Question

goal: find the mean, median, and mode of data given in a frequency chart.
every year, the students at a school are given a musical aptitude test that rates them from 0 (no musical aptitude) to 5 (high musical aptitude). this years results were:
the average (mean) aptitude score:
the median aptitude score:
question help: video 1 video 2 message instructor

Explanation:

Step1: Calculate the total number of students

The total number of students \(n\) is the sum of all frequencies.
\(n=2 + 1+3+3+3+4=16\)

Step2: Calculate the mean

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{k}x_{i}f_{i}}{n}\), where \(x_{i}\) is the aptitude score and \(f_{i}\) is the frequency.

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\(\bar{x}=\frac{48}{16}=3\)

Step3: Calculate the median

Since \(n = 16\) (an even number), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th ordered - data values.
First, find the cumulative frequencies:

• For \(x = 0\): cumulative frequency \(CF_{0}=2\)
• For \(x = 1\): cumulative frequency \(CF_{1}=2 + 1=3\)
• For \(x = 2\): cumulative frequency \(CF_{2}=3+3 = 6\)
• For \(x = 3\): cumulative frequency \(CF_{3}=6+3 = 9\)

The \(\frac{n}{2}=8\)th and \((\frac{n}{2}+1) = 9\)th values. Both the 8th and 9th values fall in the \(x = 3\) group. So the median is \(3\)

Answer:

The average (mean) aptitude score: \(3\)
The median aptitude score: \(3\)