/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q27E When a dart is thrown at a circu... [FREE SOLUTION] | 91影视

91影视

When a dart is thrown at a circular target, consider the location of the landing point relative to the bull鈥檚 eye. Let \({\rm{X}}\) be the angle in degrees measured from the horizontal, and assume that \({\rm{X}}\) is uniformly distributed on \(\left( {{\rm{0, 360}}} \right){\rm{.}}\)Define\({\rm{Y}}\)to be the transformed variable \({\rm{Y = h(X) = (2\pi /360)X - \pi ,}}\) so \({\rm{Y}}\) is the angle measured in radians and\({\rm{Y}}\)is between \({\rm{ - \pi and \pi }}\). Obtain \({\rm{E(Y)}}\)and\({{\rm{\sigma }}_{\rm{y}}}\)by first obtaining E(X) and \({{\rm{\sigma }}_{\rm{X}}}\), and then using the fact that \({\rm{h(X)}}\) is a linear function of \({\rm{X}}\).

Short Answer

Expert verified

\(\begin{array}{*{20}{c}}{{\rm{E(X) = 180;}}{{\rm{\sigma }}_{\rm{x}}}{\rm{ = 103}}{\rm{.9230}}}\\{{\rm{E(Y) = 0;}}{{\rm{\sigma }}_{\rm{y}}}{\rm{ = (0}}{\rm{.57735)\pi }}}\end{array}\)

Step by step solution

01

Definition of probability

The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

 Calculating when a dart when it’s thrown at a circular target, consider the location of the landing point relative to the bull’s eye

Because \({\rm{X}}\)is uniformly distributed on \({\rm{(0,360)}}\), its pdf \({\rm{(x)}}\) can be written as follows for all \({\rm{x}}\) in this interval:

\({\rm{f(x) = }}\frac{{\rm{1}}}{{{\rm{360 - 0}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{360}}}}\)

Overall, \({\rm{f(x)}}\) can be written as: \({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{1}}}{{{\rm{360}}}}{\rm{0\poundsx\pounds360}}}\\{{\rm{\;0 otherwise\;}}}\end{array}} \right.\)

\({\rm{X}}\)'s expected value can be expressed as: \(\begin{array}{*{20}{c}}{{\rm{E(X) = \`o }}_{{\rm{ - \currency}}}^{\rm{\currency}}{\rm{nx \times f(x) \times dx}}}\\{{\rm{ = \`o }}_{\rm{0}}^{{\rm{360}}}{\rm{nx \times }}\frac{{\rm{1}}}{{{\rm{360}}}}{\rm{ \times dx}}}\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{360}}}}{\rm{\`o }}_{\rm{0}}^{{\rm{360}}}{\rm{nx \times dx}}}\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{360}}}}\left( {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{0}}^{{\rm{360}}}}\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{720}}}}\left( {{\rm{36}}{{\rm{0}}^{\rm{2}}}{\rm{ - }}{{\rm{0}}^{\rm{2}}}} \right)}\\{{\rm{E(X) = 180}}}\end{array}\)

Definition: A continuous \({\rm{rv's}}\) expected or mean value. \({\rm{f(x)}}\) is \({\rm{X}}\)with pdf

\({\rm{E}}\left( {{{\rm{X}}^{}}} \right)\)\({\rm{ = }}\mathop {\rm{\`o }}\nolimits_{{\rm{ - \currency}}}^{\rm{\currency}} {{\rm{x}}^{\rm{2}}}{\rm{ \times f(x) \times dx}}\)

Now consider the following argument:

The standard deviation \({{\rm{\sigma }}_{\rm{x}}}\) and variance \({\rm{V(X)}}\)of a \({\rm{rv's}}\) \({\rm{X}}\)with a specified pdf can be stated as: \(\begin{array}{*{20}{c}}{{\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}}\\{{{\rm{\sigma }}_{\rm{x}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} }\end{array}\)

We begin by calculating \({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\).

\(\begin{array}{*{20}{c}}{{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)}&{{\rm{ = }}\mathop {\rm{\`o }}\nolimits_{{\rm{ - \currency}}}^{\rm{\currency}} {{\rm{x}}^{\rm{2}}}{\rm{ \times f(x) \times dx}}}\\{}&{{\rm{ = }}\mathop {\rm{\`o }}\nolimits_{\rm{0}}^{{\rm{360}}} {{\rm{x}}^{\rm{2}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{360}}}}{\rm{ \times dx}}}\\{}&{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{360}}}}\mathop {\rm{\`o }}\nolimits_{\rm{0}}^{{\rm{360}}} {{\rm{x}}^{\rm{2}}}{\rm{ \times dx}}}\\{}&{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{360}}}}\left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{\rm{0}}^{{\rm{360}}}}\\{}&{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{1080}}}}\left( {{\rm{36}}{{\rm{0}}^{\rm{3}}}{\rm{ - }}{{\rm{0}}^{\rm{2}}}} \right)}\\{{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)}&{{\rm{ = 43200}}}\end{array}\)

03

Determining the probability using proprsition

Using the above proposition, we can now write:

\(\begin{array}{*{20}{c}}{{\sigma _x}}&{ = \sqrt {V(X)} }\\{}&{ = \sqrt {E\left( {{X^2}} \right) - {{(E(X))}^2}} }\\{}&{\sqrt {43200 - {{(180)}^2}} }\end{array}\)

\(\sqrt {43200 - 32400} \)

\(\sqrt {10800} \)

\({\sigma _x} = 103.9230\)

We can apply the following proposition because \({\rm{Y( = h(X)}}\) is a linear function of \({\rm{X}}\):

The expected value and standard deviation of \({\rm{h(X)}}\) meet the following conditions for \({\rm{h(X) = aX + b}}\).

