Chapter 2.2, Problem 30E

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. a. Construct a frequency distribution using a class width of 5, and using 55 as the lower class limit for the first class.
2. b. Construct a frequency histogram from the frequency distribution in part (a).
3. c. Construct a relative frequency distribution using the same class width and lower limit for the first class.
4. d. Construct a relative frequency histogram.
5. e. Are the histograms skewed to the left, skewed to the right, or approximately symmetric?
6. f. Repeat parts (a)–(d), using a class width of 10, and using 50 as the lower class limit for the first class.
7. g. 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

Construct the frequency distribution with a class width of 5 and a lower limit of 55 for the first class.

Answer to Problem 30E

The frequency distribution is,

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
Total	60

Explanation of Solution

Calculation:

The given information is that a table representing the duration of 60 dormancy periods that occurred during a recent year.

Frequency:

The frequencies are calculated by using the tally mark and the range of the data is from 55 to 119.9.

• Based on the given information, the class intervals are 55-59.9, 60-64.9, 65-69.9, 70-74.9, 75-79.9, 80-84.9, 85-89.9, 90-94.9, 95-99.9, 100-104.9, 105-109.9, 110-114.9, 115-119.9.
• Make a tally mark for each value in the corresponding dormancy period and continue for all values in the data.
• The number of tally marks in each class represents the frequency, f of that class.

Similarly, the frequency of remaining classes for the dormancy period is given below:

Dormancy Period	Tally	Frequency
55-59.9	$|$	1
60-64.9	$|$	1
65-69.9	$|$	1
70-74.9	$|$	1
75-79.9		0
80-84.9	$\left|\right|\left|\right|$	4
85-89.9	$\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}|$	11
90-94.9	$\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}\left|\right||$	23
95-99.9	$\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}\left|\right|$	12
100-104.9	$\left|\right||$	3
105-109.9	$\left|\right|$	2
110-114.9		0
115-119.9	$|$	1
Total		60

To determine

Construct the frequency histogram based on the frequency distribution.

Answer to Problem 30E

Output obtained from MINITAB software for the dormancy period is:

Explanation of Solution

Calculation:

Frequency Histogram:

Software procedure:

• Step by step procedure to draw the relative frequency histogram for dormancy period using MINITAB software.
• Choose Graph > Histogram.
• Choose Simple.
• Click OK.
• In Graph variables, enter the column of Dormancy Period.
• In Scale, Choose Y-scale Type as Frequency.
• Click OK.
• Select Edit Scale, Enter 55,60,65,70,75,80,85,90,95,100,105,110, 115, 120 in Positions of ticks.
• In Binning, Under Interval Type select Cutpoint and under Interval Definition select Number of intervals as 13 and Cutpoint positions as 55,60,65,70,75,80,85,90,95,100,105,110, 115, 120.
• In Labels, Enter 55,60,65,70,75,80,85,90,95,100,105,110, 115, 120 in Specified.
• Click OK.

Observation:

From the bar graph, it can be seen that maximum dormancy period is in the interval 90-95.

To determine

Construct a relative frequency distribution for the data.

Answer to Problem 30E

The relative frequency distribution for the data is:

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

Explanation of Solution

Calculation:

Relative frequency:

The general formula for the relative frequency is,

$\text{Relative}\text{Frequency = }\frac{\text{Frequency}}{\text{Total frequency}}$

Therefore,

$\begin{array}{c}\text{Relative}\text{Frequency}\text{for the class 55-59}\text{.9}=\frac{1}{60}\\ =0.017\end{array}$

Similarly, the relative frequencies for the remaining dormancy periods are obtained below:

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	$\frac{1}{60}=0.017$
Total	60

To determine

Construct the relative frequency histogram based on the relative frequency distribution.

Answer to Problem 30E

Output obtained from MINITAB software for the dormancy period is:

Explanation of Solution

Calculation:

Relative Frequency Histogram:

Software procedure:

• Step by step procedure to draw the relative frequency histogram for the dormancy period using MINITAB software.
• Choose Graph > Histogram.
• Choose Simple.
• Click OK.
• In Graph variables, enter the column of Dormancy Period.
• In Scale, Choose Y-scale Type as Percent.
• Click OK.
• Select Edit Scale, Enter 55,60,65,70,75,80,85,90,95,100,105,110, 115, 120 in Positions of ticks.
• In Binning, Under Interval Type select Cutpoint and under Interval Definition select Number of intervals as 13 and Cutpoint positions as 55,60,65,70,75,80,85,90,95,100,105,110, 115, 120.
• In Labels, Enter 30,40,50,60,70,80,90,100,110,120,130,140 in Specified.
• Click OK.

