Chapter 2.2, Problem 32E

Geysers: The geyser Old Faithful in Yellowstone National Park alternates periods of eruption, which typically last from 1.5 to 4 minutes, with periods of dormancy, which are considerably longer. The following table presents the durations, in minutes, of 60 dormancy periods that occurred during a recent year.

1. Construct a frequency distribution using a class width of 5, and using 55 as the lower class limit for the first class.
2. Construct a frequency histogram from the frequency distribution in part (a).
3. Construct a relative frequency distribution using the same class width and lower limit for the first class.
4. Construct a relative frequency histogram.
5. Are the histograms skewed to the left, skewed to the right, or approximately symmetric?
6. Repeat parts (a)—(d), using a class width of 10, and using 50 as the lower class limit for the first class.
7. Do you think that class widths of 5 and 10 are both reasonably good choices for these data, or do you think that one choice is much better than the other? Explain your reasoning.

To determine

To construct:A frequency distribution using a class width of 5, and using 55 as the lower class limit for the first class.

Explanation of Solution

Given information:The following table presents the durations, in minutes, of 60 dormancy periods that occurred during a recent year.

91	99	99	83	99	85	90	96	88	93
88	88	92	116	59	101	90	71	103	97
82	91	89	89	94	94	61	96	66	105
90	93	88	92	86	93	95	83	90	99
89	94	90	95	93	105	96	92	101	91
94	92	94	86	88	99	90	99	84	92

Definition used: Frequency distributions for quantitative data are just like those for qualitative data, except the data are divided into classes rather categories.

Solution:

The class width is 5. The minimum and maximum values of the ratings are 55 and 119.9.

The table of frequency distribution is given by

Dormancy period	Frequency
55-59.9	1
60-64.9	1
65-69.9	1
70-74.9	1
75-79.9	0
80-84.9	4
85-89.9	11
90-94.9	23
95-99.9	12
100-104.9	3
105-109.9	2
110-114.9	0
115-119.9	1

To determine

To construct:A frequency histogram from the frequency distribution.

Explanation of Solution

Given information:The table of frequency distribution is given by

Dormancy period	Frequency
55-59.9	1
60-64.9	1
65-69.9	1
70-74.9	1
75-79.9	0
80-84.9	4
85-89.9	11
90-94.9	23
95-99.9	12
100-104.9	3
105-109.9	2
110-114.9	0
115-119.9	1

Definition used: Histograms based on frequency distributions are called frequency histogram.

Solution:

The frequency histogram for the given data is given by

  

To determine

To construct: A relative frequency distribution.

Explanation of Solution

Given information:The table of frequency distribution is given by

Dormancy period	Frequency
55-59.9	1
60-64.9	1
65-69.9	1
70-74.9	1
75-79.9	0
80-84.9	4
85-89.9	11
90-94.9	23
95-99.9	12
100-104.9	3
105-109.9	2
110-114.9	0
115-119.9	1

Formula used: $Relativefrequency=\frac{Frequency}{ Sum\text{ of all }Frequency}$

Solution:

From the given table,

The sum of all frequency is $1+1+1+1+0+4+11+23+12+3+2+0+1=60$

The table of relative frequency is given by

Dormancy period	Frequency	Relative frequency
55-59.9	1	$\frac{1}{60}=0.017$
60-64.9	1	$\frac{1}{60}=0.017$
65-69.9	1	$\frac{1}{60}=0.017$
70-74.9	1	$\frac{1}{60}=0.017$
75-79.9	0	$\frac{0}{60}=0.000$
80-84.9	4	$\frac{4}{60}=0.067$
85-89.9	11	$\frac{11}{60}=0.183$
90-94.9	23	$\frac{23}{60}=0.383$
95-99.9	12	$\frac{12}{60}=0.200$
100-104.9	3	$\frac{3}{60}=0.050$
105-109.9	2	$\frac{2}{60}=0.033$
110-114.9	0	$\frac{0}{60}=0.000$
115-119.9	1	$\frac{1}{60}=0.017$

To determine

To construct: A relative frequency histogram.

Explanation of Solution

Given information:The following table presents the durations, in minutes, of 60 dormancy periods that occurred during a recent year.