\(\begin{array}{*{20}{c}}{{\rm{E(h(X)) = aE(X) + b}}}\\{{{\rm{\sigma }}_{{\rm{h(x)}}}}{\rm{ = a}}{{\rm{\sigma }}_{\rm{x}}}}\end{array}\)

We are informed that:

\({\rm{Y = }}\frac{{{\rm{2\pi }}}}{{{\rm{360}}}}{\rm{ \times X - \pi }}\)

As a result, the predicted value of \({\rm{Y}}\) is:

\(\begin{array}{*{20}{c}}{{\rm{E(Y)}}}&{{\rm{ = }}\frac{{{\rm{2\pi }}}}{{{\rm{360}}}}{\rm{ \times E(X) - \pi }}}\\{}&{{\rm{ = }}\frac{{{\rm{2\pi }}}}{{{\rm{360}}}}{\rm{ \times 180 - \pi }}}\\{}&{{\rm{ = }}\frac{{{\rm{360}}}}{{{\rm{360}}}}{\rm{ \times \pi - \pi }}}\\{{\rm{E(Y)}}}&{{\rm{ = 0}}}\end{array}\)

The standard deviation of \({\rm{Y}}\) is calculated as follows:

\(\begin{array}{*{20}{c}}{{{\rm{\sigma }}_{\rm{y}}}{\rm{ = }}\frac{{{\rm{2\pi }}}}{{{\rm{360}}}}{\rm{ \times }}{{\rm{\sigma }}_{\rm{x}}}}\\{{\rm{ = }}\frac{{{\rm{2\pi }}}}{{{\rm{360}}}}{\rm{ \times (103}}{\rm{.923)}}}\\{{{\rm{\sigma }}_{\rm{y}}}{\rm{ = (0}}{\rm{.57735)\pi }}}\end{array}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Example \({\rm{4}}{\rm{.5}}\) introduced the concept of time headway in traffic flow and proposed a particular distribution for \({\rm{X = }}\) the headway between two randomly selected consecutive cars (sec). Suppose that in a different traffic environment, the distribution of time headway has the form

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{k}}}{{{{\rm{x}}^{\rm{4}}}}}}&{{\rm{x > 1}}}\\{\rm{0}}&{{\rm{x}} \le {\rm{1}}}\end{array}} \right.\)

a. Determine the value of \({\rm{k}}\) for which \({\rm{f(x)}}\) is a legitimate pdf. b. Obtain the cumulative distribution function. c. Use the cdf from (b) to determine the probability that headway exceeds \({\rm{2}}\) sec and also the probability that headway is between \({\rm{2}}\) and \({\rm{3}}\) sec. d. Obtain the mean value of headway and the standard deviation of headway. e. What is the probability that headway is within \({\rm{1}}\) standard deviation of the mean value?

Suppose the proportion \({\rm{X}}\) of surface area in a randomly selected quadrat that is covered by a certain plant has a standard beta distribution with \({\rm{\alpha = 5}}\)and \({\rm{\beta = 2}}\).

a. Compute \({\rm{E(X)}}\) and \({\rm{V(X)}}\).

b. Compute \({\rm{P(X}} \le {\rm{.2)}}\).

c. Compute \({\rm{P(}}{\rm{.2}} \le {\rm{X}} \le {\rm{.4)}}\).

d. What is the expected proportion of the sampling region not covered by the plant?

Show that the relationship between a general normal percentile and the corresponding z percentile is as stated in this section.

The article 鈥淪econd Moment Reliability Evaluation vs. Monte Carlo Simulations for Weld Fatigue Strength鈥 (Quality and Reliability Engr. Intl., \({\rm{2012: 887 - 896}}\)) considered the use of a uniform distribution with \({\rm{A = }}{\rm{.20}}\) and \({\rm{B = 4}}{\rm{.25}}\) for the diameter \({\rm{X}}\) of a certain type of weld (mm).

a. Determine the pdf of \({\rm{X}}\) and graph it.

b. What is the probability that diameter exceeds \({\rm{3 mm}}\)?

c. What is the probability that diameter is within \({\rm{1 mm}}\) of the mean diameter?

d. For any value \({\rm{a}}\) satisfying \({\rm{.20 < a < a + 1 < 4}}{\rm{.25}}\), what is \({\rm{P(a < X < a + 1)}}\)?

Let \({\rm{X}}\) have a standard beta density with parameters \({\rm{\alpha }}\) and \({\rm{\beta }}\).

a. Verify the formula for \({\rm{ E}}\left( {\rm{X}} \right)\) given in the section.

b. Compute \({\rm{E}}\left( {{{\left( {{\rm{1 - X}}} \right)}^{\rm{m}}}} \right){\rm{.}}\) If \({\rm{X}}\) represents the proportion of a substance consisting of a particular ingredient, what is the expected proportion that does not consist of this ingredient?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.