We have that the absolute deviation is given by the following expression:
[tex]\frac{1}{n}\sum ^n_{i=1}|x_i-m(X)|[/tex]Here m(X) is the average value of the data set, n is the number of data values & xi represents the data values in the set.
We calculate as follows:
[tex]\frac{1}{5}\sum ^5_{i=1}|x_i-90|=3.2[/tex]So, the mean absolute deviation for Kal's test is 3.2.
