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In contrast to probability and possibility, which both point to objective reality, plausibility is a totally subjective concept: plausibility can only exist because it is borne by human reasoning. To put it another way, something is only feasible if someone believes it is.

It is given that X is the time it takes a read/write head to locate the desired record on a computer disk memory device once the head has been positioned over the correct track.

It is also given that X is uniformly distributed on the interval\((0,25)\). Hence value of\(f(x)\)in this interval is equal to:

\(f(x) = \frac{1}{{25 - 0}} = \frac{1}{{25}} = 0.04\)

and zero otherwise. Then\(f(x)\)can be written as:

\(f(x) = \left\{ {\begin{array}{*{20}{l}}{0.04}&{0 < x < 5}\\0&{{\rm{ otherwise }}}\end{array}} \right.\)

(a) We have to compute\(P(10 \le X \le 20)\)

\(\begin{array}{l}P(10 \le X \le 20) = \int_{10}^{20} 0 .04 \cdot dx\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.04(x)_{10}^{20}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.04(20 - 10)\\P(10 \le X \le 20) = 0.4\end{array}\)

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

\(P(a \le X \le b) = \int_a^b f (x) \cdot dx\)

(b) We have to compute\(P(10 \le X)\)

\(\begin{array}{l}P(10 \le X) = \int_{10}^\infty f (x) \cdot dx\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_{10}^{25} {(0.04)} \cdot dx + \int_{25}^\infty {(0)} \cdot dx\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.04(x)_{10}^{25}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.04(25 - 10)\\P(10 \le X) = 0.6\end{array}\)

Proposition: Let X be a continuous rv with\(pdff(x)\)and\(cdf\;F(x)\). Then for any number a,

\(P(a \le X) = \int_a^\infty f (x) \cdot dx\)

(c) We recall the definition of cdf of a continuous variable.

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

\(F(x) = P(X \le x) = \int_{ - \infty }^x f (y) \cdot dy\)

we have already derived pdf\(f(x)\)as:

\(f(x) = \left\{ {\begin{array}{*{20}{l}}{0.04}&{0 < x < 25}\\0&{{\rm{ otherwise }}}\end{array}} \right.\)

For any number x between 0 and 25

\(\begin{array}{l}F(X) = \int_0^x {(0.04)} \cdot dy\\\,\,\,\,\,\,\,\,\,\,\,\, = (0.04)\int_0^x d y\\\,\,\,\,\,\,\,\,\,\,\,\, = 0.04(y)_0^x\\\,\,\,\,\,\,\,\,\,\,\,\, = 0.04(x - 0)\\F(X) = 0.04x\end{array}\)

Thus,\(F(X)\)can be given as:

\(F(x) = \left\{ {\begin{array}{*{20}{l}}0&{x < 0}\\{0.04x}&{0 \le x \le 25}\\1&{x > 25}\end{array}} \right.\)

(d) We have already derived pdf\(f(x)\)as:

\(f(x) = \left\{ {\begin{array}{*{20}{l}}{0.04}&{0 < x < 25}\\0&{{\rm{ otherwise }}}\end{array}} \right.\)

The mean value of the given distribution can be given as:

\(\begin{array}{l}E(X) = \int_{ - \infty }^\infty x \cdot f(x) \cdot dx\\\,\,\,\,\,\,\,\,\,\,\,\, = \int_0^{25} x \cdot (0.04) \cdot dx\\\,\,\,\,\,\,\,\,\,\,\,\, = 0.04\int_0^{25} x \cdot dx\\\,\,\,\,\,\,\,\,\,\,\,\, = 0.04\left( {\frac{{{x^2}}}{2}} \right)_0^{25}\\\,\,\,\,\,\,\,\,\,\,\,\, = 0.04\left( {\frac{{{{(25)}^2}}}{2} - \frac{{{{(0)}^2}}}{2}} \right)\\E(X) = 12.5\end{array}\)

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

\(\mu = E(X) = \int_{ - \infty }^\infty x \cdot f(x) \cdot dx\)

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

\(\begin{array}{l}E\left( {{X^2}} \right) = \int_{ - \infty }^\infty {{x^2}} \cdot f(x) \cdot dx\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_0^{25} {{x^2}} \cdot (0.04) \cdot dx\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = (0.04)\int_0^{25} {{x^2}} \cdot dx\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = (0.04)\left( {\frac{{{x^3}}}{3}} \right)_0^{25}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.04}}{3}\left( {{{(25)}^3} - {{(0)}^3}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.04}}{3}(15625)\\E\left( {{X^2}} \right) = 208.33\end{array}\)

As we have already calculated $E(X)$, hence we use following proposition:

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

Using this, we can write:

\(\begin{array}{l}{\sigma _X} = \sqrt {208.33 - {{(12.5)}^2}} \\{\sigma _X} = 7.22\end{array}\)

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

\(E(h(x)) = \int_{ - \infty }^\infty h (x) \cdot f(x) \cdot dx\)