/*! 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} Q107SE New car crash tests.聽Refer to t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Q107SE (page 204)

New car crash tests.Refer to the National Highway TrafficSafety Administration (NHTSA) crash tests of new car models, Exercise 2.153 (p. 143). Recall that the NHTSA has developed a 鈥渟tar鈥� scoring system, with results ranging from one star (*) to five stars (). The more stars in the rating, the better the level of crash protection in a head-on collision. A summary of the driver-side star ratings for 98 cars is reproduced in the accompanying Minitab

Printout. Assume that one of the 98 cars is selected at random. State whether each of the following is true or false.

a.The probability that the car has a rating of two stars is 4.

b.The probability that the car has a rating of four or five stars is .7857.

c.The probability that the car has a rating of one star is 0.

d.The car has a better chance of having a two-star rating than of having a five-star rating.

Short Answer

Expert verified
  1. The solution is false.
  2. The solution is true.
  3. The solution is true
  4. The solution is false.

Step by step solution

01

The probability that the car has a rating of two stars is 4.(a)

The solution is false because P(2star)=4.08%=0.048.

Thus, the solution is false.

02

The probability that the car has a rating of four or five stars is .7857.(b)

The solution is true because

P(4鈥塻迟补谤鈥夆赌塷谤鈥夆赌�5鈥塻迟补谤)=60.20%+18.37%=6.020+1.837=7.857

Accordingly, the solution is true.

03

The probability that the car has a rating of one star is 0.(c)

The solution is true, because the data of one star is not given, so the probability is 0.

Hence, the solution is true.

04

The car has a better chance of having a two-star rating than of having a five-star rating.(d)

The solution is false because the probability of two star rating is 0.048 and five star rating is 1.837.

Therefore, the solution is false.

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

Drug testing in athletes.When Olympic athletes are tested for illegal drug use (i.e., doping), the results of a single positive test are used to ban the athlete from competition. Chance(Spring 2004) demonstrated the application of Bayes鈥檚 Rule for making inferences about testosterone abuse among Olympic athletes using the following example: In a population of 1,000 athletes, suppose 100 are illegally using testosterone. Of the users, suppose 50 would test positive for testosterone. Of the nonusers, suppose 9 would test positive.

  1. Given that the athlete is a user, find the probability that a drug test for testosterone will yield a positive result. (This probability represents the sensitivity of the drug test.)
  2. Given the athlete is a nonuser, find the probability that a drug test for testosterone will yield a negative result. (This probability represents the specificityof the drug test.)
  3. If an athlete tests positive for testosterone, use Bayes鈥檚 Rule to find the probability that the athlete is really doping. (This probability represents the positive predictive value of the drug test.)

A number between 1 and 10, inclusive, is randomly chosen, and the events A and B are defined as follows:

A: [The number is even.]

B: [The number is less than 7.]

a. Identify the sample points in the event A鈭�B.

b. Identify the sample points in the event A鈭�B.

c. Which expression represents the event that the number is even or less than 7 or both?

d. Which expression represents the event that the number is both even and less than 7?

Cell phone handoff behaviour. A 鈥渉andoff鈥� is a term used in wireless communications to describe the process of a cell phone moving from the coverage area of one base station to that of another. Each base station has multiple channels (called color codes) that allow it to communicate with the cell phone. The Journal of Engineering, Computing and Architecture (Vol. 3., 2009) published a cell phone handoff behavior study. During a sample driving trip that involved crossing from one base station to another, the different color codes accessed by the cell phone were monitored and recorded. The table below shows the number of times each color code was accessed for two identical driving trips, each using a different cell phone model. (Note: The table is similar to the one published in the article.) Suppose you randomly select one point during the combined driving trips.

Color code

0

5

b

c

Total

Model 1

20

35

40

0

85

Model 2

15

50

6

4

75

Total

35

85

46

4

160

a. What is the probability that the cell phone was using color code 5?

b. What is the probability that the cell phone was using color code 5 or color code 0?

c. What is the probability that the cell phone used was Model 2 and the color code was 0?

Consider the Venn diagram below, were

P(E1)=P(E2)=P(E3)=15,P(E4)=P(E5)=120P(E6)=110,andP(E7)=15

Find each of the following probabilities:

a.P(A)b.P(B)c.P(A鈭�B)d.P(A鈭�B) e.P(Ac)f.P(Bc)g.P(A鈭�Ac)h.P(Ac鈭�B)

Suppose the events B1and B2are mutually exclusive and complementary events, such thatP(B1)=.75andP(B2)=.25 Consider another event A such that role="math" localid="1658212959871" P(AB1)=.3, role="math" localid="1658213029408" P(AB2)=.5.

  1. FindP(B1鈭�A).
  2. FindP(B2鈭�A)
  3. Find P(A) using part a and b.
  4. Findrole="math" localid="1658213127512" P(B1A).
  5. Findrole="math" localid="1658213164846" P(B2A).
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Geometry

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.