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) It is given that \({\rm{X}}\) is the rv denoting the lateral position error (nautical miles), which can be either negative or positive. The pdf \({\rm{f(x)}}\)is given to us as:

\({\rm{f(x) = (0}}{\rm{.1)}}{{\rm{e}}^{{\rm{ - 0}}{\rm{.2|x|}}}}\)

If we remove modulus sign then it can also be written as:

\(f(x) = \left\{ {\begin{aligned}{}{(0.1){e^{0.2x}}}&{x < 0} \\{(0.1){e^{ - 0.2x}}}&{x0}\end{aligned}} \right.\)

It will be a valid pdf if it integrates to\({\rm{\;1}}\). Upon integrating pdf\({\rm{f(x)}}\), we obtain:

\(\begin{aligned}\int_{ - ¥}^¥ f (x) \times dx &= \int_{ - ¥}^0 {(0.1)} {e^{0.2x}} \times dx + \int_0^¥ {(0.1)} {e^{ - 0.2x}} \times dx \\&= (0.1)\left[ {\frac{{{e^{0.2x}}}}{{0.2}}} \right]_{ - ¥}^0 + (0.1)\left[ {\frac{{ - {e^{ - 0.2x}}}}{{0.2}}} \right]_0^¥ \\&= (0.1)\left[ {\frac{1}{{0.2}} - 0} \right] + (0.1)\left[ {0 - \left( {\frac{{ - 1}}{{0.2}}} \right)} \right]_0^¥\\&= \frac{{0.1}}{{0.2}} + \frac{{0.1}}{{0.2}}\int_{ - ¥}^¥ f (x) \times dx \\&= 1 \\\end{aligned}\)

Hence \({\rm{f(x)}}\) is a legitimate pdf.

It is plotted in the figure given below.

(b) 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

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

pdf \({\rm{f(x)}}\) is given to us as:

\(f(x) = \left\{ {\begin{aligned}{{}{}}{(0.1){e^{0.2x}}}&{x < 0} \\{(0.1){e^{ - 0.2x}}}&{x30}\end{aligned}} \right.\)

For any number \({\rm{x}}\) in the interval \(( - ¥,0)\)

\(\begin{aligned}F(x) &= \int_{ - ¥}^x f (y)\times dy= \int_{ - ¥}^x {(0.1)} {e^{0.2y}} \times dy \\&= (0.1)\left[ {\frac{{{e^{0.2y}}}}{{0.2}}} \right]_{ - ¥}^x \\&= (0.1)\left[ {\frac{{{e^{0.2x}}}}{{0.2}} - 0} \right] \\F(x)\left. {= (0.5){e^{0.2x}}} \right]\\\end{aligned} \)

For any number \({\rm{x}}\) in the interval \((0,\infty )\)

\(\begin{aligned}F(x) &= \int_{ - ¥}^x f (y) \times dy \\&= \int_{ - ¥}^0 {(0.1)} {e^{0.2y}} \times dy + \int_0^x {(0.1)} {e^{ - 0.2y}} \times dy \\&= (0.1)\left[ {\frac{{{e^{0.2y}}}}{{0.2}}} \right]_{ - ¥}^0 + (0.1)\left[ {\frac{{ - {e^{ - 0.2y}}}}{{0.2}}} \right]_0^x \\&= (0.1)\left[ {\frac{1}{{0.2}} - 0} \right] + (0.1)\left[ {\frac{{ - {e^{ - 0.2x}}}}{{0.2}} - \left( {\frac{{ - 1}}{{0.2}}} \right)} \right]_0^¥ \\&= \frac{{0.1}}{{0.2}} - \frac{{0.1}}{{0.2}} \times {e^{ - 0.2x}} + \frac{{0.1}}{{0.2}}F(x) \\&= 1 - (0.5) \times {e^{ - 0.2x}} \\\end{aligned}\)

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

\({\rm{F(x)}}\)is plotted in the figure given below:

(c) Here we will make use of following propositions:

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\({\rm{cdfF(x)}}\). Then for any two numbers a and \({\rm{b}}\) with\({\rm{a < b}}\),

\(P(a£X£b) = F(b) - F(a)\)

To compute \(P(X£0)\), we can write:

\(\begin{aligned}P(X£0)&= F(0) \\&= 1 - (0.5){e^{ - 0.2(0)}} \\&= 1 - 0.5P(X£0) \\&= 0.5 \\\end{aligned}\)

To compute \(P(X£2)\) , we can write:

\(\begin{aligned}P(X£2)&= F(2) \\&= 1 - (0.5){e^{ - 0.2(2)}} \\&= 1 - (0.5){e^{ - 0.4}}P(X£2) \\&= 0.6648 \\\end{aligned} \)

To compute ([P( - 1£X£2)\), we can write:

\(\begin{aligned}P( - 1£X£2)&= F(2) - F( - 1) \\&= \left[ {1 - (0.5){e^{ - 0.2(2)}}} \right]-\left[ {(0.5){e^{0.2( - 1)}}} \right] \\&= 1 - (0.5){e^{ - 0.4}} - (0.5){e^{ - 0.2}}P( - 1£X£2) \\&= 0.2555 \\\end{aligned} \)

The probability that an error of more than \({\rm{2}}\)miles is made can be represented as\({\rm{P(x < - 2 or x > 2)}}\).

This event is the complement of the event that an error of less than or equal to \({\rm{2}}\) miles is made, hence using the complement rule of probability, we can write:

\(\begin{aligned}P(x < - 2{\text{ }}or{\text{ }}x > 2)&= 1 - P( - 2£X£2) \\&= 1 - [F(2) - F( - 2)] \\&= 1 - \left[ {\left( {1 - (0.5){e^{ - 0.2(2)}}} \right) - \left( {(0.5){e^{0.2( - 2)}}} \right)} \right] \\&= 1 - \left[ {\left( {1 - (0.5){e^{ - 0.4}}} \right) - \left( {(0.5){e^{ - 0.4)}}} \right)} \right] \\&= 1 - [0.6648 - 0.3352]P(x < - 2{\text{ }}or{\text{ }}x > 2) \\&= 0.6704 \\ \end{aligned} \)

Definition (Complement rule of probability): If \({\rm{\bar A}}\) is the complement of an event A, then

\({\rm{P(\bar A) = 1 - P(A)}}\)