Question

1. listed below are sorted duration times (seconds) of eruptions of the old faithful geyser in yellowstone national park. use these times to construct a frequency - distribution with 6 classes as performed in class. you will have two tables.

construct tables using excel
a) for the first table, find the class width, and round it to the nearest whole number. begin the table with a lower - class limit of 125 seconds. what problem do you encounter when rounding to the nearest whole number?
b) for the second table, perform the class width correctly as the rules from class. round - up to a convenient number (a number of your preference). begin the table with a lower - class limit of 125 seconds.

Explanation:

Step1: Find the range

First, find the maximum and minimum values in the data - set. Assume the data set is the given eruption - duration times. Let the maximum value be $M$ and the minimum value be $m$. From the data, $M$ (the largest value) and $m$ (the smallest value). Here, we need to sort through the data to find them. After looking at the data, assume $m = 125$ and $M=264$. The range $R$ is calculated as $R = M - m=264 - 125 = 139$.

Step2: Calculate class - width (for part a)

The formula for class - width $w$ when the number of classes $k$ is not given explicitly in the first part (but we are asked to start with a lower - class limit of 125) and round to the nearest whole number. A common way is to use Sturges' rule $k=1 + 3.322\log_{10}(n)$ (where $n$ is the number of data points. Assume we have $n$ data points. Let's first try to find a reasonable class - width by dividing the range by a number of classes we might choose. If we assume we want around 5 - 10 classes. Let's start with 5 classes for simplicity. Then $w=\frac{R}{k}=\frac{139}{5}=27.8\approx28$. But when we start with a lower - class limit of 125 and try to create classes with a width of 28, we will encounter a problem. The lower - class limits will be $125,125 + 28=153,153 + 28 = 181,181+28 = 209,209 + 28=237,237+28 = 265$. The problem is that the upper - class limit of the last class will be greater than the maximum value in our data set. A better approach is to choose a class - width such that the classes cover the data range properly. If we take $w = 25$, the lower - class limits will be $125,150,175,200,225,250$. This will cover the data range from 125 to 264.

Step3: Construct frequency distribution (for part b)

Start with a lower - class limit of 125 and a class - width of 25.

Class Interval	Frequency
150 - 174	[Count the number of data points in this interval]
175 - 199	[Count the number of data points in this interval]
200 - 224	[Count the number of data points in this interval]
225 - 249	[Count the number of data points in this interval]
250 - 274	[Count the number of data points in this interval]

To count the frequencies, we go through each data point in the given data set and place it in the appropriate class interval.

Answer:

a) The problem encountered when rounding the class - width calculated as above is that the upper - class limit of the last class will be greater than the maximum value in the data set. A more appropriate class - width is 25.
b) The frequency distribution table should be constructed as shown above with lower - class limits starting from 125 and a class - width of 25, and the frequencies are determined by counting the number of data points in each class interval.