![Elementary Statistics (Text Only)](https://www.bartleby.com/isbn_cover_images/9780077836351/9780077836351_largeCoverImage.gif)
Concept explainers
Batting average: The following frequency distribution presents the batting averages of Major League Baseball players in both the American League and the National League who had 300 or more plate appearances during a recent season.
- Construct a frequency histogram for the American League.
- Construct a frequency histogram for the National League.
- Construct a relative frequency distribution for the American League.
- Construct a relative frequency distribution for the National League.
- Construct a relative frequency histogram for the American League.
- Construct a relative frequency histogram for the National League.
- What percentage of American League players had batting averages of 0.300 or more?
- What percentage of National League players had batting averages of 0.300 or more?
- Compare the relative frequency histograms. What is the main difference between the distributions of batting averages in the two leagues?
a.
![Check Mark](/static/check-mark.png)
To construct: A frequency histogram for American League.
Explanation of Solution
Given information:The following frequency distribution presents the batting averages of Major League Baseball players in both the American League and the National League who had 300 or more plate appearances during a recent season.
Batting average | American LeagueFrequency | National LeagueFrequency |
0.180-0.199 | 2 | 2 |
0.200-0.219 | 7 | 7 |
0.220-0.239 | 21 | 20 |
0.240-0.259 | 30 | 29 |
0.260-0.279 | 26 | 32 |
0.280-0.299 | 21 | 30 |
0.300-0.319 | 12 | 17 |
0.320-0.339 | 5 | 5 |
0.340-0.359 | 0 | 1 |
Definition used: Histograms based on frequency distributions are called frequency histogram.
Solution:
The following frequency distribution presents the batting averages of Major League Baseball players in the American Leaguewho had 300 or more plate appearances during a recent season.
Batting average | American LeagueFrequency |
0.180-0.199 | 2 |
0.200-0.219 | 7 |
0.220-0.239 | 21 |
0.240-0.259 | 30 |
0.260-0.279 | 26 |
0.280-0.299 | 21 |
0.300-0.319 | 12 |
0.320-0.339 | 5 |
0.340-0.359 | 0 |
The frequency histogram for American League is given by
b.
![Check Mark](/static/check-mark.png)
To construct: A frequency histogram for National League.
Explanation of Solution
Given information:The following frequency distribution presents the batting averages of Major League Baseball players in both the American League and the National League who had 300 or more plate appearances during a recent season.
Batting average | American LeagueFrequency | National LeagueFrequency |
0.180-0.199 | 2 | 2 |
0.200-0.219 | 7 | 7 |
0.220-0.239 | 21 | 20 |
0.240-0.259 | 30 | 29 |
0.260-0.279 | 26 | 32 |
0.280-0.299 | 21 | 30 |
0.300-0.319 | 12 | 17 |
0.320-0.339 | 5 | 5 |
0.340-0.359 | 0 | 1 |
Definition used: Histograms based on frequency distributions are called frequency histogram.
Solution:
The following frequency distribution presents the batting averages of Major League Baseball players in the American Leaguewho had 300 or more plate appearances during a recent season.
Batting average | National LeagueFrequency |
0.180-0.199 | 2 |
0.200-0.219 | 7 |
0.220-0.239 | 20 |
0.240-0.259 | 29 |
0.260-0.279 | 32 |
0.280-0.299 | 30 |
0.300-0.319 | 17 |
0.320-0.339 | 5 |
0.340-0.359 | 1 |
The frequency histogram for American League is given by
c.
![Check Mark](/static/check-mark.png)
To construct: A relative frequency distribution for American League.
Explanation of Solution
Given information:The following frequency distribution presents the batting averages of Major League Baseball players in both the American League and the National League who had 300 or more plate appearances during a recent season.
