Question

Frequency Polygon. In Exercises 15 and 16, construct the frequency polygons.

Old Faithful Use the frequency distribution from Exercise 11 in Section 2-1 on page 49 to construct a frequency polygon. Does the graph suggest that the distribution is skewed? If so, how?

Short answer

The following frequency polygon is constructed:

There are low frequencies toward the left of the graph. This indicates that the graph has a tail toward the left.

Thus, the graph suggests that the distribution is skewed to the left.

Step-by-step solution

Given information

A frequency distribution is constructed showing thetime duration (in seconds) of the eruptions of Old Faithful geyser in Yellowstone National Park.

Refer to Exercise 11 in Section 2-1. The frequency distribution is given as:

Duration Times (in seconds)	Frequency
125-149	1
150-174	0
175-199	0
200-224	3
225-249	34
250-274	12

Define frequency polygon

A frequency polygon is based on the midpoints of the classes and the associated frequencies forming a closed polygon. The midpoints of the class intervals are plotted on the x-axis, and the y-axis shows the frequencies.

Points are plotted and then connected using straight lines to form a polygon-like structure.

Define skewness

If the graph appears to have anelongated tail on one of the sides (left or right) of a distribution, then the distribution is said to be skewed.

A right skewed distribution has a longer right tail, while a left skewed as a longer left tail.

Form the continuous frequency distribution

To construct a frequency polygon, first, convert the classes into a continuous form by subtracting 0.5 from the lower limit of each of the classes and adding 0.5 to the upper limit of each of the classes. So, the obtained classes are as shown in the following table:

Duration Times (in seconds)	Frequency
124.5-149.5	1
149.5-174.5	0
174.5-199.5	0
199.5-224.5	3
224.5-249.5	34
249.5-274.5	12

The midpoints of the classes are obtained by taking the average of the lower and upper limits of the respective classes.

The midpoints of the classes are computed as follows:

$$\begin{gathered} {\text{Midpoint}}_1=\frac{124.5+149.5}{2} \\ =137 \\ {\text{Midpoint}}_2=\frac{124.5+174.5}{2} \\ =162 \\ {\text{Midpoint}}_3=\frac{174.5+199.5}{2} \\ =187 \\ {\text{Midpoint}}_4=\frac{199.5+224.5}{2} \\ =212 \\ {\text{Midpoint}}_5=\frac{224.5+249.5}{2} \\ =237 \\ {\text{Midpoint}}_6=\frac{249.5+274.5}{2} \\ =262 \end{gathered}$$

To form a polygon (a closed figure), the two ends of the line should touch the horizontal axis. For that, add a class at the beginning and end of the frequency distribution with frequency 0. They are 99.5-124.5 and 274.5-298.5.

The midpoint of the class interval preceding the first interval is equal to:

$\frac{99.5+124.5}{2}=112$

The midpoint of the class interval succeeding the last interval is equal to:

$\frac{274.5+299.5}{2}=287$

Thus, the new frequency distribution becomes:

Duration Times (in seconds)	Midpoints	Frequency
99.5-124.5	112	0
124.5-149.5	137	1
149.5-174.5	162	0
174.5-199.5	187	0
199.5-224.5	212	3
224.5-249.5	237	34
249.5-274.5	262	12
274.5-299.5	287	0

Sketch the frequency polygon

The given steps are used to plot the frequency polygon:

• Plot the midpoints on the horizontal axis starting from 112, 137, 162, and so on up to 312.
• Plot the frequencies on the vertical axis starting from 0, 5, 10, and so on up to 40.
• Mark the points on the graph for each midpoint corresponding to the frequency.
• Join the points using a straight line.

The frequency polygon is plotted as shown below:

Examine the skewness

It can be observed that the graph appears to have a longer tail on the left side. This suggests that the given distribution is skewed to the left.