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 2.1, Problem 9E

(a)

To determine

To estimate: the interquartile range of the distribution

(a)

Expert Solution
Check Mark

Answer to Problem 9E

The Inter-Quartile range is $27.

Explanation of Solution

Given:

  The Practice of Statistics for AP - 4th Edition, Chapter 2.1, Problem 9E , additional homework tip  1

Calculation:

The figure below is a cumulative relative frequency graph of the amount spent by 50consecutive grocery shoppers at a store.

  The Practice of Statistics for AP - 4th Edition, Chapter 2.1, Problem 9E , additional homework tip  2

We are supposed to find the Inter-Quartile range of the amount spent by 50 consecutive grocery shoppers at a store basing on the above cumulative relative frequency graph.

Inter-Quartile range is defined as follows:

  IQR=Q3Q1

Where Q1 corresponds to the 25th  percentile and Q3 corresponds to the 75th  percentile.

Procedure to find the quartiles is given as follows:

Step1: Draw a horizontal line from the value 25 on the vertical axis and determine the point onthe ogive where the line intersects the ogive.

Step2: Draw a vertical line from this point to the horizontal axis. The line intersects the axis atapproximately $19

Thus, $19 is the estimate of the Q1

Step3: Draw a horizontal line from the value 75 on the vertical axis and determine the point onthe ogive where the line intersects the ogive.

Step4: Draw a vertical line from this point to the horizontal axis. The line intersects the axis at approximately $46

Thus, $46is the estimate of the Q3

The following graph shows the above procedure:

  The Practice of Statistics for AP - 4th Edition, Chapter 2.1, Problem 9E , additional homework tip  3

Hence, the Inter-Quartile range is calculated as follows:

  IQR =Q3Q1=4619=27

Therefore, the Inter-Quartile range of the amount spent by 50 consecutive grocery shoppers at a store is $27.

Conclusion:

Therefore, the Inter-Quartile range is $27.

(b)

To determine

To find: the percentile for the shopper who spent $19.50

(b)

Expert Solution
Check Mark

Answer to Problem 9E

Here, $19.50 is at about the 26th  percentile.

Explanation of Solution

Calculation:

From the above figure we can clearly see that shopper who spent $19.50 is just above the 25th  percentile. Hence, we can say that $19.50 is at about the 26th  percentile.

Conclusion:

Hence, $19.50 is at about the 26th  percentile.

(c)

To determine

To draw: the histogram that correspondsto the given graph

(c)

Expert Solution
Check Mark

Answer to Problem 9E

  The Practice of Statistics for AP - 4th Edition, Chapter 2.1, Problem 9E , additional homework tip  4

Explanation of Solution

Calculation:

From the above cumulative graph, we find the cumulative frequencies corresponding todifferent mid values.

From the cumulative graph, the obtained cumulative frequencies are 5,25,55,62,80,87,88 90,96,98 corresponding to mid values 10,20,30,40,50,60,70,80,90,100 .The corresponding frequencies are obtained as 5,20,30,7,18,7,1,2,6,2

Using frequencies, we construct histogram for the given data. The obtained graph is as follows:

  The Practice of Statistics for AP - 4th Edition, Chapter 2.1, Problem 9E , additional homework tip  5

From the above histogram, we can see that the data of the amount spent by 50 consecutivegrocery shoppers at a store is spread towards right side. Hence, we can say that the abovedata is skewed towards right.

Conclusion:

Therefore, the histogram is drawn.

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
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