ap stats chapter 25 homework answers

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

AP®︎/College Statistics

Unit 1: exploring categorical data, unit 2: exploring one-variable quantitative data: displaying and describing, unit 3: exploring one-variable quantitative data: summary statistics, unit 4: exploring one-variable quantitative data: percentiles, z-scores, and the normal distribution, unit 5: exploring two-variable quantitative data, unit 6: collecting data, unit 7: probability, unit 8: random variables and probability distributions, unit 9: sampling distributions, unit 10: inference for categorical data: proportions, unit 11: inference for quantitative data: means, unit 12: inference for categorical data: chi-square, unit 13: inference for quantitative data: slopes, unit 14: prepare for the 2022 ap®︎ statistics exam.

• AP Statistics Syllabus
• AP Statistics PowerPoints
• AP Statistics Additional Notes

AP Statistics Answers

Unit 1 answers.

• Homework Unit 1 – 1st part
• Homework Unit 1 – 2nd part
• Homework Unit 1 – 3rd part

Unit 2 Answers

• Unit 2 Homework – Part 1
• Unit 2 Homework – Part 2
• Unit 2 Homework – Part 3
• Unit 2 Homework – Part 4

Unit 3 Answers

• Chapter 5, Section 1
• Chapter 5, Section 2
• Chapter 5, Section 3
• Chapter 5 answers

Unit 4 Answers

• Go to google classroom

Unit 5 Answers

• TRM-Section 8.1 Full Solutions
• TRM-Section 8.2 Full Solutions
• TRM-Section 8.3 Full Solutions

Unit 6 Answers

• Homework Unit 6 – 2019-2020

Unit 7 Answers

Comments are closed.

• HomeLogic Log-in
• Mr. Gallo's Website
• NHHS Home Page
• NHHS Moodle
• NHV District Page
• Wordpress Log-in

mean = 4 hours, standard deviation = 1.2 hours, sample size = 16

a. Check student's solution. b. 3.5, 4.25, 0.2441

The fact that the two distributions are different accounts for the different probabilities.

The sample size, n , gets larger.

• U (24, 26), 25, 0.5774
• N (25, 0.0577)
• N (2500, 5.7735)

N ( 10,  10 8 ) ( 10,  10 8 )

• Χ = amount of change students carry
• Χ ~ Exp (1/0.88) or approximately Χ ~ Exp (1.1364)
• X ¯ X ¯ = average amount of change carried by a sample of 25 students.
• X ¯ X ¯ ~ N (0.88, 0.176)
• The probability in part (e) represents the probability of an individual value. In part (f), the probability describes the mean of a sample of 25. Part (f) relies on the central limit theorem, so the distributions are different. Part (e) is exponential and part (f) is normal.
• length of time for an individual to complete IRS form 1040, in hours
• mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours
• N ( 10 .53,  1 3 ) ( 10 .53,  1 3 )
• Yes, I would be surprised, because the probability is almost 0.
• No, I would not be totally surprised because the probability is 0.2312.
• the length of a song, in minutes, in the collection
• the average length, in minutes, of the songs from a sample of five albums from the collection
• N (2.75, 0.0220)
• 2.74 minutes
• 0.03 minutes
• True. The mean of a sampling distribution of the means is approximately the mean of the data distribution.
• True. According to the central limit theorem, the larger the sample, the closer the sampling distribution of the means becomes normal.
• The standard deviation of the sampling distribution of the means will decrease, making it approximately the same as the standard deviation of X as the sample size increases.
• X = the yearly income of someone in a Third World country
• the average salary from samples of 1,000 residents of a Third World country
• X ¯ X ¯ ∼ N ( 2,000,  8,000 1,000 ) ( 2,000,  8,000 1,000 )
• Very wide differences in data values can have averages smaller than standard deviations.
• The distribution of the sample mean will have higher probabilities closer to the population mean. P (2,000 < X ¯ X ¯ < 2,100) = 0.1537 P (2,100 < X ¯ X ¯ < 2,200) = 0.1317
• the total length of time for nine criminal trials
• N (189, 21)
• 162.09; 90 percent of the total nine trials of this type will last 162 days or more.
• X = the salary of one elementary school teacher in the district
• X ~ N (44000, 6500)
• ΣX ~ sum of the salaries of 10 elementary school teachers in the sample
• ΣX ~ N (44,000, 20,554.80)
• Sampling 70 teachers instead of 10 would cause the distribution to be more spread out. It would be a more symmetrical normal curve.
• If every teacher received a $3,000 raise, the distribution of X would shift to the right by $3,000. In other words, it would have a mean of $47,000.
• X = the closing stock prices for U.S. semiconductor manufacturers
• i. $20.71, ii. $17.31, iii. 35
• exponential distribution, Χ ~ Exp ( 1 20.71 ) ( 1 20.71 )
• Answers will vary.
• i. $20.71, ii. $11.14
• N ( 20 .71,  17.31 5 ) ( 20 .71,  17.31 5 )
• Check student’s solution.
• X ¯ X ¯ ~ N ( 60,  9 25 ) ( 60,  9 25 )
• N (1500, 45)
• We have μ = 17, σ = 0.8, x ¯ x ¯ = 16.7, and n = 30. To calculate the probability, we use normalcdf (lower, upper, μ , σ n σ n ) = normalcdf ( E – 99,16 .7,17, 0. 8 30 ) ( E – 99,16 .7,17, 0. 8 30 ) = 0.0200.
• If the process is working properly, then the probability that a sample of 30 batteries would have at most 16.7 life span hours is only 2%. Therefore, the class was justified to question the claim.
• For the sample, we have n = 100, x ¯ x ¯ = 0.862, and s = 0.05.
• Σ x ¯ Σ x ¯ = 85.65, Σs = 5.18
• normalcdf (396.9, E 99,(465)(0.8565),(0.05)( 465 465 )) ≈ 1
• Because the probability of a sample of size of 465 having at least a mean sum of 396.9 is appproximately 1, we can conclude that the company is correctly labeling their candy packages.

