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
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Question
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Chapter 1.2, Problem 54E

(a)

To determine

To make: a histogram using classes of width 2, starting at 0.

(b) Describe the shape, center, and spread of the distribution. Which countries are outliers?

(a)

Expert Solution
Check Mark

Answer to Problem 54E

  The Practice of Statistics for AP - 4th Edition, Chapter 1.2, Problem 54E , additional homework tip  1

Explanation of Solution

Given:

    CountryCO2CountryCO2CountryCO2CountryCO2
    Algeria2.6Egypt2.0Italy7.8Romania4.2
    Argentina3.6Ethiopia0.1Japan9.5Russia10.8
    Australia18.4France6.2Kenya0.3Saudi Arabia13.8
    Bangladesh0.3Germany9.9Korea, North3.3South Africa7.0
    Brazil1.8Ghana0.3Korea, South9.3Spain7.9
    Canada17.0India1.1Malaysia5.5Sudan0.3
    China3.9Indonesia1.6Mexico3.7Tanzania0.1
    Colombia1.3Iran6.0Morocco1.4Thailand3.3
    Congo0.2Iraq2.9Myanmar0.2Turkey3.0
    CountryCO2CountryCO2
    Italy7.8Romania4.2
    Japan9.5Russia10.8
    Kenya0.3Saudi Arabia13.8
    Korea, North3.3South Africa7.0
    Korea, South9.3Spain7.9
    Malaysia5.5Sudan0.3
    Mexico3.7Tanzania0.1
    Morocco1.4Thailand3.3
    Myanmar0.2Turkey3.0
    Nepal0.1Ukraine6.3
    Nigeria0.4United Kingdom8.8
    Pakistan0.8United States19.6
    Peru1.0Uzbekistan4.2
    Philippines0.9Venezuela5.4
    Poland7.8Vietnam1.0

Calculation:

The Histogram of Carbon dioxide emission per person from countries with population of at least 20 million6 is below:

  The Practice of Statistics for AP - 4th Edition, Chapter 1.2, Problem 54E , additional homework tip  2

Conclusion:

Thus, a histogram is made for the given data.

(b)

To determine

To describe: the shape, center, spread and outliers

(b)

Expert Solution
Check Mark

Answer to Problem 54E

Shape: skewed to the right

Center: between 2 and 4 metric tons per person

Spread: 0.1 to 19.5metric tons per person

Outliers: Australia, Canada and the US

Explanation of Solution

The shape is irregular and asymmetric.lt is skewed towards one side. It seems closer to the right skewed.

The centre of data is approximately between 4 to 5 metric tons emission per person. There is three outlier that is 17, 18.4,and 19.6

  range =Minimum value  Maximum value=19.60.1=19.5

Therefore, the data is approximately follows skewed distribution.

With 17 metric tons emission per person, with 18.4 metric tons emission per person and US with 19.6 metric tons emission per person are the outliers in the data.

