The Practice of Statistics for AP - 4th Edition
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
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 2.2, Problem 63E

(a)

To determine

To describe: the shape, center, and spread of the distribution of shark lengths

(a)

Expert Solution
Check Mark

Answer to Problem 63E

Shape: symmetric and roughly bell-shaped.

Center: Mean is 15.5884 and median is 15.75

Spread: standard deviation is 2.5499

Explanation of Solution

Given:

    18.712.318.616.415.718.314.615.814.917.612.1
    16.416.717.816.212.617.813.812.215.214.712.4
    13.215.814.316.69.418.213.213.615.316.113.5
    19.116.222.816.813.613115.719.718.713.216.8

Lengths of sharks are in feet of 44

Calculation:

Descriptive statistics and a histogram are provided below:

    VariableNMeanSt.Dev.MinQ1MQ3Max
    Shark length4415.5862.5509.4013.52515.7517.4022.80

  The Practice of Statistics for AP - 4th Edition, Chapter 2.2, Problem 63E , additional homework tip  1

Shark lengths are roughly symmetric with a peak at 16 and vary from minimum value of 9.40 feet to maximum value 22.8 feet.

Shape: symmetric and roughly bell-shaped.

Center: Mean is 15.5884 and median is 15.75

Spread: standard deviation is 2.5499

Conclusion:

Shape: symmetric and roughly bell-shaped.

Center: Mean is 15.5884 and median is 15.75

Spread: standard deviation is 2.5499

(b)

To determine

To describe: the percent of observations that fall within one, two, and three standard deviations of the mean.

(b)

Expert Solution
Check Mark

Answer to Problem 63E

One standard deviations = 68.2%

Two standard deviations = 95.5%

Three standard deviations = 100%

Explanation of Solution

Calculation:

From the above output,

Mean, x¯=15.586

Sample standard deviation, s=2.550

  x¯s=13.036x¯+s=18.136

Percent of interval

  (68.26%)          68.2%

  x¯2s=10.487x¯+2s=20.686

Percent of interval

  (95.44%)          95.5%

  x¯3s=7.937x¯+3s=23.236

Percent of interval

  (99.73%)          100.0%

According to the 68-95-99.7 rule, approximately 68.2% of the data values fall within one standard deviation. 95.5% of the data values fall within two standard deviations, and 100% of the data values fall within three standard deviations of the mean.

Conclusion:

Therefore,

One standard deviations = 68.2%

Two standard deviations = 95.5%

Threestandard deviations = 100%

(c)

To determine

To construct: a Normal probability plot and interpret the plot.

(c)

Expert Solution
Check Mark

Answer to Problem 63E

The pattern in the Normal probability plot is fairly linear and it is roughly normally distributed.

  The Practice of Statistics for AP - 4th Edition, Chapter 2.2, Problem 63E , additional homework tip  2

Explanation of Solution

Calculation:

A Normal probability plot from Minitab is shown below:

  The Practice of Statistics for AP - 4th Edition, Chapter 2.2, Problem 63E , additional homework tip  3

The plot is fairly linear except from one small shark and one large shark lengths, indicating that the Normal distribution is appropriate.

Conclusion:

Therefore, a Normal probability plot is plotted. The pattern in the Normal probability plot is fairly linear and it is roughly normally distributed.

(c)

To determine

To explain:whether the data are approximately Normal

(c)

Expert Solution
Check Mark

Answer to Problem 63E

The data are approximately Normal

Explanation of Solution

Calculation:

Results from the parts (a), (b), and (c) indicates that shark lengths are approximately normal.

Approximately normal, because the normal probability plot was roughlynormal and the histogram were roughly bell-shaped.

Conclusion:

Therefore, the data are approximately Normal

Chapter 2 Solutions

The Practice of Statistics for AP - 4th Edition

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
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.
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
Statistics 4.1 Point Estimators; Author: Dr. Jack L. Jackson II;https://www.youtube.com/watch?v=2MrI0J8XCEE;License: Standard YouTube License, CC-BY
Statistics 101: Point Estimators; Author: Brandon Foltz;https://www.youtube.com/watch?v=4v41z3HwLaM;License: Standard YouTube License, CC-BY
Central limit theorem; Author: 365 Data Science;https://www.youtube.com/watch?v=b5xQmk9veZ4;License: Standard YouTube License, CC-BY
Point Estimate Definition & Example; Author: Prof. Essa;https://www.youtube.com/watch?v=OTVwtvQmSn0;License: Standard Youtube License
Point Estimation; Author: Vamsidhar Ambatipudi;https://www.youtube.com/watch?v=flqhlM2bZWc;License: Standard Youtube License