/*! 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} Q71E In Exercise \({\rm{66}}\), the w... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Q71E (page 243)

In Exercise \({\rm{66}}\), the weight of the beam itself contributes to the bending moment. Assume that the beam is of uniform thickness and density so that the resulting load is uniformly distributed on the beam. If the weight of the beam is random, the resulting load from the weight is also random; denote this load by \({\rm{W}}\) (kip-ft).

a. If the beam is\({\rm{12 }}\)ft long, \({\rm{W}}\)has mean \({\rm{1}}{\rm{.5}}\)and standard deviation \({\rm{.25}}\), and the fixed loads are as described in part (a) of Exercise \({\rm{66}}\), what are the expected value and variance of the bending moment? (Hint: If the load due to the beam were w kip-ft, the contribution to the bending moment would be \(W\grave{o} _0^{12}nxdx\)

b. If all three variables (\({{\rm{X}}_{\rm{1}}}\),\({{\rm{X}}_{\rm{2}}}\), and \({\rm{W}}\)) are normally distributed, what is the probability that the bending moment will be at most \({\rm{200}}\) kip-ft?

Short Answer

Expert verified

\({\rm{\;a}}{\rm{.\;E(B) = 158;V(B) = 430}}{\rm{.25;\;b}}{\rm{.\;P(B\pounds200) = 0}}{\rm{.9788}}{\rm{.\;}}\)

Step by step solution

01

Definition of standard deviation

The square root of the variance is the standard deviation of a random variable, sample, statistical population, data collection, or probability distribution. It is less resilient in practice than the average absolute deviation, but it is algebraically easier.

02

Determining the expected value and variance of the bending moment

Assign \({\rm{B}}\)to the random variable that describes the bending moment. The bending moment is denoted by the symbol

(\({\rm{B = }}{{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}{\rm{ + W\grave{o} }}_{\rm{0}}^{{\rm{12}}}{\rm{nxdx = }}{{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}{\rm{ + 72W}}\))

where the integral is calculated as

(\({\rm{\grave{o}}}_{\rm{0}}^{{\rm{12}}}{\rm{nxdx = }}\left. {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right|_{\rm{0}}^{{\rm{12}}}{\rm{ = 72}}\))

The values \({{\rm{a}}_{\rm{1}}}{\rm{,}}{{\rm{a}}_{\rm{2}}}\)are presented in the exercise's portion \({\rm{(a)}}\).Therefore,

\({\rm{B = 5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}{\rm{ + 72W}}\)

The bending moment's predicted values are

\(\begin{array}{*{20}{c}}{{\rm{E(B) = E}}\left( {{\rm{5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}{\rm{ + 72W}}} \right){\rm{ = 5E}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ + 10E}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ + 72E(W)}}}\\{\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{5 \times 2 + 10 \times 4 + 72 \times 1}}{\rm{.5 = 158}}}\end{array}{\rm{ }}\)

(1): the mean values are given in the exercise, as well as \({\rm{E(W) = 1}}{\rm{.5}}\)

The bending moment variance is

\(\begin{array}{*{20}{c}}{{\rm{V(B) = V}}\left( {{\rm{5}}{{\rm{X}}_{\rm{1}}}{\rm{ + 10}}{{\rm{X}}_{\rm{2}}}{\rm{ + 72W}}} \right)\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {{\rm{5}}^{\rm{2}}}{\rm{V}}\left( {{{\rm{X}}_{\rm{1}}}} \right){\rm{ + 1}}{{\rm{0}}^{\rm{2}}}{\rm{V}}\left( {{{\rm{X}}_{\rm{2}}}} \right){\rm{ + 7}}{{\rm{2}}^{\rm{2}}}{\rm{V(W)}}}\\{\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {{\rm{5}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.}}{{\rm{5}}^{\rm{2}}}{\rm{ + 1}}{{\rm{0}}^{\rm{2}}}{\rm{ \times }}{{\rm{1}}^{\rm{2}}}{\rm{ + 7}}{{\rm{2}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.2}}{{\rm{5}}^{\rm{2}}}{\rm{ = 430}}{\rm{.25}}}\end{array}\)

(2): the random variables are unrelated, (3): the standard deviations are provided in the exercise, \({\rm{ V(W) = 0}}{\rm{.2}}{{\rm{5}}^{\rm{2}}}\)

03

Determining the probability that the bending moment will be at most \({\rm{200}}\) kip-ft

Assume that the random variables you've been given are regularly distributed. The following statement is correct:

