The Practice of Statistics for AP - 4th Edition
4th Edition
ISBN: 9781429245593
Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.
Publisher: Macmillan Higher Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 2.2, Problem 43E
(a)
To determine
To find: the percent of men who are taller than 74 inches
(a)
Expert Solution
Answer to Problem 43E
Approximately 2.5 % of men are taller than 74 inches.
Explanation of Solution
Calculation:
The distribution of heights is approximately N (69, 2.5). Figure below is a sketch of the Normaldensity curve with the desired points labeled.
By the
Conclusion:
Therefore,2.5% of men are taller than 74 inches
(b)
To determine
To find: the height
(b)
Expert Solution
Answer to Problem 43E
Approximately 95% of men have heights between 64 inches and 74 inches.
Explanation of Solution
Calculation:
By the 68-95-99.7 rule, approximately 95% of men have heights between 64 inches and 74inches which fall within 2s of µ.
Conclusion:
Therefore, 95% of men have heights between 64 inches and 74inches
(c)
To determine
To find: the percent of men who arebetween 64 and 66.5 inches tall
(c)
Expert Solution
Answer to Problem 43E
Approximately16% -2.5% = 13.5% of men have heights between 64 inches and 66.5 inches.
Explanation of Solution
Calculation:
By the 68-95-99.7 rule, approximately 16% of men are shorter than 66.5 inches, because66.5 is one standard deviation below the mean. Approximately 2.5% are shorter than 64 inches,because 64 inches is two standard deviations below the mean. So approximately 16% - 2.5% = 13.5% of men have heights between 64 inches and 66.5 inches.
Conclusion:
Therefore, 16% -2.5% = 13.5% of men have heights between 64 inches and 66.5 inches.
(d)
To determine
To find: the percentile of adult male American heights corresponding to 71.5 inches height.
(d)
Expert Solution
Answer to Problem 43E
Here, 71.5 is the 84th percentile of adult male American heights.
Explanation of Solution
Calculation:
By the 68-95-99 .7 rule, the value 71.5 is one standard deviation above the mean. Thus, the area to the left of 71.5is 0.68 +0.16 = 0.84. In other words, 71.5 is the 84th percentile of adult male American heights.
Conclusion:
Therefore, 71.5 is the 84th percentile of adult male American heights.
Chapter 2 Solutions
The Practice of Statistics for AP - 4th Edition
Ch. 2.1 - Prob. 1.1CYUCh. 2.1 - Prob. 1.2CYUCh. 2.1 - Prob. 1.3CYUCh. 2.1 - Prob. 1.4CYUCh. 2.1 - Prob. 2.1CYUCh. 2.1 - Prob. 2.2CYUCh. 2.1 - Prob. 2.3CYUCh. 2.1 - Prob. 3.1CYUCh. 2.1 - Prob. 3.2CYUCh. 2.1 - Prob. 3.3CYU