Batting average | American LeagueFrequency | National LeagueFrequency |
0.180-0.199 | 2 | 2 |
0.200-0.219 | 7 | 7 |
0.220-0.239 | 21 | 20 |
0.240-0.259 | 30 | 29 |
0.260-0.279 | 26 | 32 |
0.280-0.299 | 21 | 30 |
0.300-0.319 | 12 | 17 |
0.320-0.339 | 5 | 5 |
0.340-0.359 | 0 | 1 |
Formula used:
Solution:
From the given table,
The sum of all frequency for American League is
The table of relative frequency is given by
Batting average | American LeagueFrequency | American LeagueRelative frequency |
0.180-0.199 | 2 | |
0.200-0.219 | 7 | |
0.220-0.239 | 21 | |
0.240-0.259 | 30 | |
0.260-0.279 | 26 | |
0.280-0.299 | 21 | |
0.300-0.319 | 12 | |
0.320-0.339 | 5 | |
0.340-0.359 | 0 |
The relative frequency for the American League is given by
Batting average | American LeagueRelative frequency |
0.180-0.199 | 0.016 |
0.200-0.219 | 0.056 |
0.220-0.239 | 0.169 |
0.240-0.259 | 0.242 |
0.260-0.279 | 0.210 |
0.280-0.299 | 0.169 |
0.300-0.319 | 0.097 |
0.320-0.339 | 0.040 |
0.340-0.359 | 0.000 |
d.
![Check Mark](/static/check-mark.png)
To construct: A relative frequency distribution for National League.
Explanation of Solution
Given information:The following frequency distribution presents the batting averages of Major League Baseball players in both the American League and the National League who had 300 or more plate appearances during a recent season.
Batting average | American LeagueFrequency | National LeagueFrequency |
0.180-0.199 | 2 | 2 |
0.200-0.219 | 7 | 7 |
0.220-0.239 | 21 | 20 |
0.240-0.259 | 30 | 29 |
0.260-0.279 | 26 | 32 |
0.280-0.299 | 21 | 30 |
0.300-0.319 | 12 | 17 |
0.320-0.339 | 5 | 5 |
0.340-0.359 | 0 | 1 |
Formula used:
Solution:
From the given table,
The sum of all frequency for National League is
The table of relative frequency is given by
Batting average | National LeagueFrequency | National LeagueRelative frequency |
0.180-0.199 | 2 | |
0.200-0.219 | 7 | |
0.220-0.239 | 20 | |
0.240-0.259 | 29 | |
0.260-0.279 | 32 | |
0.280-0.299 | 30 | |
0.300-0.319 | 17 | |
0.320-0.339 | 5 | |
0.340-0.359 | 1 |
The relative frequency for the NationalLeague is given by
Batting average | National LeagueRelative frequency |
0.180-0.199 | 0.014 |
0.200-0.219 | 0.049 |
0.220-0.239 | 0.140 |
0.240-0.259 | 0.203 |
0.260-0.279 | 0.224 |
0.280-0.299 | 0.210 |
0.300-0.319 | 0.119 |
0.320-0.339 | 0.035 |
0.340-0.359 | 0.007 |
e.
![Check Mark](/static/check-mark.png)
To construct: A relative frequency histogram for American League.
Explanation of Solution
Given information:The following frequency distribution presents the batting averages of Major League Baseball players in both the American League and the National League who had 300 or more plate appearances during a recent season.
Batting average | American LeagueFrequency | National LeagueFrequency |
0.180-0.199 | 2 | 2 |
0.200-0.219 | 7 | 7 |
0.220-0.239 | 21 | 20 |
0.240-0.259 | 30 | 29 |
0.260-0.279 | 26 | 32 |
0.280-0.299 | 21 | 30 |
0.300-0.319 | 12 | 17 |
0.320-0.339 | 5 | 5 |
0.340-0.359 | 0 | 1 |
Definition used: Histograms based on relative frequency distributions are called relative frequency histogram.