Observation:

From the bar graph, it can be seen that maximum dormancy period is in the interval 90-95.

To determine

Check whether the histograms are skewed to the left, skewed to the right or approximately symmetric.

Answer to Problem 30E

The histogram is skewed to the left.

Explanation of Solution

From the given histogram, the left hand tail is larger than the right hand tail from the maximum frequency value. Hence, the shape of the distribution is left skewed.

To determine

Repeat parts (a) to (d), using a class width of 10 and a lower limit of 50 for the first class.

Answer to Problem 30E

The frequency distribution is,

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
Total	60

Output obtained from MINITAB software for the dormancy period is:

The relative frequency distribution for the data is:

Dormancy Period	Relative frequency
50-59.9	0.017
60-69.9	0.033
70-79.9	0.017
80-89.9	0.250
90-99.9	0.583
100-109.9	0.083
110-119.9	0.017

Output obtained from MINITAB software for the dormancy period is:

Explanation of Solution

Calculation:

Frequency:

The frequencies are calculated by using the tally mark and the range of the data is from 50 to 119.9.

• Based on the given information, the class intervals are 50-59.9, 60-69.9, 70-79.9, 80-89.9, 90-99.9, 100-109.9, 110-119.9.
• Make a tally mark for each value in the corresponding duration periods and continue for all values in the data.
• The number of tally marks in each class represents the frequency, f of that class.

Similarly, the frequency of remaining classes for the dormancy period is given below:

Dormancy Period	Tally	Frequency
50-59.9	$|$	1
60-69.9	$\left|\right|$	2
70-79.9	$|$	1
80-89.9	$\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}$	15
90-99.9	$\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}\bcancel{\left|\right|\left|\right|}$	35
100-109.9	$\bcancel{\left|\right|\left|\right|}$	5
110-119.9	$|$	1
Total		60

Frequency Histogram:

Software procedure:

• Step by step procedure to draw the relative frequency histogram for the dormancy periods using MINITAB software.
• Choose Graph > Histogram.
• Choose Simple.
• Click OK.
• In Graph variables, enter the column of Dormancy Periods.
• In Scale, Choose Y-scale Type as Frequency.
• Click OK.
• Select Edit Scale, Enter 50, 60, 70, 80, 90, 100, 110, 120 in Positions of ticks.
• In Binning, Under Interval Type select Cutpoint and under Interval Definition select Number of intervals as 7 and Cutpoint positions as 50, 60, 70, 80, 90, 100, 110, 120.
• In Labels, Enter 50, 60, 70, 80, 90, 100, 110, 120 in Specified.
• Click OK.

Observation:

From the bar graph, it can be seen that maximum dormancy period is in the interval 90-100.

$\begin{array}{c}\text{Relative}\text{Frequency}\text{for the class 50-60}=\frac{1}{60}\\ =0.017\end{array}$

Similarly, the relative frequencies for the remaining dormancy periods are obtained below:

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	$\frac{1}{60}=0.017$
Total	60

Relative Frequency Histogram:

Software procedure:

• Step by step procedure to draw the relative frequency histogram for the dormancy period using MINITAB software.
• Choose Graph > Histogram.
• Choose Simple.
• Click OK.
• In Graph variables, enter the column of Dormancy Period.
• In Scale, Choose Y-scale Type as Percent.
• Click OK.
• Select Edit Scale, Enter 50, 60, 70, 80, 90, 100, 110, 120 in Positions of ticks.
• In Binning, Under Interval Type select Cutpoint and under Interval Definition select Number of intervals as 7 and Cutpoint positions as 50, 60, 70, 80, 90, 100, 110, 120.
• In Labels, Enter 50, 60, 70, 80, 90, 100, 110, 120 in Specified.
• Click OK.

Observation:

From the bar graph, it can be seen that maximum dormancy period is in the interval 90-100.

To determine

Check whether the class widths of both 5 and 10 are reasonable or one choice is much better than the other and explain the reason.

Answer to Problem 30E

Both the class width of 5 and 10 are reasonable.

Explanation of Solution

Answer will vary.

The number of classes can be obtained by the following two points:

• The number of classes should be between 5 and 20.
• A larger number of classes will be appropriate for a very large data set.

Here number of classes for the histogram with class width 5 and 10 are 13 and 7 respectively. Thus, the number of classes should be between 5 and 20. From the histograms, it can be observed that the histogram of class width 5 gives more information in the middle of the histogram but histogram of class width 10 is better in tails.

Hence, the class widths of both 5 and 10 are reasonable.