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This module introduces the concepts of frequency, relative frequency, and cumulative relative frequency, and the relationship between these measures. Students will have the opportunity to interpret data through the sample problems provided.

Twenty students were asked how many hours they worked per day. Their responses, in hours, are listed below:

  • 5
  • 6
  • 3
  • 3
  • 2
  • 4
  • 7
  • 5
  • 2
  • 3
  • 5
  • 6
  • 5
  • 4
  • 4
  • 3
  • 5
  • 2
  • 5
  • 3

Below is a frequency table listing 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

A frequency is the number of times a given datum occurs in a data set. According to the table above, there are three students who work 2 hours, five students who work 3 hours, etc. The total of the frequency column, 20, represents the total number of students included in the sample.

A relative frequency is the fraction or proportion of times an answer occurs. 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.

Frequency table of student work hours w/ relative frequency
DATA VALUE FREQUENCY RELATIVE FREQUENCY
2 3 3 20 or 0.15
3 5 5 20 or 0.25
4 3 3 20 or 0.15
5 6 6 20 or 0.30
6 2 2 20 or 0.10
7 1 1 20 or 0.05

The sum of the relative frequency column is 20 20 , or 1.

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.

Frequency table of student work hours w/ relative and cumulative relative frequency
DATA VALUE FREQUENCY RELATIVE
FREQUENCY
CUMULATIVE RELATIVE
FREQUENCY
2 3 3 20 or 0.15 0.15
3 5 5 20 or 0.25 0.15 + 0.25 = 0.40
4 3 3 20 or 0.15 0.40 + 0.15 = 0.55
5 6 6 20 or 0.30 0.55 + 0.30 = 0.85
6 2 2 20 or 0.10 0.85 + 0.10 = 0.95
7 1 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.

Because of rounding, the relative frequency column may not always sum to one and the last entry in the cumulative relative frequency column may not be one. However, they each should be close to one.

The following table represents the heights, in inches, of a sample of 100 male semiprofessional soccer players.

Frequency table of soccer player height
HEIGHTS
(INCHES)
FREQUENCY RELATIVE
FREQUENCY
CUMULATIVE
RELATIVE
FREQUENCY
Total = 100 Total = 1.00
59.95 - 61.95 5 5 100 = 0.05 0.05
61.95 - 63.95 3 3 100 = 0.03 0.05 + 0.03 = 0.08
63.95 - 65.95 15 15 100 = 0.15 0.08 + 0.15 = 0.23
65.95 - 67.95 40 40 100 = 0.40 0.23 + 0.40 = 0.63
67.95 - 69.95 17 17 100 = 0.17 0.63 + 0.17 = 0.80
69.95 - 71.95 12 12 100 = 0.12 0.80 + 0.12 = 0.92
71.95 - 73.95 7 7 100 = 0.07 0.92 + 0.07 = 0.99
73.95 - 75.95 1 1 100 = 0.01 0.99 + 0.01 = 1.00

The data in this table has been grouped into the following intervals:

  • 59.95 - 61.95 inches
  • 61.95 - 63.95 inches
  • 63.95 - 65.95 inches
  • 65.95 - 67.95 inches
  • 67.95 - 69.95 inches
  • 69.95 - 71.95 inches
  • 71.95 - 73.95 inches
  • 73.95 - 75.95 inches
This example is used again in the Descriptive Statistics chapter, where the method used to compute the intervals will be explained.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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cm
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what is physics
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
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what is chemistry
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what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
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answer
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progressive wave
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Mohammed
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Who can show me the full solution in this problem?
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Source:  OpenStax, Principles of business statistics. OpenStax CNX. Aug 05, 2009 Download for free at http://cnx.org/content/col10874/1.5
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