Solution:
Batting average | American LeagueRelative frequency |
0.180-0.199 | 0.016 |
0.200-0.219 | 0.056 |
0.220-0.239 | 0.169 |
0.240-0.259 | 0.242 |
0.260-0.279 | 0.210 |
0.280-0.299 | 0.169 |
0.300-0.319 | 0.097 |
0.320-0.339 | 0.040 |
0.340-0.359 | 0.000 |
The relative frequency histogram for the given data is given by
f.
![Check Mark](/static/check-mark.png)
To construct: A relative frequency histogram for National League.
Explanation of Solution
Given information:The following frequency distribution presents the batting averages of Major League Baseball players in both the American League and the National League who had 300 or more plate appearances during a recent season.
Batting average | American LeagueFrequency | National LeagueFrequency |
0.180-0.199 | 2 | 2 |
0.200-0.219 | 7 | 7 |
0.220-0.239 | 21 | 20 |
0.240-0.259 | 30 | 29 |
0.260-0.279 | 26 | 32 |
0.280-0.299 | 21 | 30 |
0.300-0.319 | 12 | 17 |
0.320-0.339 | 5 | 5 |
0.340-0.359 | 0 | 1 |
Definition used: Histograms based on relative frequency distributions are called relative frequency histogram.
Solution:
Batting average | NationalLeagueRelative frequency |
0.180-0.199 | 0.014 |
0.200-0.219 | 0.049 |
0.220-0.239 | 0.140 |
0.240-0.259 | 0.203 |
0.260-0.279 | 0.224 |
0.280-0.299 | 0.210 |
0.300-0.319 | 0.119 |
0.320-0.339 | 0.035 |
0.340-0.359 | 0.007 |
The relative frequency histogram for the given data is given by
g.
![Check Mark](/static/check-mark.png)
To find: The percentage of American League players who had batting averages of 0.300 or more.
Answer to Problem 26E
The percentage of American League players who had batting averages of 0.300 or more is 13.7%.
Explanation of Solution
Given information:The following frequency distribution presents the batting averages of Major League Baseball players in both the American League and the National League who had 300 or more plate appearances during a recent season.
Batting average | American LeagueFrequency | National LeagueFrequency |
0.180-0.199 | 2 | 2 |
0.200-0.219 | 7 | 7 |
0.220-0.239 | 21 | 20 |
0.240-0.259 | 30 | 29 |
0.260-0.279 | 26 | 32 |
0.280-0.299 | 21 | 30 |
0.300-0.319 | 12 | 17 |
0.320-0.339 | 5 | 5 |
0.340-0.359 | 0 | 1 |
Calculation:
The relative frequency for American League is given by
Batting average | American LeagueRelative frequency |
0.180-0.199 | 0.016 |
0.200-0.219 | 0.056 |
0.220-0.239 | 0.169 |
0.240-0.259 | 0.242 |
0.260-0.279 | 0.210 |
0.280-0.299 | 0.169 |
0.300-0.319 | 0.097 |
0.320-0.339 | 0.040 |
0.340-0.359 | 0.000 |
From the above data, the relative frequencies of batting averages of 0.300 or more are 0.097, 0.040 and 0.000
The sum of all the above relative frequencies is
Then, its percentage is 13.7%
Hence, the percentage of American League players who had batting averages of 0.300 or more is 13.7%.
h.
![Check Mark](/static/check-mark.png)
To find: The percentage of National League players who had batting averages of 0.300 or more.
Answer to Problem 26E
The percentage of National League players who had batting averages of 0.300 or more is 16.1%.
Explanation of Solution
Given information:The following frequency distribution presents the batting averages of Major League Baseball players in both the American League and the National League who had 300 or more plate appearances during a recent season.
