LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access)
LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access)
8th Edition
ISBN: 9781464133404
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 1.4, Problem 119E

(a)

To determine

To find: The distribution using 689599.7 rule.

(a)

Expert Solution
Check Mark

Answer to Problem 119E

Solution: The calculated values are (8489,20,919)_, (2274,27,134)_, and (3941,33,349)_.

Explanation of Solution

Calculation: The 689599.7 rule states that in the normal distribution, 68% of data falls under (μ±σ), 95% of the data falls under (μ±2σ), and 99.7% of the data falls under (μ±3σ).

With the given values of mean and standard deviation, the range of the number of words can be obtained by the following steps:

Step 1: For calculating 68% of data,

(μ±σ)=(μσ,μ+σ)=(14,7046215,14,704+6215)=(8489,20,919)

Step 2: For calculating 95% of data,

(μ±2σ)=(μ2σ,μ+2σ)=(14,704(2×6215),14,704+(2×6215))=(2274,27,134)

Step 3: For calculating 99.7% of data,

(μ±3σ)=(μ3σ,μ+3σ)=(14,704(3×6215),14,704+(3×6215))=(3941,33,349)

To determine

To explain: Whether the above rule is applicable or not.

Expert Solution
Check Mark

Answer to Problem 119E

Solution: This rule is not applicable.

Explanation of Solution

From the above calculation, it is clear that this rule is not applicable here because it is giving negative values and words spoken cannot be negative. The normal distribution cannot be used in the provided case. This rule is applicable only in the case of normal distribution.

To determine

To find: The distribution using 689599.7 rule.

Expert Solution
Check Mark

Answer to Problem 119E

Solution: The calculated values are (7158,22,886)_, (706,30,750)_, and (8570,38,614)_.

Explanation of Solution

Calculation: The 689599.7 rule states that in the normal distribution, 68% of data falls under (μ±σ), 95% of the data falls under (μ±2σ), and 99.7% of the data falls under (μ±3σ).

With the given values of mean and standard deviation, the range of scores for the students can be obtained by the following steps shown below:

Step 1: For calculating 68% of data,

(μ±σ)=(μσ,μ+σ)=(15,0227864,15,022+7864)=(7158,22,886)

Step 2: For calculating 95% of data,

(μ±2σ)=(μ2σ,μ+2σ)=(15,022(2×7864),15,022+(2×7864))=(706,30,750)

Step 3: For calculating 99.7% of data,

(μ±3σ)=(μ3σ,μ+3σ)=(15,022(3×7864),15,022+(3×7864))=(8570,38,614)

To determine

To explain: Whether the above rule is applicable or not.

Expert Solution
Check Mark

Answer to Problem 119E

Solution: This rule is also not applicable.

Explanation of Solution

From the above calculation, it is clear that this rule is not applicable here because it is also providing negative values which cannot be possible.

To determine

To explain: Whether the data supports conventional wisdom or not.

Expert Solution
Check Mark

Answer to Problem 119E

Solution: Data does not support conventional wisdom.

Explanation of Solution

It can be seen from the question that the mean number of words spoken per day by the men is greater than the mean number of words spoken per day by the women, and the standard deviation of women is less than the standard deviation of men. There is a more concentrated distribution of the number of words spoken by women than men.

(b)

To determine

To explain: The conclusion from the observation.

(b)

Expert Solution
Check Mark

Answer to Problem 119E

Solution: Both men and women from Mexico tend to speak more than those from United states.

