Q. 24 - Elementary Statistics

Question

On-Time Arrivals. The Air Travel Consumer Report is a monthly product of the Department of Transportation's Office of Aviation Enforcement and Proceedings. The report is designed to assist consumers with information on the quality of services provided by the airlines. Following are the percentages of on-time arrivals for June 2013 by the 16 reporting airlines.

93.1    87.5    77.9    77.3

76.1    73.8    73.0    72.5

71.1    70.6    69.8    69.6

69.0    66.0    65.9    61.8

a. Identify the classes for grouping these data, using cutpoint grouping with classes of equal width 5 and a first lower class cutpoint of 60.
 b. Identify the class midpoints of the classes found in part (a).
c. Construct frequency and relative frequency distributions of the data based on your classes from part (a).
d. Draw a frequency histogram of the data based on your classes from part (a).
e. Round each observation to the nearest whole number, and then construct a stem-and-leaf diagram with two lines per stem.
f. Obtain the greatest integer in each observation, and then construct a stem-and-leaf diagram with two lines per stem.
g. Which of the stem-and-leaf diagrams in parts (e) and (f) corresponds to the frequency histogram in part (d)? Explain why.

Step-by-Step Solution

Answer

Ans:  

(a)  Given that the first cutting point is 60 and the same width is 5. Therefore, the first class is $60-less than 65$ and the second class is $65-less than 70$ and so on. Also, the highest visibility is 93.1. Therefore, the final stage is $90-less than 95$.

(b)  

Cut point grouping classes	Frequency	Class cut point
60 -under 65	1	62.5
65 -under 70	2	67.5
70 -under 75	5	72.5
75 -under 80	3	77.5
80 -under 85	0	82.5
85 -under 90	2	87.5
90 -under 95	1	92.5
Total	16

(c)  

(d) 

Step 1. Given information.

given,

93.1    87.5    77.9    77.3

76.1    73.8    73.0    72.5

71.1    70.6    69.8    69.6

69.0    66.0    65.9    61.8

Step 2. (a) Identify classes using cutpoint grouping with first-class cutpoints like 60   a n d   5 equal widths.

Given that the first cutting point is 60 and the same width is 5. Therefore, the first class is $60 - less than 65$ and the second class is $65 - less than 70$ and so on. Also, the highest visibility is 93.1. Therefore, the final stage is $90-less than 95$.

Step 3. (b) Find the class midpoints for the classes obtained in part (a).

The class midpoints for the classes are found using the formula,
$Class Midpoint=\frac{Upper cutpoint}{Lower cutpoint}$

Thus, the class midpoints for the classes found in part (a) are given below:

   

Cutpoint grouping classes	Frequency	Class cutpoint
60 -under 65	1	62.5
65 -under 70	2	67.5
70 -under 75	5	72.5
75 -under 80	3	77.5
80 -under 85	0	82.5
85 -under 90	2	87.5
90 -under 95	1	92.5
Total	16

Step 4. (c) Find frequency distribution and related frequency distribution using cutpoint grouping with first-class cutting areas such as 60 and 5 equal widths.

Thus, frequency distribution and related-frequency distribution using cutpoint grouping are provided below:

Cutpoint grouping classes	Frequency	Relative Frequency
60 -under 65	1	0.0625
65 -under 70	5	0.3125
70 -under 75	5	0.3125
75 -under 80	3	0.1875
80 -under 85	0	0.0000
85 -under 90	2	0.0625
90 -under 95	1	0.0625
Total	16	1.000

Step 5. (d) Use Minitab to create a histogram of data provided based on the results of subsection (a).

The points represent a horizontal line while forming a histogram.

MINITAB process:

Step 1: Select Graph> Histogram.

Step 2: Select Easy, then click OK.

Step 3: For graph variables, enter the corresponding PERCENTAGE column.

Step 4: Click OK.

Minitab output:

According to Minitab output, the height of the bar represents the frequency obtained in part (a).

Step 6. (e)