# 1.5: Frequency & Frequency Tables

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Twenty students were asked how many hours they worked per day. Their responses, in hours, are as follows:
5, 6, 3, 3, 2, 4, 7, 5, 2, 3, 5, 6, 5, 4, 4, 3, 5, 2, 5, 3.

The following table lists the different data values in ascending order and their frequencies.

Frequency Table of Student Work Hours
DATA VALUE FREQUENCY
2 3
3 5
4 3
5 6
6 2
7 1

In this research, 3 students studied for 2 hours. 5 students studies for 3 hours.

A frequency is the number of times a value of the data occurs. According to the table, there are three students who work two hours, five students who work three hours, and so on. The sum of the values in the frequency column, 20, represents the total number of students included in the sample.

A relative frequency is the ratio (fraction or proportion) of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes. To find the relative frequencies, divide each frequency by the total number of students in the sample–in this case, 20. Relative frequencies can be written as fractions, percents, or decimals.

Relative frequency = $\frac{\text{frequency of the class}}{\text{total}}$

Cumulative relative frequency is the accumulation of the previous relative frequencies. To find the cumulative relative frequencies, add all the previous relative frequencies to the relative frequency for the current row, as shown in the table below.

Cumulative relative frequency = sum of previous relative frequencies + current class frequency

## Example 1

Frequency Table of Student Work Hours with Relative and Cumulative Relative Frequencies
DATA VALUE FREQUENCY RELATIVE

FREQUENCY

CUMULATIVE RELATIVE

FREQUENCY

2 3 $\frac{3}{20}$ or 0.15 0.15
3 5 $\frac{5}{20}$ or 0.25 0.15 + 0.25 = 0.40
4 3 $\frac{3}{20}$ or 0.15 0.40 + 0.15 = 0.55
5 6 $\frac{6}{20}$ or 0.30 0.55 + 0.30 = 0.85
6 2 $\frac{2}{20}$ or 0.10 0.85 + 0.10 = 0.95
7 1 $\frac{1}{20}$ or 0.05 0.95 + 0.05 = 1.00

The last entry of the cumulative relative frequency column is one, indicating that one hundred percent of the data has been accumulated.

## Example 2

We sample the height of 100 soccer players. The result is shown below.

 Height (inches) Frequency 59.95 – 61.95 5 61.95 – 63.95 3 63.95 – 65.95 15 65.95 – 67.95 40 67.95 – 69.95 17 69.95 – 71.95 12 71.95 – 73.95 7 73.95 – 75.95 1 Total = 100

Find:

a. the relative frequency for each class.

 Height (Inches) Frequency Relative Frequency Cumulative Relative Frequency 59.95 – 61.95 5 $\frac{5}{100}$ or 0.05 0.05 61.95 – 63.95 3 $\frac{3}{100}$ or 0.03 0.05 + 0.03 = 0.08 63.95 – 65.95 15 $\frac{15}{100}$ or 0.15 0.08 + 0.15 = 0.23 65.95 – 67.95 40 $\frac{4}{100}$ or 0.04 0.23 + 0.40 = 0.63 67.95 – 69.95 17 $\frac{17}{100}$ or 0.17 0.63 + 0.17 = 0.80 69.95 – 71.95 12 $\frac{12}{100}$ or 0.12 0.80 + 0.12 = 0.92 71.95 – 73.95 7 $\frac{7}{100}$ or 0.07 0.92 + 0.07 = 0.99 73.95 – 75.95 1 $\frac{1}{100}$ or 0.01 0.99 + 0.01 = 1.00 Total = 100 Total = 1

b. the percentage for height that is less than 63.95 inches.
[hidden-answer a=”26068″] $\frac{5+3}{100}$ = 0.08 = 8%[/hidden-answer]

c. the percentage for height that is between 69.95 inches and 73.95 inches.
[hidden-answer a=”839825″] $\frac{12}{100}$ + $\frac{9}{100}$ = 0.12 + 0.07 = 0.19[/hidden-answer]

In this sample, there are five players whose heights fall within the interval 59.95–61.95 inches, three players whose heights fall within the interval 61.95–63.95 inches, 15 players whose heights fall within the interval 63.95–65.95 inches, 40 players whose heights fall within the interval 65.95–67.95 inches, 17 players whose heights fall within the interval 67.95–69.95 inches, 12 players whose heights fall within the interval 69.95–71.95, seven players whose heights fall within the interval 71.95–73.95, and one player whose heights fall within the interval 73.95–75.95. All heights fall between the endpoints of an interval and not at the endpoints.