Use normalcdf ( E – 99,1 .1,1, 1 70 ) ( E – 99,1 .1,1, 1 70 ) = 0.7986. This means that there is an 80 percent chance that the service time will be less than 1.1 hours. It may be wise to schedule more time because there is an associated 20 percent chance that the maintenance time will be greater than 1.1 hours.

Because we have normalcdf ( 5. 111,5 .291,5 .201, 0. 065 280 ) ( 5. 111,5 .291,5 .201, 0. 065 280 ) ≈ 1, we can conclude that practically all the coins are within the limits; therefore, there should be no rejected coins out of a well-selected sample size of 280.

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute Texas Education Agency (TEA). The original material is available at: https://www.texasgateway.org/book/tea-statistics . Changes were made to the original material, including updates to art, structure, and other content updates.

Access for free at https://openstax.org/books/statistics/pages/1-introduction

• Authors: Barbara Illowsky, Susan Dean
• Publisher/website: OpenStax
• Book title: Statistics
• Publication date: Mar 27, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/statistics/pages/1-introduction
• Section URL: https://openstax.org/books/statistics/pages/7-solutions

© Jan 23, 2024 Texas Education Agency (TEA). The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Helping math teachers bring statistics to life

Chapter 5 - Day 8

Students struggled with….

P(At least 1) problems when the probability was given in a story problem and not in a table.

Finding a conditional probability when given necessary probabilities in a story problem.

Remedies…

As soon as you read “at least 1” immediately write “1-P(None)”.  Instead of using a formula to find the probability of none, think about what you need to happen and list it as separate pieces.

For example, if a person has a 10% chance of being late to work, what is the probability that they are late at least 1 out of 5 days? To find the probability that they are late noneof the days we need them to be on time (90%) for 5 days in a row (0.9 x 0.9 x 0.9 x 0.9 x 0.9). It may feel like overkill to write out all of these probabilities but it really does help with the conceptual understanding.

Have students rewrite the story problem in probability notation after they have read the problem.  This helps them to identify which formulas may be helpful.

• AP Statistics Summer Assignment (Introduction to Chapter 1)
• AP Class Notes and HW Answers
• Algebra I Assignment Sheets
• Algebra 1 Notes and Review Answer Keys
• Honors Algebra I Assignment Sheets
• Honors Notes and Review Answer Keys
• Honors Algebra 1 Summer Assignment

AP Chapter 5 (TPS4e)

What AP Stat Students need to know about probability !!!!!

AP STAT Probability Flow Chart

Text Book Reading AP Ch5 Guided Notes for Reading Textbook (TPS4e )         <–word document AP Ch5 Guided Notes for Reading Textbook (TPS4e )     <–PDF-FORMAT AP Stats Chapter 5 – Glossary of Important Definitions

• AP Stats Chapter 5 PowerPoint (TPS4e)

HW.      AP Stats Chapter 5 Assignments (2023)

• AP Stats HW 5.1 answers (TPS4e)
• AP Stats HW 5.2 answers (TPS4e)      
• AP Stats HW 5.3 answers (TPS4e)
• 5.1 Nascar Simulation (key)
• 5.2 Dice Activities (ANSWER KEY)
• AP Stats 5.1 & 5.2 Self-Assessment Quiz (key)  
• 5.3 Conditional Probability and Independence Activity  (Part 1  KEY)
• 5.3 Conditional Probability and Independence Activity (Part 2 KEY)
• Titantic Review Handout (KEY)
• AP Stats HW 5R answers (TPS4e)
• AP Stats Chapter 5 Review WS #1

YouTube Videos Intro to Probabilities in Statistics by ProfRobBob

• Search for: