 # During contract negotiations, a company seeks to change the number of sick days employees may take, saying that the annual “average” is 7 days of absence per employee. The union negotiators counter that the “average” employee misses only 3 days of work each year. Explain how both sides might be correct, identifying the measure of center you think each side is using and why the difference might exist. The aim of this question is to understand the key concepts of mean and median that form the basis of statistical calculations.

The mean of a given sample of data is defined as the average numerical value (or arithmetic mean) of all the values. Mathematically:

$Mean \ = \ \dfrac{ \text{ sum of all values of the sample data } }{ \text{ total no. of samples } }$

$\Rightarrow Mean \ = \ \dfrac{ x_1 \ + \ x_2 \ + \ x_3 \ + \ … \ + \ x_n }{ n }$

Where $x_1, \ x_2, \ x_3, \ … \ , \ x_5$ are the values of sample data and $n$ is the total no. of samples or sample size.

Mean can be used to calculate important statistical characteristics of data such as variance, standard deviation, and other moments / central moments.

The median of a given sample of data is an order property. It is defined as the middle value of all the values given in the sample after sorting all values in ascending order. Mathematically:

$Median \ = \ \left \{ \begin{array}{ll} X[ \frac{ n }{ 2 } ] & \text{ if n is odd } \\ \dfrac{ X[ \frac{ n \ – \ 1 }{ 2 } ] \ + \ X[ \frac{ n \ + \ 1 }{ 2 } ] }{ 2 } & \text{ if n is even } \end{array} \right.$

Where $X$ is the ordered list of sample values and $n$ is the total no. of samples or sample size.

In the given question, the company’s stance is that the average value of absences per employee is 7 days. They are actually talking about the sample mean here. They have summed the total no. of leaves of all employees and divided it by the total no. of employees.

The union negotiator’s stance is that the average employee takes a maximum leave of 3 days. They are actually talking about the median of the same data.

Both the company and union have the correct figures but their point of view is different. Statistically, the company is talking about the mean whereas the union negotiators are considering the median.

## Numerical Result

Both are correct.

$Mean \ = \ 7 \ days$

$Median \ = \ 3 \ days$

## Example

Let’s say that for a given company, there are 9 employees. Here are the leaves taken in the past year:

$\{ \ 1, \ 2, \ 4, \ 6, \ 0, \ 2, \ 9, \ 1, \ 20 \ \}$

Calculate the mean and median of the sample data.

$\Rightarrow Mean \ = \ \dfrac{ 1 + 2 + 4 + 6 + 0 + 2 + 9 + 1 + 20 }{ 10 } \ = \ \dfrac{ 45 }{ 9 } \ = \ 5 \ days$

Sorting the given data in ascending order:

$\{ \ 0, \ 1, \ 1, \ 2, \ \boldsymbol{ 2 }, \ 4, \ 6, \ 9, \ 20 \ \}$

$Median \ = \ \text{ Middle Value } \ = \ \text{ 5th value } \ = \ 2 \ days$