Ch. 2.1 - Prob. 4.1CYUCh. 2.1 - Prob. 4.2CYUCh. 2.1 - Prob. 4.3CYUCh. 2.1 - Prob. 1ECh. 2.1 - Prob. 2ECh. 2.1 - Prob. 3ECh. 2.1 - Prob. 4ECh. 2.1 - Prob. 5ECh. 2.1 - Prob. 6ECh. 2.1 - Prob. 7ECh. 2.1 - Prob. 8ECh. 2.1 - Prob. 9ECh. 2.1 - Prob. 10ECh. 2.1 - Prob. 11ECh. 2.1 - Prob. 12ECh. 2.1 - Prob. 13ECh. 2.1 - Prob. 14ECh. 2.1 - Prob. 15ECh. 2.1 - Prob. 16ECh. 2.1 - Prob. 17ECh. 2.1 - Prob. 18ECh. 2.1 - Prob. 19ECh. 2.1 - Prob. 20ECh. 2.1 - Prob. 21ECh. 2.1 - Prob. 22ECh. 2.1 - Prob. 23ECh. 2.1 - Prob. 24ECh. 2.1 - Prob. 25ECh. 2.1 - Prob. 26ECh. 2.1 - Prob. 27ECh. 2.1 - Prob. 28ECh. 2.1 - Prob. 29ECh. 2.1 - Prob. 30ECh. 2.1 - Prob. 31ECh. 2.1 - Prob. 32ECh. 2.1 - Prob. 33ECh. 2.1 - Prob. 34ECh. 2.1 - Prob. 35ECh. 2.1 - Prob. 36ECh. 2.1 - Prob. 37ECh. 2.1 - Prob. 38ECh. 2.1 - Prob. 39ECh. 2.1 - Prob. 40ECh. 2.2 - Prob. 1.1CYUCh. 2.2 - Prob. 1.2CYUCh. 2.2 - Prob. 1.3CYUCh. 2.2 - Prob. 2.1CYUCh. 2.2 - Prob. 2.2CYUCh. 2.2 - Prob. 2.3CYUCh. 2.2 - Prob. 2.4CYUCh. 2.2 - Prob. 2.5CYUCh. 2.2 - Prob. 3.1CYUCh. 2.2 - Prob. 3.2CYUCh. 2.2 - Prob. 3.3CYUCh. 2.2 - Prob. 41ECh. 2.2 - Prob. 42ECh. 2.2 - Prob. 43ECh. 2.2 - Prob. 44ECh. 2.2 - Prob. 45ECh. 2.2 - Prob. 46ECh. 2.2 - Prob. 47ECh. 2.2 - Prob. 48ECh. 2.2 - Prob. 49ECh. 2.2 - Prob. 50ECh. 2.2 - Prob. 51ECh. 2.2 - Prob. 52ECh. 2.2 - Prob. 53ECh. 2.2 - Prob. 54ECh. 2.2 - Prob. 55ECh. 2.2 - Prob. 56ECh. 2.2 - Prob. 57ECh. 2.2 - Prob. 58ECh. 2.2 - Prob. 59ECh. 2.2 - Prob. 60ECh. 2.2 - Prob. 61ECh. 2.2 - Prob. 62ECh. 2.2 - Prob. 63ECh. 2.2 - Prob. 64ECh. 2.2 - Prob. 65ECh. 2.2 - Prob. 66ECh. 2.2 - Prob. 67ECh. 2.2 - Prob. 68ECh. 2.2 - Prob. 69ECh. 2.2 - Prob. 70ECh. 2.2 - Prob. 71ECh. 2.2 - Prob. 72ECh. 2.2 - Prob. 73ECh. 2.2 - Prob. 74ECh. 2.2 - Prob. 75ECh. 2.2 - Prob. 76ECh. 2 - Prob. 1CRECh. 2 - Prob. 2CRECh. 2 - Prob. 3CRECh. 2 - Prob. 4CRECh. 2 - Prob. 5CRECh. 2 - Prob. 6CRECh. 2 - Prob. 7CRECh. 2 - Prob. 8CRECh. 2 - Prob. 9CRECh. 2 - Prob. 10CRECh. 2 - Prob. 11CRECh. 2 - Prob. 12CRECh. 2 - Prob. 1PTCh. 2 - Prob. 2PTCh. 2 - Prob. 3PTCh. 2 - Prob. 4PTCh. 2 - Prob. 5PTCh. 2 - Prob. 6PTCh. 2 - Prob. 7PTCh. 2 - Prob. 8PTCh. 2 - Prob. 9PTCh. 2 - Prob. 10PTCh. 2 - Prob. 11PTCh. 2 - Prob. 12PTCh. 2 - Prob. 13PT
Additional Math Textbook Solutions
Find more solutions based on key concepts
Fill in each blank so that the resulting statement is true. The quadratic function f(x)=a(xh)2+k,a0, is in ____...
Algebra and Trigonometry (6th Edition)
Identify f as being linear, quadratic, or neither. If f is quadratic, identify the leading coefficient a and ...
College Algebra with Modeling & Visualization (5th Edition)
The table by using the given graph of h.
Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)
Voting A random sample of likely voters showed that 49 planned to support Measure X. The margin of error is 3 p...
Introductory Statistics
Continuous Uniform Distribution. In Exercises 5–8, refer to the continuous uniform distribution depicted in Fig...
