Concept explainers
Decimal Data: Batting Averages The following data represent baseball batting averages for a random sample of National League players neat the end of the baseball season. the data are from the baseball statistics section of the Denver Post.
0.194 | 0.258 | 0.190 | 0.291 | 0.158 | 0.295 | 0.261 | 0.250 | 0.181 |
0.125 | 0.107 | 0.260 | 0.309 | 0.309 | 0.276 | 0.287 | 0.317 | 0.252 |
0.215 | 0.250 | 0.246 | 0.260 | 0.265 | 0.182 | 0.113 | 0.200 |
(a) Multiply each data value by 1000 to “clear” the decimals.
(b) Use the standard procedares of this section to make a frequency table and histogram with your whole-number data. Use five classes.
(c) Divide class limits, class boundaries, and class midpoints by 1000 to get back to your original dat.
(a)
To find: The decimal data that are multiply with 1000 for each value in the data..
Answer to Problem 22P
Solution: The data multiply with 1000 for each value in the data is as follows:
Data | Data*100 | Data | Data*100 |
0.194 | 194 | 0.309 | 309 |
0.258 | 258 | 0.276 | 276 |
0.19 | 190 | 0.287 | 287 |
0.291 | 291 | 0.317 | 317 |
0.158 | 158 | 0.252 | 252 |
0.295 | 295 | 0.215 | 215 |
0.261 | 261 | 0.25 | 250 |
0.25 | 250 | 0.246 | 246 |
0.181 | 181 | 0.26 | 260 |
0.125 | 125 | 0.265 | 265 |
0.107 | 107 | 0.182 | 182 |
0.26 | 260 | 0.113 | 113 |
0.309 | 309 | 0.2 | 200 |
Explanation of Solution
Calculation: The data represent baseball batting averages for a random sample of National League players near the end of the baseball season and there are 26 values in the data set. To find the whole number data by multiplying 1000 is obtained as follows:
Data | Data*100 | Data | Data*100 |
0.194 | 0.309 | 309 | |
0.258 | 0.276 | 276 | |
0.19 | 0.287 | 287 | |
0.291 | 291 | 0.317 | 317 |
0.158 | 158 | 0.252 | 252 |
0.295 | 295 | 0.215 | 215 |
0.261 | 261 | 0.25 | 250 |
0.25 | 250 | 0.246 | 246 |
0.181 | 181 | 0.26 | 260 |
0.125 | 125 | 0.265 | 265 |
0.107 | 107 | 0.182 | 182 |
0.26 | 260 | 0.113 | 113 |
0.309 | 309 | 0.2 | 200 |
Interpretation: Hence, the data multiply with 1000 is as follows:
Data | Data*100 | Data | Data*100 |
0.194 | 194 | 0.309 | 309 |
0.258 | 258 | 0.276 | 276 |
0.19 | 190 | 0.287 | 287 |
0.291 | 291 | 0.317 | 317 |
0.158 | 158 | 0.252 | 252 |
0.295 | 295 | 0.215 | 215 |
0.261 | 261 | 0.25 | 250 |
0.25 | 250 | 0.246 | 246 |
0.181 | 181 | 0.26 | 260 |
0.125 | 125 | 0.265 | 265 |
0.107 | 107 | 0.182 | 182 |
0.26 | 260 | 0.113 | 113 |
0.309 | 309 | 0.2 | 200 |
(b)
To find: The standard frequency table for the data set..
Answer to Problem 22P
Solution: The complete frequency table is as:
Class limits | Class boundaries | Midpoints | Freq | Relative freq | Cumulative freq |
46-85 | 45.5-85.5 | 65.5 | 4 | 0.12 | 4 |
86-125 | 85.5-125.5 | 105.5 | 5 | 0.16 | 9 |
126-165 | 125.5-165.5 | 145.5 | 10 | 0.31 | 19 |
166-205 | 165.5-205.5 | 185.5 | 5 | 0.16 | 24 |
206-245 | 205.5-245.5 | 225.5 | 5 | 0.16 | 29 |
246-285 | 245.5-285.5 | 265.5 | 3 | 0.09 | 32 |
Explanation of Solution
Calculation: To find the class width for the whole data of 26 values, it is observed that largest value of the data set is 317 and the smallest value is 107 in the data. Using 5 classes, the class width calculated in the following way:
The value is round up to the nearest whole number. Hence, the class width of the data set is 43. The class width for the data is 43 and the lowest data value (107) will be the lower class limit of the first class. Because the class width is 43, it must add 43 to the lowest class limit in the first class to find the lowest class limit in the second class. There are 5 desired classes. Hence, the class limits are 107–149, 150–192, 193–235, 236–278, and 279–321. Now, to find the class boundaries subtract 0.5 from lower limit of every class and add 0.5 to the upper limit of the every class interval. Hence, the class boundaries are 106.5–149.5, 149.5–192.5, 192.5–235.5, 235.5-278.5, and 278.5-321.5.