\(\begin{array}{*{20}{c}}{{\rm{P(B拢200)}}\mathop {\rm{ = }}\limits^{{\rm{(4)}}} {\rm{P}}\left( {\frac{{{\rm{B - E(B)}}}}{{{{\rm{\sigma }}_{\rm{B}}}}}{\rm{拢 }}\frac{{{\rm{200 - 158}}}}{{{\rm{20}}{\rm{.74}}}}} \right)}\\{{\rm{ = P(Z拢 2}}{\rm{.03)}}\mathop {\rm{ = }}\limits^{{\rm{(5)}}} {\rm{0}}{\rm{.9788}}}\end{array}{\rm{v}}\)

(4): Because the three supplied variables are regularly distributed, \({\rm{B}}\) is normally distributed. To produce a standard normal distribution, standardise \({\rm{B}}\).\({\rm{Z}}\)

(5): from the appendix's normal probability table. Software can also be used to calculate the likelihood.

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

Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable鈥擷 for the right tire and Y for the left tire, with joint pdf fsx, yd 5 5 Ksx2 1 y2 d 20 # x # 30, 20 # y # 30 0 otherwise

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{K(}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{)}}}&{,{\rm{20}} \le {\rm{x}} \le {\rm{30,20}} \le {\rm{y}} \le {\rm{30}}}\\{\rm{0}}&{,{\rm{ otherwise }}}\end{array}} \right.\)

a. What is the value of K?

b. What is the probability that both tires are underfilled?

c. What is the probability that the difference in air pressure between the two tires is at most 2 psi?

d. Determine the (marginal) distribution of air pressure in the right tire alone.

e. Are X and Y independent rv鈥檚?z

Question: The number of customers waiting for gift-wrap service at a department store is an rv X with possible values \({\rm{0,1,2,3,4}}\)and corresponding probabilities \({\rm{.1,}}{\rm{.2,}}{\rm{.3,}}{\rm{.25,}}{\rm{.15}}{\rm{.}}\)A randomly selected customer will have \({\rm{1,2}}\),or \({\rm{3}}\) packages for wrapping with probabilities \({\rm{.6,}}{\rm{.3,}}\)and \({\rm{.1,}}\)respectively. Let \({\rm{Y = }}\)the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer).

a. Determine \({\rm{P(X = 3,Y = 3)}}\), i.e., \({\rm{P(3,3)}}\).

b. Determine \({\rm{p(4,11)}}\).

A stockroom currently has \({\rm{30}}\) components of a certain type, of which \({\rm{8}}\) were provided by supplier \({\rm{1,10}}\)by supplier \({\rm{2}}\) , and \({\rm{12}}\) by supplier \({\rm{3}}\). Six of these are to be randomly selected for a particular assembly. Let \({\rm{X = }}\) the number of supplier l's components selected, \({\rm{Y = }}\) the number of supplier \({\rm{2}}\) 's components selected, and \({\rm{p(x,y)}}\) denote the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\).

a. What is \({\rm{p(3,2)}}\) ? (Hint: Each sample of size \({\rm{6}}\) is equally likely to be selected. Therefore, \({\rm{p(3,2) = }}\) (number of outcomes with \({\rm{X = 3}}\) and \({\rm{Y = 2)/(}}\) total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.)

b. Using the logic of part (a), obtain \({\rm{p(x,y}}\) ). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories.)

Answer

Reconsider the minicomputer component lifetimes \({\rm{X}}\) and\({\rm{Y}}\). Determine\({\rm{E(XY)}}\). What can be said about \({\rm{Cov(X,Y)}}\) and\({\rm{\rho }}\)?

A particular brand of dishwasher soap is sold in three sizes: \({\rm{25oz,40oz}}\), and\({\rm{65oz}}\). Twenty percent of all purchasers select a\({\rm{25 - 0z}}\)box,\({\rm{50\% }}\)select a\({\rm{40 - 0z}}\)box, and the remaining\({\rm{30\% }}\)choose a\({\rm{65}}\)-oz box. Let\({{\rm{X}}_{\rm{1}}}\)and\({{\rm{X}}_{\rm{2}}}\)denote the package sizes selected by two independently selected purchasers.

a. Determine the sampling distribution of\({\rm{\bar X}}\), calculate\({\rm{E(\bar X)}}\), and compare to\({\rm{\mu }}\).

b. Determine the sampling distribution of the sample variance\({{\rm{S}}^{\rm{2}}}\), calculate\({\rm{E}}\left( {{{\rm{S}}^{\rm{2}}}} \right)\), and compare to\({{\rm{\sigma }}^{\rm{2}}}\).
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.