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 1.2, Problem 37E

(a)

To determine

To make: A dotplot to display the given data.

(a)

Expert Solution
Check Mark

Answer to Problem 37E

  The Practice of Statistics for AP - 4th Edition, Chapter 1.2, Problem 37E , additional homework tip  1

Explanation of Solution

Given:

    96868866.5679434
    5611636610784.5977

We have a data on the sleeping hours of the students.

Dot plot is one of the simplest graph that shows each data value as a dot above its location on number line. The dot plot of the given data is as follows:

  The Practice of Statistics for AP - 4th Edition, Chapter 1.2, Problem 37E , additional homework tip  2

Conclusion:

Thus, a dotplot is drawn.

(b)

To determine

To describe: The overall pattern of the distribution and deviations from that pattern.

(b)

Expert Solution
Check Mark

Answer to Problem 37E

Shape: the data is roughly symmetric.

Centre: about 6 hours, because the highest dots are at 6 in the dotplot.

Spread: it is between 3 hours and 11 hours.

Outliers: there do not appear to be deviations from the pattern.

Explanation of Solution

The overall pattern of the distribution can be described by its shape, center, spread and outliers.

Shape: the Dot plot has the peak at 6 this means that maximum students have slept for 6 hour’s night before the survey was done. Thus 6 is the mode. Most of the students have slept between 6 to 9 hours. However, distribution has a long tall towards right.

Centre: We can describe centre by finding a value that devides the observations. So that about half take larger values and about half take smaller values. This is called Median of the distribution which is 6 for this data. We can also calculate the mean no of hours that is 6.64 for this data.

Spread: The spread of given data is from 3 hours to 11 hours. The range can be calculated by subtracting smallest value from largest value 11hrs3hrs=8hrs .

Outliers: The values that differ somewhat from the overall pattern. However they do stand clearly apart from the rest of the distribution. We do not have an outlier in the given distribution.

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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