![Intro Stats](https://www.bartleby.com/isbn_cover_images/9780321825278/9780321825278_largeCoverImage.gif)
Convert the table to relative frequency table.
![Check Mark](/static/check-mark.png)
Answer to Problem 1E
The relative frequency table is as follows,
Car type |
Relative frequency |
Subcompact and Mini | 11.8% |
Compact | 31.5% |
Intermediate | 33.5% |
Full | 21.8% |
Unknown | 1.4% |
Total | 100% |
Explanation of Solution
Calculation:
The given information is that, the data represents the car types with number of passenger car occupants killed in accidents in 2009.
Relative frequency:
The general formula for the relative frequency is,
Substituting the type of Subcompact and Mini car and the number of passenger car occupants killed “1,545” as the frequency and total of passenger car occupants killed as “13,095” as total frequency in relative frequency.
Similarly, the relative frequencies for the remaining car types are obtained below:
Car type | Frequency |
Relative frequency |
Subcompact and Mini | 1,545 | 11.8% |
Compact | 4,128 | 31.5% |
Intermediate | 4,393 | 33.5% |
Full | 2,849 | 21.8% |
Unknown | 180 | 1.4% |
Total | 13,095 | 100% |
Want to see more full solutions like this?
Chapter 2 Solutions
Intro Stats
- I need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forward310015 K Question 9, 5.2.28-T Part 1 of 4 HW Score: 85.96%, 49 of 57 points Points: 1 Save of 6 Based on a poll, among adults who regret getting tattoos, 28% say that they were too young when they got their tattoos. Assume that six adults who regret getting tattoos are randomly selected, and find the indicated probability. Complete parts (a) through (d) below. a. Find the probability that none of the selected adults say that they were too young to get tattoos. 0.0520 (Round to four decimal places as needed.) Clear all Final check Feb 7 12:47 US Oarrow_forwardhow could the bar graph have been organized differently to make it easier to compare opinion changes within political partiesarrow_forward
- 30. An individual who has automobile insurance from a certain company is randomly selected. Let Y be the num- ber of moving violations for which the individual was cited during the last 3 years. The pmf of Y isy | 1 2 4 8 16p(y) | .05 .10 .35 .40 .10 a.Compute E(Y).b. Suppose an individual with Y violations incurs a surcharge of $100Y^2. Calculate the expected amount of the surcharge.arrow_forward24. An insurance company offers its policyholders a num- ber of different premium payment options. For a ran- domly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows: F(x)=0.00 : x < 10.30 : 1≤x<30.40 : 3≤ x < 40.45 : 4≤ x <60.60 : 6≤ x < 121.00 : 12≤ x a. What is the pmf of X?b. Using just the cdf, compute P(3≤ X ≤6) and P(4≤ X).arrow_forward59. At a certain gas station, 40% of the customers use regular gas (A1), 35% use plus gas (A2), and 25% use premium (A3). Of those customers using regular gas, only 30% fill their tanks (event B). Of those customers using plus, 60% fill their tanks, whereas of those using premium, 50% fill their tanks.a. What is the probability that the next customer will request plus gas and fill the tank (A2 B)?b. What is the probability that the next customer fills the tank?c. If the next customer fills the tank, what is the probability that regular gas is requested? Plus? Premium?arrow_forward
- 38. Possible values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 with corresponding probabilities .15, .35, .35, and .15, respectively. a. Calculate E(X) and then E(5 - X).b. Would the repair facility be better off charging a flat fee of $75 or else the amount $[150/(5 - X)]? [Note: It is not generally true that E(c/Y) = c/E(Y).]arrow_forward74. The proportions of blood phenotypes in the U.S. popula- tion are as follows:A B AB O .40 .11 .04 .45 Assuming that the phenotypes of two randomly selected individuals are independent of one another, what is the probability that both phenotypes are O? What is the probability that the phenotypes of two randomly selected individuals match?arrow_forward53. A certain shop repairs both audio and video compo- nents. Let A denote the event that the next component brought in for repair is an audio component, and let B be the event that the next component is a compact disc player (so the event B is contained in A). Suppose that P(A) = .6 and P(B) = .05. What is P(BA)?arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman
![Text book image](https://www.bartleby.com/isbn_cover_images/9781119256830/9781119256830_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305251809/9781305251809_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305504912/9781305504912_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9780134683416/9780134683416_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781319042578/9781319042578_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781319013387/9781319013387_smallCoverImage.gif)