Statistics Through Applications
Statistics Through Applications
2nd Edition
ISBN: 9781429219747
Author: Daren S. Starnes, David Moore, Dan Yates
Publisher: Macmillan Higher Education
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Chapter 2.2, Problem 2.30E
To determine

To construct: the box plot on basis of given data and describe the distribution.

Expert Solution & Answer
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Explanation of Solution

Given:

3 0 12 10 8 4 2 7 10 11 6 7 0 35 12 3 10 2 7 9 15 5 14 15 6

Formula used:

  Q1=1×(N+14)th postionQ2=2×(N+14)th postionQ3=3×(N+14)th postionIQR=Q3Q1

  Q3+1.5(IQR)

  Q11.5(IQR)

Calculation:

Order the numbers in ascending order

0 0 2 2 3 3 4 5 6 6 7 7 7 8 9 10 10 10 11 12 12 14 15 15 35

  Q1=1×(N+14)th postion=1×(25+14)th postion=6.5=6th position +0.5(7th6th)=3+0.5(43)=3.5

  Q2=2×(N+14)th postion=2×(25+14)th postion=13th postionM=Q2=7

  Q3=3×(N+14)th postion=3×(25+14)th postion=19.5=19th position+0.5(20th19th)=11+0.5(1211)=11.5Q3=7

  IQR=Q3Q1=11.53.5=8

For outliers

  Q3+1.5(IQR)=11.5+1.5(8)=23.5Q11.5(IQR)=3.51.5(8)=8.5

Graph:

  Statistics Through Applications, Chapter 2.2, Problem 2.30E

By seeing the stem plot centre is at 7 and 1st quartile and 3rd quartile are 3.5 and 11.5 respectively. Inter quartile range is 8 and the minimum and maximum values are 0 and 35 respectively so the spread of the data is between 0 and 35 and the mode which is 10 because of highest frequency of the number. There is one outlier and distribution is positive skewed or right skewed which looking forth.

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Statistics Through Applications

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