Elementary Statistics (13th Edition)
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.Similar questions
- What is one sample T-test? Give an example of business application of this test? What is Two-Sample T-Test. Give an example of business application of this test? .What is paired T-test. Give an example of business application of this test? What is one way ANOVA test. Give an example of business application of this test? 1. One Sample T-Test: Determine whether the average satisfaction rating of customers for a product is significantly different from a hypothetical mean of 75. (Hints: The null can be about maintaining status-quo or no difference; If your alternative hypothesis is non-directional (e.g., μ≠75), you should use the two-tailed p-value from excel file to make a decision about rejecting or not rejecting null. If alternative is directional (e.g., μ < 75), you should use the lower-tailed p-value. For alternative hypothesis μ > 75, you should use the upper-tailed p-value.) H0 = H1= Conclusion: The p value from one sample t-test is _______. Since the two-tailed p-value…arrow_forwardUsing the accompanying Accounting Professionals data to answer the following questions. a. Find and interpret a 90% confidence interval for the mean years of service. b. Find and interpret a 90% confidence interval for the proportion of employees who have a graduate degree. view the Accounting Professionals data. Employee Years of Service Graduate Degree?1 26 Y2 8 N3 10 N4 6 N5 23 N6 5 N7 8 Y8 5 N9 26 N10 14 Y11 10 N12 8 Y13 7 Y14 27 N15 16 Y16 17 N17 21 N18 9 Y19 9 N20 9 N Question content area bottom Part 1 a. A 90% confidence interval for the mean years of service is (Use ascending order. Round to two decimal places as needed.)arrow_forwardIf, based on a sample size of 900,a political candidate finds that 509people would vote for him in a two-person race, what is the 95%confidence interval for his expected proportion of the vote? Would he be confident of winning based on this poll? Question content area bottom Part 1 A 9595% confidence interval for his expected proportion of the vote is (Use ascending order. Round to four decimal places as needed.)arrow_forward
- Questions An insurance company's cumulative incurred claims for the last 5 accident years are given in the following table: Development Year Accident Year 0 2018 1 2 3 4 245 267 274 289 292 2019 255 276 288 294 2020 265 283 292 2021 263 278 2022 271 It can be assumed that claims are fully run off after 4 years. The premiums received for each year are: Accident Year Premium 2018 306 2019 312 2020 318 2021 326 2022 330 You do not need to make any allowance for inflation. 1. (a) Calculate the reserve at the end of 2022 using the basic chain ladder method. (b) Calculate the reserve at the end of 2022 using the Bornhuetter-Ferguson method. 2. Comment on the differences in the reserves produced by the methods in Part 1.arrow_forwardA population that is uniformly distributed between a=0and b=10 is given in sample sizes 50( ), 100( ), 250( ), and 500( ). Find the sample mean and the sample standard deviations for the given data. Compare your results to the average of means for a sample of size 10, and use the empirical rules to analyze the sampling error. For each sample, also find the standard error of the mean using formula given below. Standard Error of the Mean =sigma/Root Complete the following table with the results from the sampling experiment. (Round to four decimal places as needed.) Sample Size Average of 8 Sample Means Standard Deviation of 8 Sample Means Standard Error 50 100 250 500arrow_forwardA survey of 250250 young professionals found that two dash thirdstwo-thirds of them use their cell phones primarily for e-mail. Can you conclude statistically that the population proportion who use cell phones primarily for e-mail is less than 0.720.72? Use a 95% confidence interval. Question content area bottom Part 1 The 95% confidence interval is left bracket nothing comma nothing right bracket0.60820.6082, 0.72510.7251. As 0.720.72 is within the limits of the confidence interval, we cannot conclude that the population proportion is less than 0.720.72. (Use ascending order. Round to four decimal places as needed.)arrow_forward
- I need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forwardA survey of 250 young professionals found that two-thirds of them use their cell phones primarily for e-mail. Can you conclude statistically that the population proportion who use cell phones primarily for e-mail is less than 0.72? Use a 95% confidence interval. Question content area bottom Part 1 The 95% confidence interval is [ ], [ ] As 0.72 is ▼ above the upper limit within the limits below the lower limit of the confidence interval, we ▼ can cannot conclude that the population proportion is less than 0.72. (Use ascending order. Round to four decimal places as needed.)arrow_forwardI need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forward
- I need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forwardI need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forwardQuestions An insurance company's cumulative incurred claims for the last 5 accident years are given in the following table: Development Year Accident Year 0 2018 1 2 3 4 245 267 274 289 292 2019 255 276 288 294 2020 265 283 292 2021 263 278 2022 271 It can be assumed that claims are fully run off after 4 years. The premiums received for each year are: Accident Year Premium 2018 306 2019 312 2020 318 2021 326 2022 330 You do not need to make any allowance for inflation. 1. (a) Calculate the reserve at the end of 2022 using the basic chain ladder method. (b) Calculate the reserve at the end of 2022 using the Bornhuetter-Ferguson method. 2. Comment on the differences in the reserves produced by the methods in Part 1.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
Probability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage Learning
Statistics 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:PEARSON
The Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. Freeman
Introduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
Hypothesis Testing - Solving Problems With Proportions; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=76VruarGn2Q;License: Standard YouTube License, CC-BY
Hypothesis Testing and Confidence Intervals (FRM Part 1 – Book 2 – Chapter 5); Author: Analystprep;https://www.youtube.com/watch?v=vth3yZIUlGQ;License: Standard YouTube License, CC-BY