Stats: Modeling the World Nasta Edition Grades 9-12
Stats: Modeling the World Nasta Edition Grades 9-12
3rd Edition
ISBN: 9780131359581
Author: David E. Bock, Paul F. Velleman, Richard D. De Veaux
Publisher: PEARSON
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Chapter 4, Problem 50E
To determine

To Explain: the appropriate graphical displays and summary statistics.

Expert Solution & Answer
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Explanation of Solution

Given:

  Stats: Modeling the World Nasta Edition Grades 9-12, Chapter 4, Problem 50E , additional homework tip  1

Calculation:

Median: First ascending the data

-1.6, -1, -0.8, -0.3, 0, 1.1, 1.4, 1.9, 2, 2.2, 2.3, 2.5, 3.3, 3.5, 4.4, 5.5, 5.6, 6, 6.5, 7.9, 11.9

These are 21 values so median is 11th value. So median will be

Median = 2.3%

Mean: Sum of these 21 values is 64.3% so mean will be

  mean=64.321=3.1%

So standard deviation would be

  S.D=(xx¯)n1=220.9820=11.04=3.3%

So standard deviation is 3.3%

IQR

To find IQR let us first find Q1 and Q3 . Lower quartile is middle value of lower half.

Lower half of data has 11 values (including median). Lower half of information

-1.6, -1, -0.8, -0.3, 0, 1.1, 1.4, 1.9, 2, 2.2, 2.3

The lower quartile is 6th value. So

  Q1=1.1

Upper half of data has 25 values. upper half of data is

2.3, 2.5, 3.3, 3.5, 4.4, 5.5, 5.6, 6, 6.5, 7.9, 11.9

The upper quartile is 6th value. So

  Q3=5.5

Therefore IQR is

  IQR=Q3Q1=5.51.1=4.4%

Graph:

  Stats: Modeling the World Nasta Edition Grades 9-12, Chapter 4, Problem 50E , additional homework tip  2

Median and mean of the data is 2.3% and 3.1% respectively. Distribution is skewed to right. So median works as a good measure of centre of distribution in comparison of mean. There is an outlier also means in one state percent change is around 11.9% which is higher than any other state.

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