Chapter 2, Problem 7RE

More freshmen: For the data in Exercise 6:

1. Construct a frequency polygon.
2. Construct a relative frequency polygon.
3. Construct a frequency ogive.
4. Construct a relative frequency ogive.

To determine

To construct:A frequency polygon.

Explanation of Solution

Given information: The following frequency distribution presents the number of freshmen elected in each of the past 53 elections from 1912 to 2016.

Number of Freshmen	Frequency
20-39	2
40-59	15
60-79	10
80-99	14
100-119	7
120-139	3
140-159	1
160-179	1

Formula used:

  $ClassMidpoint=\frac{Lowerlimit+Lowerlimitofnextclass}{2}$

Calculation:

Number of Freshmen	Class Midpoints	Frequency
20-39	$$$\frac{20+40}{2}=30$	2
40-59	$\frac{40+60}{2}=50$	15
60-79	$\frac{60+80}{2}=70$	10
80-99	$\frac{80+100}{2}=90$	14
100-119	$\frac{100+120}{2}=110$	7
120-139	$\frac{120+140}{2}=130$	3
140-159	$\frac{140+160}{2}=150$	1
160-179	$\frac{160+180}{2}=170$	1

Now plotting the points whose x-coordinates are the class midpoints and whose y-coordinates are the frequencies. Then the frequency polygon is connecting the points with straight line

  

To determine

To construct: A relative frequency polygon.

Explanation of Solution

Given information: The following frequency distribution presents the number of freshmen elected in each of the past 53 elections from 1912 to 2016.

Number of Freshmen	Frequency
20-39	2
40-59	15
60-79	10
80-99	14
100-119	7
120-139	3
140-159	1
160-179	1

Formula used:

  $ClassMidpoint=\frac{Lowerlimit+Lowerlimitofnextclass}{2}$

Calculation:

Number of Freshmen	Class Midpoints	Frequency	Relative Frequency
20-39	$$$\frac{20+40}{2}=30$	2	$0.0377$
40-59	$\frac{40+60}{2}=50$	15	$0.2830$
60-79	$\frac{60+80}{2}=70$	10	$0.1887$
80-99	$\frac{80+100}{2}=90$	14	$0.2642$
100-119	$\frac{100+120}{2}=110$	7	$0.1321$
120-139	$\frac{120+140}{2}=130$	3	$0.0566$
140-159	$\frac{140+160}{2}=150$	1	$0.0189$
160-179	$\frac{160+180}{2}=170$	1	$0.0189$

Now plotting the points whose x-coordinates are the class midpoints and whose y-coordinates are therelative frequencies. Then the relative frequency polygon is connecting the points with straight line.

s

  

To determine

To construct: A frequency ogive.

Explanation of Solution

Given information: The following frequency distribution presents the number of freshmen elected in each of the past 53 elections from 1912 to 2016.

Number of Freshmen	Frequency
20-39	2
40-59	15
60-79	10
80-99	14
100-119	7
120-139	3
140-159	1
160-179	1

Definition used: The cumulative frequency of a class is the sum of the frequencies of that class and all previous classes.

Anogive plots the cumulative frequencies.

Calculation:

The cumulative classes are given by in the following table.

Number of Freshmen	Frequency	Cumulative Frequency
20-39	2	2
40-59	15	$2+15=17$
60-79	10	$17+10=27$
80-99	14	$27+14=41$
100-119	7	$41+7=48$
120-139	3	$48+3=51$
140-159	1	$51+1=52$
160-179	1	$52+1=53$

The frequency ogive is is given by

  

To determine

To construct: A relative frequency ogive.

Explanation of Solution

Given information: The following frequency distribution presents the number of freshmen elected in each of the past 53 elections from 1912 to 2016.

Number of Freshmen	Frequency
20-39	2
40-59	15
60-79	10
80-99	14
100-119	7
120-139	3
140-159	1
160-179	1

Definition used: The cumulative relative frequency of a class is given by

  $cumulativerelativefrequency=\frac{cumulativefrequency}{Sumofallfrequency}$

A relative frequency ogive plots the cumulative relative frequencies.

Calculation:

The cumulative classes are given by in the following table.

Number of Freshmen	Frequency	Cumulative Frequency	Cumulative relative Frequency
20-39	2	2	$\frac{2}{53}=0.03774$
40-59	15	$2+15=17$	$\frac{17}{53}=0.3208$
60-79	10	$17+10=27$	$\frac{27}{53}=0.5094$
80-99	14	$27+14=41$	$\frac{41}{53}=0.7736$
100-119	7	$41+7=48$	$\frac{48}{53}=0.9057$
120-139	3	$48+3=51$	$\frac{51}{53}=0.9623$
140-159	1	$51+1=52$	$\frac{52}{53}=0.9811$
160-179	1	$52+1=53$	$\frac{53}{53}=1$

The relative frequency ogive is is given by