Explanation of Solution

The mean for both men and women is higher for the Mexican study as compared to U.S. study. It shows that both men and women from Mexico tend to speak more than those from the United States.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Suppose you are gambling on a roulette wheel. Each time the wheel is spun, the result is one of the outcomes 0, 1, and so on through 36. Of these outcomes, 18 are red, 18 are black, and 1 is green. On each spin you bet $5 that a red outcome will occur and $1 that the green outcome will occur. If red occurs, you win a net $4. (You win $10 from red and nothing from green.) If green occurs, you win a net $24. (You win $30 from green and nothing from red.) If black occurs, you lose everything you bet for a loss of $6. a.  Use simulation to generate 1,000 plays from this strategy. Each play should indicate the net amount won or lost. Then, based on these outcomes, calculate a 95% confidence interval for the total net amount won or lost from 1,000 plays of the game. (Round your answers to two decimal places and if your answer is negative value, enter "minus" sign.)   I worked out the Upper Limit, but I can't seem to arrive at the correct answer for the Lower Limit. What is the Lower Limit?…
Let us suppose we have some article reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks, lifting and using ultrasound. Assume that both sample sizes are 6, the sample mean force for lifting was 6.2 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.4 pounds with standard deviation 0.3 pounds. Assume that the standard deviations are known. Suppose that you wanted to detect a true difference in mean force of 0.25 pounds on the hands for these two activities. Under the null hypothesis, 40 0. What level of type II error would you recommend here? = Round your answer to four decimal places (e.g. 98.7654). Use α = 0.05. β = 0.0594 What sample size would be required? Assume the sample sizes are to be…
Consider the hypothesis test Ho: 0 s² = = 4.5; s² = 2.3. Use a = 0.01. = σ against H₁: 6 > σ2. Suppose that the sample sizes are n₁ = 20 and 2 = 8, and that (a) Test the hypothesis. Round your answers to two decimal places (e.g. 98.76). The test statistic is fo = 1.96 The critical value is f = 6.18 Conclusion: fail to reject the null hypothesis at a = 0.01. (b) Construct the confidence interval on 02/2/622 which can be used to test the hypothesis: (Round your answer to two decimal places (e.g. 98.76).) 035

Chapter 1 Solutions

LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access)