Chapter 1 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 1.1 - Prob. 16ECh. 1.1 - Prob. 17ECh. 1.1 - Prob. 18ECh. 1.1 - Prob. 19ECh. 1.1 - Prob. 20ECh. 1.1 - Prob. 21ECh. 1.1 - Prob. 22ECh. 1.1 - Prob. 23ECh. 1.1 - Prob. 24ECh. 1.1 - Prob. 25ECh. 1.1 - Prob. 26ECh. 1.1 - Prob. 27ECh. 1.1 - Prob. 28ECh. 1.1 - Prob. 29ECh. 1.1 - Prob. 30ECh. 1.1 - Prob. 31ECh. 1.1 - Prob. 32ECh. 1.1 - Prob. 33ECh. 1.1 - Prob. 34ECh. 1.1 - Prob. 35ECh. 1.1 - Prob. 36ECh. 1.2 - Prob. 1.1CYUCh. 1.2 - Prob. 1.2CYUCh. 1.2 - Prob. 1.3CYUCh. 1.2 - Prob. 1.4CYUCh. 1.2 - Prob. 2.1CYUCh. 1.2 - Prob. 2.2CYUCh. 1.2 - Prob. 2.3CYUCh. 1.2 - Prob. 2.4CYUCh. 1.2 - Prob. 3.1CYUCh. 1.2 - Prob. 3.2CYUCh. 1.2 - Prob. 4.1CYUCh. 1.2 - Prob. 4.2CYUCh. 1.2 - Prob. 4.3CYUCh. 1.2 - Prob. 4.4CYUCh. 1.2 - Prob. 37ECh. 1.2 - Prob. 38ECh. 1.2 - Prob. 39ECh. 1.2 - Prob. 40ECh. 1.2 - Prob. 41ECh. 1.2 - Prob. 42ECh. 1.2 - Prob. 43ECh. 1.2 - Prob. 44ECh. 1.2 - Prob. 45ECh. 1.2 - Prob. 46ECh. 1.2 - Prob. 47ECh. 1.2 - Prob. 48ECh. 1.2 - Prob. 49ECh. 1.2 - Prob. 50ECh. 1.2 - Prob. 51ECh. 1.2 - Prob. 52ECh. 1.2 - Prob. 53ECh. 1.2 - Prob. 54ECh. 1.2 - Prob. 55ECh. 1.2 - Prob. 56ECh. 1.2 - Prob. 57ECh. 1.2 - Prob. 58ECh. 1.2 - Prob. 59ECh. 1.2 - Prob. 60ECh. 1.2 - Prob. 61ECh. 1.2 - Prob. 62ECh. 1.2 - Prob. 63ECh. 1.2 - Prob. 64ECh. 1.2 - Prob. 65ECh. 1.2 - Prob. 66ECh. 1.2 - Prob. 67ECh. 1.2 - Prob. 68ECh. 1.2 - Prob. 69ECh. 1.2 - Prob. 70ECh. 1.2 - Prob. 71ECh. 1.2 - Prob. 72ECh. 1.2 - Prob. 73ECh. 1.2 - Prob. 74ECh. 1.2 - Prob. 75ECh. 1.2 - Prob. 76ECh. 1.2 - Prob. 77ECh. 1.2 - Prob. 78ECh. 1.3 - Prob. 1.1CYUCh. 1.3 - Prob. 1.2CYUCh. 1.3 - Prob. 1.3CYUCh. 1.3 - Prob. 1.4CYUCh. 1.3 - Prob. 2.1CYUCh. 1.3 - Prob. 2.2CYUCh. 1.3 - Prob. 2.3CYUCh. 1.3 - Prob. 2.4CYUCh. 1.3 - Prob. 3.1CYUCh. 1.3 - Prob. 3.2CYUCh. 1.3 - Prob. 3.3CYUCh. 1.3 - Prob. 3.4CYUCh. 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. 97ECh. 1.3 - Prob. 98ECh. 1.3 - Prob. 99ECh. 1.3 - Prob. 100ECh. 1.3 - Prob. 101ECh. 1.3 - Prob. 102ECh. 1.3 - Prob. 103ECh. 1.3 - Prob. 104ECh. 1.3 - Prob. 105ECh. 1.3 - Prob. 106ECh. 1.3 - Prob. 107ECh. 1.3 - Prob. 108ECh. 1.3 - Prob. 109ECh. 1.3 - Prob. 110ECh. 1.3 - Prob. 111ECh. 1.3 - Prob. 112ECh. 1.3 - Prob. 113ECh. 1 - Prob. 1CYUCh. 1 - Prob. 2CYUCh. 1 - Prob. 1ECh. 1 - Prob. 2ECh. 1 - Prob. 3ECh. 1 - Prob. 4ECh. 1 - Prob. 5ECh. 1 - Prob. 6ECh. 1 - Prob. 7ECh. 1 - Prob. 8ECh. 1 - Prob. 1CRECh. 1 - Prob. 2CRECh. 1 - Prob. 3CRECh. 1 - Prob. 4CRECh. 1 - Prob. 5CRECh. 1 - Prob. 6CRECh. 1 - Prob. 7CRECh. 1 - Prob. 8CRECh. 1 - Prob. 9CRECh. 1 - Prob. 10CRECh. 1 - Prob. 1PTCh. 1 - Prob. 2PTCh. 1 - Prob. 3PTCh. 1 - Prob. 4PTCh. 1 - Prob. 5PTCh. 1 - Prob. 6PTCh. 1 - Prob. 7PTCh. 1 - Prob. 8PTCh. 1 - Prob. 9PTCh. 1 - Prob. 10PTCh. 1 - Prob. 11PTCh. 1 - Prob. 12PTCh. 1 - Prob. 13PTCh. 1 - Prob. 14PTCh. 1 - Prob. 15PT
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