac{1}{{2.5}}} \right)P(X£2.5) \\ &= 0.9 \\\end{aligned} \)

Probability that the reaction time is between \({\rm{1}}{\rm{.5}}\) and \({\rm{2}}{\rm{.5sec}}\) is denoted by\({\rm{P(1}}{\rm{.5 < X < 2}}{\rm{.5)}}\). Then using the given cdf from part(a), we can write:

\(\begin{aligned}P(1.5 < X < 2.5) &= F(2.5) - F(1.5) \\ &= \frac{3}{2}\left( {1 - \frac{1}{{2.5}}} \right) - \frac{3}{2}\left( {1 - \frac{1}{{1.5}}} \right)\\&= (0.9) - (0.5)P(1.5 < X < 2.5) \\ &= 0.4 \\\end{aligned} \)

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

P(X£a) = F(a)

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

P(a£ X£ sb) = F(b) - F(a)

(c)

The expected reaction time can be given as:

\(\begin{aligned}E(X){\text{ }}\ &= \int_{ - \yen}^\yen x \times f(x) \times dx \\ &= \int_1^3 x \times \left( {\frac{3}{{2 \times {x^2}}}} \right) \times dx \\ &= \frac{3}{2} \times \int_1^3 {\frac{1}{x}} \times dx \\ &= \frac{3}{2} \times [\ln (x)]_1^3 \\ &= \frac{3}{2} \times [\ln (3) - \ln (1)] \\&= \frac{3}{2} \times [\ln (3) - 0]E(X) \\ &= 1.648 \\\end{aligned} \)

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

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

(d) For the pdf \({\rm{f(x)}}\); to calculate variance, we first calculate \({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\) :

\(\begin{aligned}{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)&=\int_{{\rm{ - \yen}}}^{\rm{\yen}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times f(x) \times dx}}\\&=\int_{\rm{1}}^{\rm{3}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times }}\left( {\frac{{\rm{3}}}{{{\rm{2 \times }}{{\rm{x}}^{\rm{2}}}}}} \right){\rm{ \times dx}}\\&=\frac{{\rm{3}}}{{\rm{2}}}{\rm{ \times }}\int_{\rm{1}}^{\rm{3}} {\rm{d}} {\rm{x}}\\&=\frac{{\rm{3}}}{{\rm{2}}}{\rm{ \times (x)}}_{\rm{1}}^{\rm{3}}\\&=\frac{{\rm{3}}}{{\rm{2}}}{\rm{ \times (3 - 1)E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\\&=3\end{aligned}\)

As we have already calculated \({\rm{E(X)}}\)in the last part: \({\rm{E(X) = 1}}{\rm{.648}}\), hence we use following proposition:

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

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

Using this, we can write:

\(\begin{aligned}{l}{{\rm{\sigma }}_{\rm{x}}}&=\sqrt {{\rm{3 - (1}}{\rm{.648}}{{\rm{)}}^{\rm{2}}}} \\{{\rm{\sigma }}_{\rm{x}}}&=0{\rm{.533}}\end{aligned}\)

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

(e)

Let \({\rm{h(X)}}\) be the time that the light is on as a function of reaction time\({\rm{X}}\). Then using the given data, we can say that:

\({\rm{h(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < 1}}{\rm{.5}}}\\{{\rm{x - 1}}{\rm{.5}}}&{{\rm{1}}{\rm{.5£ x£ 2}}{\rm{.5}}}\\{\rm{1}}&{{\rm{x > 2}}{\rm{.5}}}\end{array}} \right.\)

Then the expected amount of time that the light remains lit can be given as:

\(\begin{aligned}E(h(x))&=\int_{{\rm{ - \yen}}}^{\rm{\yen}} {\rm{h}} {\rm{(x) \times f(x) \times dx}}\\&=\int_{{\rm{ - \yen}}}^{{\rm{1}}{\rm{.5}}} {{\rm{(0)}}} {\rm{ \times dx + }}\int_{{\rm{1}}{\rm{.5}}}^{{\rm{2}}{\rm{.5}}} {{\rm{(x - 1}}{\rm{.5)}}} {\rm{ \times }}\frac{{\rm{3}}}{{{\rm{2 \times }}{{\rm{x}}^{\rm{2}}}}}{\rm{ \times dx + }}\int_{{\rm{2}}{\rm{.5}}}^{\rm{3}} {{\rm{(1)}}} {\rm{ \times }}\frac{{\rm{3}}}{{{\rm{2 \times }}{{\rm{x}}^{\rm{2}}}}}{\rm{ \times dx + }}\int_{\rm{3}}^{\rm{\yen}} {{\rm{(0)}}} {\rm{ \times dx}}\\&=\int_{{\rm{1}}{\rm{.5}}}^{{\rm{2}}{\rm{.5}}} {\frac{{\rm{3}}}{{{\rm{2 \times x}}}}} {\rm{ \times dx - 1}}{\rm{.5}}\int_{{\rm{1}}{\rm{.5}}}^{{\rm{2}}{\rm{.5}}} {\frac{{\rm{3}}}{{{\rm{2 \times }}{{\rm{x}}^{\rm{2}}}}}} {\rm{ \times dx + }}\int_{{\rm{2}}{\rm{.5}}}^{\rm{3}} {\frac{{\rm{3}}}{{{\rm{2 \times }}{{\rm{x}}^{\rm{2}}}}}} {\rm{ \times dx}}\\&=\frac{{\rm{3}}}{{\rm{2}}}{\rm{(ln(x))}}_{{\rm{1}}{\rm{.5}}}^{{\rm{2}}{\rm{.5}}}{\rm{ - 1}}{\rm{.5}}\left( {\frac{{{\rm{ - 3}}}}{{{\rm{2 \times x}}}}} \right)_{{\rm{1}}{\rm{.5}}}^{{\rm{2}}{\rm{.5}}}{\rm{ + }}\left( {\frac{{{\rm{ - 3}}}}{{{\rm{2 \times x}}}}} \right)_{{\rm{2}}{\rm{.5}}}^{\rm{3}}\left( {\frac{{\rm{1}}}{{{\rm{2}}{\rm{.5}}}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{1}}{\rm{.5}}}}} \right){\rm{ - }}\frac{{\rm{3}}}{{\rm{2}}}\left( {\frac{{\rm{1}}}{{\rm{3}}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{2}}{\rm{.5}}}}} \right)\\&=\frac{{\rm{3}}}{{\rm{2}}}{\rm{(ln(2}}{\rm{.5) - ln(1}}{\rm{.5)) + (1}}{\rm{.5)}}\frac{{\rm{3}}}{{\rm{2}}}\left( {\frac{{\rm{1}}}{{{\rm{2}}{\rm{.5}}}}} \right)\\&=0{\rm{.7662 - 0}}{\rm{.6 + 0}}{\rm{.1E(h(x))}}\\&=0{\rm{.2662}}\end{aligned}\)

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

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