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
Chapter 2.2, Problem 63E
(a)
To determine
To describe: the shape, center, and spread of the distribution of shark lengths
(a)
Expert Solution
Answer to Problem 63E
Shape: symmetric and roughly bell-shaped.
Center:
Spread: standard deviation is 2.5499
Explanation of Solution
Given:
| 18.7 | 12.3 | 18.6 | 16.4 | 15.7 | 18.3 | 14.6 | 15.8 | 14.9 | 17.6 | 12.1 |
| 16.4 | 16.7 | 17.8 | 16.2 | 12.6 | 17.8 | 13.8 | 12.2 | 15.2 | 14.7 | 12.4 |
| 13.2 | 15.8 | 14.3 | 16.6 | 9.4 | 18.2 | 13.2 | 13.6 | 15.3 | 16.1 | 13.5 |
| 19.1 | 16.2 | 22.8 | 16.8 | 13.6 | 131 | 15.7 | 19.7 | 18.7 | 13.2 | 16.8 |
Lengths of sharks are in feet of 44
Calculation:
| Variable | N | Mean | St.Dev. | Min | Q1 | M | Q3 | Max |
| Shark length | 44 | 15.586 | 2.550 | 9.40 | 13.525 | 15.75 | 17.40 | 22.80 |
Shark lengths are roughly symmetric with a peak at 16 and vary from minimum value of 9.40 feet to maximum value 22.8 feet.
Shape: symmetric and roughly bell-shaped.
Center: Mean is 15.5884 and median is 15.75
Spread: standard deviation is 2.5499
Conclusion:
Shape: symmetric and roughly bell-shaped.
Center: Mean is 15.5884 and median is 15.75
Spread: standard deviation is 2.5499
(b)
To determine
To describe: the percent of observations that fall within one, two, and three standard deviations of the mean.
(b)
Expert Solution
Answer to Problem 63E
One standard deviations = 68.2%
Two standard deviations = 95.5%
Three standard deviations = 100%
Explanation of Solution
Calculation:
From the above output,
Mean, ˉx=15.586
Sample standard deviation, s=2.550
ˉx−s=13.036ˉx+s=18.136
Percent of interval
(68.26%) 68.2%
ˉx−2s=10.487ˉx+2s=20.686
Percent of interval
(95.44%) 95.5%
ˉx−3s=7.937ˉx+3s=23.236
Percent of interval
(99.73%) 100.0%
According to the 68-95-99.7 rule, approximately 68.2% of the data values fall within one standard deviation. 95.5% of the data values fall within two standard deviations, and 100% of the data values fall within three standard deviations of the mean.
Conclusion:
Therefore,
One standard deviations = 68.2%
Two standard deviations = 95.5%
Threestandard deviations = 100%
(c)
To determine
To construct: a Normal
(c)
Expert Solution
Answer to Problem 63E
The pattern in the Normal probability plot is fairly linear and it is roughly
Explanation of Solution
Calculation:
A Normal probability plot from Minitab is shown below:
The plot is fairly linear except from one small shark and one large shark lengths, indicating that the Normal distribution is appropriate.
Conclusion:
Therefore, a Normal probability plot is plotted. The pattern in the Normal probability plot is fairly linear and it is roughly normally distributed.
(c)
To determine
To explain:whether the data are approximately Normal
(c)
Expert Solution
Answer to Problem 63E
The data are approximately Normal
Explanation of Solution
Calculation:
Results from the parts (a), (b), and (c) indicates that shark lengths are approximately normal.
Approximately normal, because the normal probability plot was roughlynormal and the histogram were roughly bell-shaped.
Conclusion:
Therefore, the data are approximately Normal
Chapter 2 Solutions
The Practice of Statistics for AP - 4th Edition
Ch. 2.1 - Prob. 1.1CYUCh. 2.1 - Prob. 1.2CYUCh. 2.1 - Prob. 1.3CYUCh. 2.1 - Prob. 1.4CYUCh. 2.1 - Prob. 2.1CYUCh. 2.1 - Prob. 2.2CYUCh. 2.1 - Prob. 2.3CYUCh. 2.1 - Prob. 3.1CYUCh. 2.1 - Prob. 3.2CYUCh. 2.1 - Prob. 3.3CYU
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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