Batting average | American LeagueFrequency | National LeagueFrequency |
0.180-0.199 | 2 | 2 |
0.200-0.219 | 7 | 7 |
0.220-0.239 | 21 | 20 |
0.240-0.259 | 30 | 29 |
0.260-0.279 | 26 | 32 |
0.280-0.299 | 21 | 30 |
0.300-0.319 | 12 | 17 |
0.320-0.339 | 5 | 5 |
0.340-0.359 | 0 | 1 |
Calculation:
The relative frequency for National League is given by
Batting average | NationalLeagueRelative frequency |
0.180-0.199 | 0.014 |
0.200-0.219 | 0.049 |
0.220-0.239 | 0.140 |
0.240-0.259 | 0.203 |
0.260-0.279 | 0.224 |
0.280-0.299 | 0.210 |
0.300-0.319 | 0.119 |
0.320-0.339 | 0.035 |
0.340-0.359 | 0.007 |
From the above data, the relative frequencies of batting averages of 0.300 or more are 0.119, 0.035 and 0.007.
The sum of all the above relative frequencies is
Then, its percentage is 16.1%
Hence, the percentage of National League players who had batting averages of 0.300 or more is 16.1%.
i.
![Check Mark](/static/check-mark.png)
To find: The percentage of players who had batting averages less than 0.220.
Answer to Problem 26E
The batting averages tend to be a bit higher in the National League.
Explanation of Solution
Given information:The following frequency distribution presents the batting averages of Major League Baseball players in both the American League and the National League who had 300 or more plate appearances during a recent season.
Batting average | American LeagueFrequency | National LeagueFrequency |
0.180-0.199 | 2 | 7 |
0.200-0.219 | 7 | 7 |
0.220-0.239 | 21 | 20 |
0.240-0.259 | 30 | 29 |
0.260-0.279 | 26 | 32 |
0.280-0.299 | 21 | 30 |
0.300-0.319 | 12 | 17 |
0.320-0.339 | 5 | 5 |
0.340-0.359 | 0 | 1 |
Solution:
The relative frequency histogram for American League is given by
The relative frequency histogram for the National League is given by
From the above two histograms, we can see that the relative frequency for National league is bit higher than the National League.
Hence, the batting averages tend to be a bit higher in the National League.
Want to see more full solutions like this?
Chapter 2 Solutions
Elementary Statistics (Text Only)
Additional Math Textbook Solutions
Precalculus: A Unit Circle Approach (3rd Edition)
College Algebra (Collegiate Math)
Intermediate Algebra (13th Edition)
Introductory Statistics
Basic College Mathematics
- 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
- 26. A certain system can experience three different types of defects. Let A;(i = 1,2,3) denote the event that the sys- tem has a defect of type i. Suppose thatP(A1) = .12 P(A) = .07 P(A) = .05P(A, U A2) = .13P(A, U A3) = .14P(A2 U A3) = .10P(A, A2 A3) = .011Rshelfa. What is the probability that the system does not havea type 1 defect?b. What is the probability that the system has both type 1 and type 2 defects?c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?arrow_forwardThe following are suggested designs for group sequential studies. Using PROCSEQDESIGN, provide the following for the design O’Brien Fleming and Pocock.• The critical boundary values for each analysis of the data• The expected sample sizes at each interim analysisAssume the standardized Z score method for calculating boundaries.Investigators are evaluating the success rate of a novel drug for treating a certain type ofbacterial wound infection. Since no existing treatment exists, they have planned a one-armstudy. They wish to test whether the success rate of the drug is better than 50%, whichthey have defined as the null success rate. Preliminary testing has estimated the successrate of the drug at 55%. The investigators are eager to get the drug into production andwould like to plan for 9 interim analyses (10 analyzes in total) of the data. Assume thesignificance level is 5% and power is 90%.Besides, draw a combined boundary plot (OBF, POC, and HP)arrow_forwardPlease provide the solution for the attached image in detailed.arrow_forward
- Big Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin HarcourtHolt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGALGlencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill
![Text book image](https://www.bartleby.com/isbn_cover_images/9781680331141/9781680331141_smallCoverImage.jpg)
![Text book image](https://www.bartleby.com/isbn_cover_images/9780547587776/9780547587776_smallCoverImage.jpg)
![Text book image](https://www.bartleby.com/isbn_cover_images/9780079039897/9780079039897_smallCoverImage.jpg)