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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Chapter 1.3, Problem 94E

(a)

To determine

To make: a boxplot of the given data

(a)

Expert Solution
Check Mark

Answer to Problem 94E

  The Practice of Statistics for AP - 4th Edition, Chapter 1.3, Problem 94E , additional homework tip  1

Explanation of Solution

Given:

  The Practice of Statistics for AP - 4th Edition, Chapter 1.3, Problem 94E , additional homework tip  2

Calculation:

Sort the data values:

3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9,10,10, 10, 10, 11, 11, 11, 11, 12, 13, 15, 15, 15, 17,20,21,21,27,31, 34,55

Since the number of data values is odd, the median is the middle value of the sorted data set:

MEDIAN = 8

The first quartile is the median of all data values below the median. Since the data set contains 25 values below the median, the first quartile is then the 13th data value.

  Q1=4

The third quartile is the median of all data values above the median. Since the data set contains 12 values above the median, the third quartile is then the 39th data value.

  Q3=12

Outliers are observations that are more than 1.5 times the IQR above Q3 or below Ql. The interquartile range is the difference between the third and first quartile.

  Q3+1.5IQR=12+1.5(124)=24Q1+1.5IQR=41.5(124)=9

The minimum is 3 and the maximum is 55. We then note that 27, 31, 34 and 55 are outliers.

Create a boxplot

The outliers are represented by an X. The box has boundaries the quartiles and a line has been drawn at the median. The whiskers represent the minimum and maximum of the data points (that are not outliers).

  The Practice of Statistics for AP - 4th Edition, Chapter 1.3, Problem 94E , additional homework tip  3

Conclusion:

Therefore, a boxplot of the given data is drawn.

(b)

To determine

To find: to summarize the distribution which one is used

(b)

Expert Solution
Check Mark

Answer to Problem 94E

It is best to summarize the distribution by using the median and the IQR

Explanation of Solution

In exercise 94a, we found that the data contains 4 outliers.

It is best to summarize the distribution by using the median and the IQR, because they are not influenced by theoutliers, while the mean and the standard deviation are influenced by the outliers.

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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