Next to find the midpoint of the class is calculated by using formula,
Midpoint of first class is calculated as:
The frequencies for respective classes are 3, 4, 3, 10, and 6.
Relative frequency is calculated by using the formula
The frequency for first class is 3 and total frequencies are 26 so the relative frequency is
The calculated frequency table is as follows:
Class limits | Class boundaries | Midpoints | freq | relative freq |
107-149 | 106.5-149.5 | 128 | 3 | 0.12 |
150-192 | 149.5-192.5 | 171 | 4 | 0.15 |
193-235 | 192.5-235.5 | 214 | 3 | 0.12 |
236-278 | 235.5-278.5 | 257 | 10 | 0.38 |
279-321 | 278.5-321.5 | 300 | 6 | 0.23 |
Graph: To construct the histogram by using the MINITAB, the steps are as follows:
Step 1: Enter the class boundaries in C1 and frequency in C2.
Step 2: Go to Graph > Histogram > Simple.
Step 3: Enter C1 in Graph variable then go to Data options > Frequency > C2.
Step 4: Click on OK.
The obtained histogram is
Interpretation: Hence, the complete frequency table is as follows:
Class limits | Class boundaries | Midpoints | Freq | Relative freq |
107-149 | 106.5-149.5 | 128 | 3 | 0.12 |
150-192 | 149.5-192.5 | 171 | 4 | 0.15 |
193-235 | 192.5-235.5 | 214 | 3 | 0.12 |
236-278 | 235.5-278.5 | 257 | 10 | 0.38 |
279-321 | 278.5-321.5 | 300 | 6 | 0.23 |
(c)
To find: The class limits, class boundaries, and midpoints in the frequency table by dividing 1000..
Answer to Problem 22P
Solution: The frequency table of original data is as follows:
Class limits | Class boundaries | Midpoints | |
0.107-0.149 | 0.1065-0.1495 | 0.128 | |
0.149-0.192 | 0.1495-0.1925 | 0.171 | |
0.193-0.235 | 0.1925-0.2355 | 0.214 | |
0.236-0.278 | 0.2355-0.2785 | 0.257 | |
0.279-0.321 | 0.2785-0.3215 | 0.3 |
Explanation of Solution
Calculation: The frequency table for whole number is obtained in above part. It is the data that multiply each value by 1000 to ‘clear’ decimals from the data. The frequency table for whole number is as follows:
Class limits | Class boundaries | Midpoints | Freq | Relative freq |
107–149 | 106.5–149.5 | 128 | 3 | 0.12 |
150–192 | 149.5–192.5 | 171 | 4 | 0.15 |
193–235 | 192.5–235.5 | 214 | 3 | 0.12 |
236–278 | 235.5–278.5 | 257 | 10 | 0.38 |
279–321 | 278.5–321.5 | 300 | 6 | 0.23 |
To find the decimal or original data, divide the class limits, class boundaries, and midpoints by 1000. The calculation as follows:
Class limits | Class boundaries | Midpoints | |
0.107–0.149 | 0.1065–0.1495 | 0.128 | |
0.149–0.192 | 0.1495–0.1925 | 0.171 | |
0.193–0.235 | 0.1925–0.2355 | 0.214 | |
0.236–0.278 | 0.2355–0.2785 | 0.257 | |
0.279–0.321 | 0.2785–0.3215 | 0.3 |
Interpretation: Hence, the data divide by 1000 is as follows:
Class limits | Class boundaries | Midpoints | |
0.107–0.149 | 0.1065–0.1495 | 0.128 | |
0.149–0.192 | 0.1495–0.1925 | 0.171 | |
0.193–0.235 | 0.1925–0.2355 | 0.214 | |
0.236–0.278 | 0.2355–0.2785 | 0.257 | |
0.279–0.321 | 0.2785–0.3215 | 0.300 |
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Chapter 2 Solutions
Understanding Basic Statistics
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillBig Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin Harcourt