Ch. 1.1 - Prob. 11ECh. 1.1 - Prob. 12ECh. 1.1 - Prob. 13ECh. 1.1 - Prob. 14ECh. 1.1 - Prob. 15ECh. 1.2 - Prob. 16UYKCh. 1.2 - Prob. 17UYKCh. 1.2 - Prob. 18UYKCh. 1.2 - Prob. 19UYKCh. 1.2 - Prob. 20UYKCh. 1.2 - Prob. 21UYKCh. 1.2 - Prob. 22UYKCh. 1.2 - Prob. 23UYKCh. 1.2 - Prob. 24UYKCh. 1.2 - Prob. 25ECh. 1.2 - Prob. 26ECh. 1.2 - Prob. 27ECh. 1.2 - Prob. 28ECh. 1.2 - Prob. 29ECh. 1.2 - Prob. 30ECh. 1.2 - Prob. 31ECh. 1.2 - Prob. 32ECh. 1.2 - Prob. 33ECh. 1.2 - Prob. 34ECh. 1.2 - Prob. 35ECh. 1.2 - Prob. 42ECh. 1.2 - Prob. 36ECh. 1.2 - Prob. 43ECh. 1.2 - Prob. 37ECh. 1.2 - Prob. 44ECh. 1.2 - Prob. 38ECh. 1.2 - Prob. 45ECh. 1.2 - Prob. 46ECh. 1.2 - Prob. 39ECh. 1.2 - Prob. 40ECh. 1.2 - Prob. 41ECh. 1.3 - Prob. 47UYKCh. 1.3 - Prob. 48UYKCh. 1.3 - Prob. 49UYKCh. 1.3 - Prob. 50UYKCh. 1.3 - Prob. 51UYKCh. 1.3 - Prob. 52UYKCh. 1.3 - Prob. 53UYKCh. 1.3 - Prob. 54UYKCh. 1.3 - Prob. 55UYKCh. 1.3 - Prob. 56UYKCh. 1.3 - Prob. 57UYKCh. 1.3 - Prob. 58UYKCh. 1.3 - Prob. 59UYKCh. 1.3 - Prob. 60UYKCh. 1.3 - Prob. 67ECh. 1.3 - Prob. 69ECh. 1.3 - Prob. 61ECh. 1.3 - Prob. 62ECh. 1.3 - Prob. 63ECh. 1.3 - Prob. 64ECh. 1.3 - Prob. 65ECh. 1.3 - Prob. 66ECh. 1.3 - Prob. 74ECh. 1.3 - Prob. 75ECh. 1.3 - Prob. 76ECh. 1.3 - Prob. 71ECh. 1.3 - Prob. 68ECh. 1.3 - Prob. 70ECh. 1.3 - Prob. 77ECh. 1.3 - Prob. 78ECh. 1.3 - Prob. 79ECh. 1.3 - Prob. 80ECh. 1.3 - Prob. 81ECh. 1.3 - Prob. 82ECh. 1.3 - Prob. 83ECh. 1.3 - Prob. 84ECh. 1.3 - Prob. 85ECh. 1.3 - Prob. 86ECh. 1.3 - Prob. 87ECh. 1.3 - Prob. 88ECh. 1.3 - Prob. 89ECh. 1.3 - Prob. 90ECh. 1.3 - Prob. 91ECh. 1.3 - Prob. 92ECh. 1.3 - Prob. 93ECh. 1.3 - Prob. 94ECh. 1.3 - Prob. 95ECh. 1.3 - Prob. 96ECh. 1.3 - Prob. 72ECh. 1.3 - Prob. 97ECh. 1.3 - Prob. 98ECh. 1.3 - Prob. 99ECh. 1.3 - Prob. 100ECh. 1.3 - Prob. 73ECh. 1.4 - Prob. 101UYKCh. 1.4 - Prob. 102UYKCh. 1.4 - Prob. 103UYKCh. 1.4 - Prob. 104UYKCh. 1.4 - Prob. 105UYKCh. 1.4 - Prob. 106UYKCh. 1.4 - Prob. 107UYKCh. 1.4 - Prob. 108UYKCh. 1.4 - Prob. 109ECh. 1.4 - Prob. 110ECh. 1.4 - Prob. 111ECh. 1.4 - Prob. 112ECh. 1.4 - Prob. 113ECh. 1.4 - Prob. 114ECh. 1.4 - Prob. 115ECh. 1.4 - Prob. 116ECh. 1.4 - Prob. 117ECh. 1.4 - Prob. 118ECh. 1.4 - Prob. 119ECh. 1.4 - Prob. 120ECh. 1.4 - Prob. 121ECh. 1.4 - Prob. 122ECh. 1.4 - Prob. 123ECh. 1.4 - Prob. 124ECh. 1.4 - Prob. 125ECh. 1.4 - Prob. 126ECh. 1.4 - Prob. 127ECh. 1.4 - Prob. 128ECh. 1.4 - Prob. 129ECh. 1.4 - Prob. 130ECh. 1.4 - Prob. 131ECh. 1.4 - Prob. 132ECh. 1.4 - Prob. 133ECh. 1.4 - Prob. 134ECh. 1.4 - Prob. 135ECh. 1.4 - Prob. 136ECh. 1.4 - Prob. 137ECh. 1.4 - Prob. 138ECh. 1.4 - Prob. 139ECh. 1.4 - Prob. 140ECh. 1.4 - Prob. 141ECh. 1.4 - Prob. 142ECh. 1.4 - Prob. 143ECh. 1.4 - Prob. 144ECh. 1.4 - Prob. 145ECh. 1.4 - Prob. 146ECh. 1.4 - Prob. 147ECh. 1.4 - Prob. 148ECh. 1.4 - Prob. 149ECh. 1.4 - Prob. 150ECh. 1.4 - Prob. 151ECh. 1.4 - Prob. 152ECh. 1.4 - Prob. 153ECh. 1.4 - Prob. 154ECh. 1.4 - Prob. 155ECh. 1 - Prob. 156ECh. 1 - Prob. 157ECh. 1 - Prob. 158ECh. 1 - Prob. 159ECh. 1 - Prob. 160ECh. 1 - Prob. 161ECh. 1 - Prob. 162ECh. 1 - Prob. 163ECh. 1 - Prob. 164ECh. 1 - Prob. 165ECh. 1 - Prob. 166ECh. 1 - Prob. 167ECh. 1 - Prob. 168ECh. 1 - Prob. 169ECh. 1 - Prob. 170ECh. 1 - Prob. 171ECh. 1 - Prob. 172ECh. 1 - Prob. 173ECh. 1 - Prob. 174ECh. 1 - Prob. 175ECh. 1 - Prob. 176ECh. 1 - Prob. 177E
Knowledge Booster
Background pattern image
Statistics
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
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
The Shape of Data: Distributions: Crash Course Statistics #7; Author: CrashCourse;https://www.youtube.com/watch?v=bPFNxD3Yg6U;License: Standard YouTube License, CC-BY
Shape, Center, and Spread - Module 20.2 (Part 1); Author: Mrmathblog;https://www.youtube.com/watch?v=COaid7O_Gag;License: Standard YouTube License, CC-BY
Shape, Center and Spread; Author: Emily Murdock;https://www.youtube.com/watch?v=_YyW0DSCzpM;License: Standard Youtube License