91	99	99	83	99	85	90	96	88	93
88	88	92	116	59	101	90	71	103	97
82	91	89	89	94	94	61	96	66	105
90	93	88	92	86	93	95	83	90	99
89	94	90	95	93	105	96	92	101	91
94	92	94	86	88	99	90	99	84	92

Definition used: Histograms based on relative frequency distributions are called relative frequency histogram.

Solution:

Dormancy period	Relative frequency
55-59.9	0.017
60-64.9	0.017
65-69.9	0.017
70-74.9	0.017
75-79.9	0.000
80-84.9	0.067
85-89.9	0.183
90-94.9	0.383
95-99.9	0.200
100-104.9	0.050
105-109.9	0.033
110-114.9	0.000
115-119.9	0.017

Therelative frequency histogram for the given data is given by

  

To determine

To find: Whether the histograms are skewed to the right, skewed to the left , or approximately symmetric.

Answer to Problem 32E

The histogram is skewed to the left.

Explanation of Solution

Given information:The following table presents the durations, in minutes, of 60 dormancy periods that occurred during a recent year.

91	99	99	83	99	85	90	96	88	93
88	88	92	116	59	101	90	71	103	97
82	91	89	89	94	94	61	96	66	105
90	93	88	92	86	93	95	83	90	99
89	94	90	95	93	105	96	92	101	91
94	92	94	86	88	99	90	99	84	92

Definition used:

A histogram which has a long right-hand tail is said to be skewed to the right.

A histogram which has a long left-hand tail is said to be skewed to the left.

A histogram is symmetric if its right half is a minor image of its left half.

Solution:

The frequency histogram for the given data is given by

  

The above histogram has a long left-hand tail; therefore, it is skewed to the left.

Hence, the histogram is skewed to the left.

To determine

To construct: A frequency distribution using a class width of 10, and using 50 as the lower class limit for the first class, a frequency histogram, relative frequency distribution and relative frequency histogram.

Explanation of Solution

Given information: The following table presents the durations, in minutes, of 60 dormancy periods that occurred during a recent year.

91	99	99	83	99	85	90	96	88	93
88	88	92	116	59	101	90	71	103	97
82	91	89	89	94	94	61	96	66	105
90	93	88	92	86	93	95	83	90	99
89	94	90	95	93	105	96	92	101	91
94	92	94	86	88	99	90	99	84	92

Solution:

The class width is 10. The minimum and maximum values of the ratings are 50 and 119.9.

The table of frequency distribution is given by

Dormancy period	Frequency
50-59.9	1
60-69.9	2
70-79.9	1
80-89.9	15
90-99.9	35
100-109.9	5
110-119.9	1

The frequency histogram for the given data is given by

  

The sum of all frequency is $1+2+1+15+35+5+1=60$

The relative frequency distribution table is given by

Dormancy period	Frequency	Relative frequency
50-59.9	1	$\frac{1}{60}=0.017$
60-69.9	2	$\frac{2}{60}=0.033$
70-79.9	1	$\frac{1}{60}=0.017$
80-89.9	15	$\frac{15}{60}=0.250$
90-99.9	35	$\frac{35}{60}=0.583$
100-109.9	5	$\frac{5}{60}=0.083$
110-119.9	1	$\frac{1}{60}=0.017$

The relative frequency histogram for the given data is given by

  

To determine

To explain: Whether the good choices for the data are that class width of 5 or 10.

Answer to Problem 32E

The distribution of class width of 5 is more reasonably good choice than class width of 10.

Explanation of Solution

Given information:The following table presents the durations, in minutes, of 60 dormancy periods that occurred during a recent year.

91	99	99	83	99	85	90	96	88	93
88	88	92	116	59	101	90	71	103	97
82	91	89	89	94	94	61	96	66	105
90	93	88	92	86	93	95	83	90	99
89	94	90	95	93	105	96	92	101	91
94	92	94	86	88	99	90	99	84	92

The class width of 5 provides more appropriate level of detail in the middle of the histogram, but it is very sparse in the tail.

The class width of 10 is better in the tails, but most of the data are in only two classes.

Therefore, the distribution of class width of 5 is more reasonably good choice than class width of 10.