## Example 3

The table shows the amount, in inches, of annual rainfall in a sample of towns.

 Rainfall (inches) Frequency 2.95 – 4.97 6 4.97 – 6.99 7 6.99 – 9.01 15 9.01 – 11.03 8 11.03 – 13.05 9 13.05 – 15.07 5

Find

1. the relative frequency and cumulative relative frequency for each class.
Total = sum of all frequencies = 6 + 7 + 15 + 8 + 9 + 5 = 50
 Rainfall (inches) Frequency Relative frequency Cumulative relative frequency 2.95 – 4.97 6 $\frac{6}{50}$ = 0.12 0.12 4.97 – 6.99 7 $\frac{7}{50}$ = 0.14 0.12 + 0.14 = 0.26 6.99 – 9.01 15 $\frac{15}{50}$ = 0.30 0.26 + 0.30 = 0.56 9.01 – 11.03 8 $\frac{8}{50}$ = 0.16 0.56 + 0.16 = 0.72 11.03 – 13.05 9 $\frac{9}{50}$ = 0.18 0.72 + 0.18 = 0.90 13.05 – 15.07 5 $\frac{5}{50}$ = 0.10 0.90 + 0.10 = 1.00

2. the percentage of rainfall that is less than 9.01 inches.
[hidden-answer a=”355649″]The percentage of rainfall that is less than 9.01 inches = 0.12 + 0.14 + 0.30 = 0.56[/hidden-answer]
3. the percentage of heights that fall between 61.95 and 65.95 inches.
[hidden-answer a=”94434″]The percentage of heights that fall between 6.99 inches and 11.03 inches = $\frac{15}{50}$ $\frac{8}{50}$ = 0.26[/hidden-answer]

### Try It

The table contains the total number of deaths worldwide as a result of earthquakes for the period from 2000 to 2012.

 Year Total Number of Deaths 2000 231 2001 21,357 2002 11,685 2003 33,819 2004 228,802 2005 88,003 2006 6,605 2007 712 2008 88,011 2009 1,790 2010 320,120 2011 21,953 2012 768 Total 823,356
1.  What is the frequency of deaths measured from 2006 through 2009?
2. What percentage of deaths occurred after 2009?
3. What is the relative frequency of deaths that occurred in 2003 or earlier?
[hidden-answer a=”294919″] $\frac{67,092}{823,356}$ = 0.081[/hidden-answer]
4. What is the percentage of deaths that occurred in 2004?
5. What kind of data are the numbers of deaths?
6. The Richter scale is used to quantify the energy produced by an earthquake. Examples of Richter scale numbers are 2.3, 4.0, 6.1, and 7.0. What kind of data are these numbers?

## Example 4

The table contains the total number of fatal motor vehicle traffic crashes in the United States for the period from 1994 to 2011.

Year Total Number of Crashes Year Total Number of Crashes
1994 36,254 2004 38,444
1995 37,241 2005 39,252
1996 37,494 2006 38,648
1997 37,324 2007 37,435
1998 37,107 2008 34,172
1999 37,140 2009 30,862
2000 37,526 2010 30,296
2001 37,862 2011 29,757
2002 38,491 Total 653,782
2003 38,477
1. What is the frequency of deaths measured from 2000 through 2004?
[hidden-answer a=”514149″]37,526 + 37,862 + 38,491 + 38,477 + 38,444 = 190,800 [/hidden-answer]
2. What percentage of deaths occurred after 2006?
[hidden-answer a=”107848″] $\frac{37,435 + 34,172 + 30,862 + 30,296 + 29,757}{653,782}$ or 24.9%[/hidden-answer]
3. What is the relative frequency of deaths that occurred in 2000 or before?
[hidden-answer a=”897794″] $\frac{260,086}{653,782}$ or 39.8%[/hidden-answer]
4. What is the percentage of deaths that occurred in 2011?
[hidden-answer a=”43840″] $\frac{29,757}{653,782}$ or 4.6%[/hidden-answer]