The average mark of candidates in an aptitude test was 128.5 with a standard deviation of 8.2. Three scores extracted from the test are 148,102,152. What is the average of the extracted scores that are outliers

102

Step-by-step explanation:

We have the mean (m) 128.5 and the standard deviation (sd) 8.2, we must calculate the value of z for each one and determine whether or not it is an outlier:

z = (x - m) / sd

In the first case x = 148:

z = (148 - 128.5) / 8.2

z = 2.37

In the second case x = 102:

z = (102 - 128.5) / 8.2

z = - 3.23

In the first case x = 152:

z = (152 - 128.5) / 8.2

z = 2.86

The value of this is usually between - 3 and 3, therefore when x is 102 it goes outside the range of the value of z, which means that this is the outlier.