Question

In Exercises 1–6, refer to the data below, which are total home game playing times (hours) for all Major League Baseball teams in a recent year (based on data from Baseball Prospectus).

236 237 238 239 241 241 242 245 245 245 246 247 247 248 248 249 250 250 250 251 252 252 253 253 258 258 258 260 262 264

Frequency Distribution Construct a frequency distribution. Use a class width of 5 hours and use a starting time of 235 hours.

Short answer

The following frequency distribution is constructed for the given data on playing times:

Playing Times (in hours)	Frequency
235-239	4
240-244	3
245-249	9
250-254	8
255-259	3
260-264	3

Step-by-step solution

Given information

Data are given on playing times for Major League Baseball Teams.

The first lower limit is equal to 235 hours.

The class width is equal to 5 hours.

Define frequency distribution and class width

A frequency distribution is an arrangement of data values in the form of closed intervals.

The frequencies of each class interval are tabulated by counting the number of data values that fall in each interval.

The class width is equal to the difference between two consecutive lower class limits.

Construction of frequency distribution

An inclusive type frequency distribution is constructed where the class width is equal to 5 hours, and the first value is equal to 235 hours.

The maximum value in the data is equal to 264 hours.

The lower class limits of the 6 intervals are computed below:

$1^{\text{st}}\text{lower}\text{class}\text{limit}=235$

$$\begin{gathered} 2^{\text{nd}}\text{lower}\text{class}\text{limit}=235+5 \\ =240 \\ {3}^{\text{rd}}\text{lower}\text{class}\text{limit}=240+5 \\ =245 \end{gathered}$$

$$\begin{gathered} {\text{4}}^{\text{th}}\text{lower}\text{class}\text{limit}=245+5 \\ =250 \\ {\text{5}}^{\text{th}}\text{lower}\text{class}\text{limit}=250+5 \\ =255 \end{gathered}$$

$$\begin{gathered} 6^{\text{th}}\text{lower}\text{class}\text{limit}=255+5 \\ =260 \end{gathered}$$

The following upper-class limits are constructed, considering a gap of 1 unit in between each successive interval:

$$\begin{gathered} {\text{1}}^{\text{st}}\text{upper}\text{class}\text{limit}=2^{\text{nd}}\text{lower}\text{class}\text{limit}-\text{1} \\ =240-1 \\ =239 \end{gathered}$$

$$\begin{gathered} {\text{2}}^{\text{nd}}\text{upper}\text{class}\text{limit}=3^{\text{rd}}\text{lower}\text{class}\text{limit}-\text{1} \\ =245-1 \\ =244 \end{gathered}$$

$$\begin{gathered} 3^{\text{rd}}\text{upper}\text{class}\text{limit}=4^{\text{th}}\text{lower}\text{class}\text{limit}-\text{1} \\ =250-1 \\ =249 \end{gathered}$$

$$\begin{gathered} 4^{\text{th}}\text{upper}\text{class}\text{limit}=5^{\text{th}}\text{lower}\text{class}\text{limit}-\text{1} \\ =255-1 \\ =254 \end{gathered}$$

$$\begin{gathered} 5^{\text{th}}\text{upper}\text{class}\text{limit}=6^{\text{th}}\text{lower}\text{class}\text{limit}-\text{1} \\ =260-1 \\ =259 \end{gathered}$$

$$\begin{gathered} 6^{\text{th}}\text{upper}\text{class}\text{limit}=7^{\text{th}}\text{lower}\text{class}\text{limit}-\text{1} \\ =265-1 \\ =264 \end{gathered}$$

By counting the number of hours that fall in each interval (both limits inclusive), the following frequency distribution is constructed:

Playing Times (in hours)	Frequency
235-239	4
240-244	3
245-249	9
250-254	8
255-259	3
260-264	3