/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

When a dart is thrown at a circular target, consider the location of the landing point relative to the bull’s 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 91Ó°ÊÓ!

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

The accompanying observations are precipitation values during March over a \(30\)-year period in Minneapolis-St. Paul.

\(\begin{array}{*{20}{l}}{.77\;\;1.20\;\;3.00\;\;1.62\;\;2.81\;\;2.48}\\{1.74\;\;.47\;\;3.09\;\;1.31\;\;1.87\;\;\;\;.96}\\{.81\;\;1.43\;\;1.51\;\;\;\;.32\;\;1.18\;\;1.89}\\{1.20 3.37\;\;2.10\;\;\;\;.59\;\;1.35\;\;\;\;.90}\\{1.95 2.20\;\;\;\;.52\;\;\;\;\;.81\;\;4.75\;\;2.05}\end{array}\)

a. Construct and interpret a normal probability plot for this data set. b. Calculate the square root of each value and then construct a normal probability plot based on this transformed data. Does it seem plausible that the square root of precipitation is normally distributed? c. Repeat part (b) after transforming by cube roots.

a. The event \(\left\{ {{X^2} \le y} \right\}\)is equivalent to what event involvingXitself?

b. If \(X\)has a standard normal distribution, use part (a) to write the integral that equals \(P\left( {{X^2} \le y} \right)\). Then differentiate this with respect to \(y\)to obtain the pdf of \({{\rm{X}}^{\rm{2}}}\) (the square of a \({\rm{N(0,1)}}\)variable). Finally, show that \({{\rm{X}}^{\rm{2}}}\)has a chi-squared distribution with \(\nu = 1\) df (see (4.10)). (Hint: Use the following identity.)

\(\frac{d}{{dy}}\left\{ {\int_{a(y)}^{b(y)} f (x)dx} \right\} = f(b(y)) \cdot {b^\prime }(y) - f(a(y)) \cdot {a^\prime }(y)\)

The two-parameter gamma distribution can be generalized by introducing a third parameter \(\gamma ,\)called a threshold or location parameter: replace \({\rm{x}}\)in (4.8) by \(x - \gamma \)and \(x \ge 0\)by \(x \ge \gamma \)This amounts to shifting the density curves in Figure \({\rm{4}}{\rm{.27}}\)so that they begin their ascent or descent at \(\gamma \)rather than 0. The article "Bivariate Flood Frequency Analysis with Historical Information Based on Copulas" ( \({\bf{J}}.\)of Hydrologic Engr., 2013: 1018-1030) employs this distribution to model \(X = \)3-day flood volume \(\left( {{\rm{1}}{{\rm{0}}^{\rm{8}}}{\rm{\;}}{{\rm{m}}^{\rm{3}}}} \right){\rm{.}}\)Suppose that values of the parameters are \({\rm{\alpha = 12,\beta = 7,\gamma = 40}}\) (very close to estimates in the cited article based on past data).

a. What are the mean value and standard deviation of X?

b. What is the probability that flood volume is between 100 and 150?

c. What is the probability that flood volume exceeds its mean value by more than one standard deviation?

d. What is the 95th percentile of the flood volume distribution?

The article "'Response of \({\rm{Si}}{{\rm{C}}_{\rm{i}}}{\rm{/S}}{{\rm{i}}_{\rm{3}}}{\rm{\;}}{{\rm{N}}_{\rm{4}}}\) Composites Under Static and Cyclic Loading-An Experimental and Statistical Analysis" (J. of Engr. Materials and Technology, \({\rm{1997: 186 - 193}}\)) suggests that tensile strength (MPa) of composites under specified conditions can be modeled by a Weibull distribution with \({\rm{\alpha = 9}}\) and\({\rm{\beta = 180}}\).

a. Sketch a graph of the density function.

b. What is the probability that the strength of a randomly selected specimen will exceed \({\rm{175}}\)? Will be between \({\rm{150}}\) and \({\rm{175}}\)?

c. If two randomly selected specimens are chosen and their strengths are independent of one another, what is the probability that at least one has a strength between \({\rm{150}}\) and\({\rm{175}}\)?

d. What strength value separates the weakest \({\rm{10\% }}\) of all specimens from the remaining\({\rm{90\% }}\)?

Suppose only \({\rm{75\% }}\) of all drivers in a certain state regularly wear a seat belt. A random sample of 500 drivers is selected. What is the probability that

a. Between 360 and 400 (inclusive) of the drivers in the sample regularly wear a seat belt?

b. Fewer than 400 of those in the sample regularly wear a seat belt?

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.