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Data are given on the duration of eruptions of the Old Faithful geyser in Yellowstone National Park.

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 minimum value in the data is equal to 125 seconds.

The maximum value in the data is equal to264 seconds.

The first lower limit is equal to 125 seconds.

Class width = 25 seconds.

The number of class intervals required is computed below.

Therefore, the total number of class intervals in the frequency distribution is equal to 6.

$$\begin{gathered} \text{Number}\text{of}\text{classes}=\frac{\text{Maximum}\text{value}-\text{Minimum}\text{value}}{\text{Class}\text{width}} \\ =\frac{264-125}{25} \\ =5.56 \\ 鈮�6 \end{gathered}$$

According to the given formula, the lower class limits of the 6 intervals are computed below.

$\text{Class}\text{width}=\text{Difference}\text{between}\text{2}\text{consecutive}\text{lower}\text{class}\text{limits}$

$$\begin{gathered} 1^{st}\text{lower}\text{class}\text{limit}=125 \\ {2}^{nd}\text{lower}\text{class}\text{limit}=125+25 \\ =150 \end{gathered}$$

$$\begin{gathered} 3^{rd}\text{lower}\text{class}\text{limit}=150+25 \\ =175 \\ {4}^{th}\text{lower}\text{class}\text{limit}=175+25 \\ =200 \end{gathered}$$

$$\begin{gathered} 5^{th}\text{lower}\text{class}\text{limit}=200+25 \\ =225 \\ {6}^{th}\text{lower}\text{class}\text{limit}=225+25 \\ =250 \end{gathered}$$

Considering a gap of 1 unit between each successive interval, the following upper class limits are constructed:

$$\begin{gathered} 1^{st}\text{upper}\text{class}\text{limit}=2^{nd}\text{lower}\text{class}\text{limit}-\text{1} \\ =150-1 \\ =149 \end{gathered}$$

$$\begin{gathered} 2^{nd}\text{upper}\text{class}\text{limit}=3^{rd}\text{lower}\text{class}\text{limit}-\text{1} \\ =175-1 \\ =174 \end{gathered}$$

$$\begin{gathered} 3^{rd}\text{upper}\text{class}\text{limit}=4^{th}\text{lower}\text{class}\text{limit}-\text{1} \\ =200-1 \\ =199 \end{gathered}$$

$$\begin{gathered} 4^{th}\text{upper}\text{class}\text{limit}=5^{th}\text{lower}\text{class}\text{limit}-\text{1} \\ =225-1 \\ =224 \end{gathered}$$

$$\begin{gathered} 5^{th}\text{upper}\text{class}\text{limit}=6^{th}\text{lower}\text{class}\text{limit}-\text{1} \\ =250-1 \\ =249 \end{gathered}$$

$$\begin{gathered} 6^{th}\text{upper}\text{class}\text{limit}=7^{th}\text{lower}\text{class}\text{limit}-\text{1} \\ =275-1 \\ =274 \end{gathered}$$

By counting the durationsthat fall in each interval (both limits inclusive), the following frequency distribution is constructed: