Chapter 2.1, Problem 22P

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.

To determine

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	$0.194\times 1000=194$	0.309	309
0.258	$0.258\times 1000=258$	0.276	276
0.19	$0.190\times 1000=190$	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

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

To determine

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:

$\begin{array}{l}Classwidth=\frac{Largestdatavalue-smallestdatavalue}{Desirednumberofclasses}\\ \end{array}$

$\begin{array}{c}Classwidth=\frac{317-107}{5}\\ =42\\ \approx 43\end{array}$

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,

$\begin{array}{l}\text{Midpoint=}\frac{\text{Lower class limit + Upper class limit}}{\text{2}}\\ \end{array}$

Midpoint of first class is calculated as:

$\begin{array}{c}\text{Midpoint}=\frac{\text{106}\text{.5+ 149}\text{.5}}{2}\\ =128\end{array}$

The frequencies for respective classes are 3, 4, 3, 10, and 6.

Relative frequency is calculated by using the formula

$\mathrm{Re}lative\text{ frequency}=\frac{f}{n}=\frac{frequency}{Total\text{ of all frequencies}}$

The frequency for first class is 3 and total frequencies are 26 so the relative frequency is $\frac{3}{26}=0.12$. Do it same calculation for other class.

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

To determine

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