Question

Describe an instance where a graph/chart/histogram/etc. or a median/mean given that was misleading and did not reveal the whole situation.

Answer 1

AN INSTANCE WHERE MEAN WAS MISLEADING AND DID NOT REVEAL THE WHOLE SITUATION:

Arithmetic Mean is generally employed to find the Central Tendency of a given data. That is, The Arithmetic Mean is used to represent the entire data set.

For example, suppose we are interested how much an employeee gets salary in a company. For that purpose, the usual method is to enumerate all the employees of the company, tabulate their individual salary, add them up and finally divide the total salary by the total number of employees. That will give an idea of how much an employee gets salary.

For example, consider an company with 5 employees.

Let their salary be:

100, 120, 80, 90, 110

Then, the Arithmetic Mean = (100 + 120 + 80 + 90 + 110)/5 =500/5 = 100.

So, we can conclude that an employee gets 100 in that company on the average.

This is correct as long as all the values are close to each other and there is no extreme values.

MISLEADING AND DID NOT REVEAL THE WHOLE SITUATION SCENERY:
Now, consider adding the Managing Director's salary with the other 5 employees.

As we know, the Managing Director's salary is very high: Let it be 1000.

Now, taking the average of all the 6 persons, we get:
Arithmetic Mean =(100 + 120 + 80 + 90 + 110 + 1000)/6 = 250.

We note that avrage salary of 250 is quite misleading and did not reveal the whole situation.

The reason for this misleading case is the outlier of 1000.

In such situations, Arithmetic Mean should not be used. Median plays a useful role here.